· Web view2021. 3. 14. · 3.8.1.3 Radioactive Decay. Q1. An ancient sealed flask contains a...

Colonel Frank Seely School

3.8.1.3 Radioactive Decay

Q1.An ancient sealed flask contains a liquid, assumed to be water. An archaeologist asks a scientist to determine the volume of liquid in the flask without opening the flask. The scientist decides to use a radioactive isotope of sodium ( ) that decays with a half-life of 14.8 h.

(a) She first mixes a compound that contains 3.0 × 10–10 g of sodium-24 with 1500 cm3 of water. She then injects 15 cm3 of the solution into the flask through the seal.Show that initially about 7.5 × 1010 atoms of sodium-24 are injected into the flask.

(1)

(b) Show that the initial activity of the solution that is injected into the flask is about 1 × 106 Bq.

activity = _____________Bq(3)

(c) She waits for 3.5 h to allow the injected solution to mix thoroughly with the liquid in the flask. She then extracts 15 cm3 of the liquid from the flask and measures its activity which is found to be 3600 Bq.

Calculate the total activity of the sodium-24 in the flask after 3.5 h and hence determine the volume of liquid in the flask.


(3)

(d) The archaeologist obtained an estimate of the volume knowing that similar empty flasks have an average mass of 1.5 kg and that mass of the flask and liquid was 5.2 kg. Compare the estimate that the archaeologist could obtain from these masses with the volume calculated in part 4.3 and account for any difference.

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(Total 9 marks)

Q2.Which of the following best describes the decay constant for a radioisotope?

A The reciprocal of the half-life of the radioisotope.

B The rate of decay of the radioisotope.

C The constant of proportionality which links half-

life to the rate of decay of nuclei.

D The constant of proportionality which links rate of

decay to the number of undecayed nuclei.

(Total 1 mark)

Q3.After 64 days the activity of a radioactive nuclide has fallen to one sixteenth of its original value. The half-life of the radioactive nuclide is

A 2 days.


B 4 days.

C 8 days.

D 16 days.

(Total 1 mark)

Q4.The carbon content of living trees includes a small proportion of carbon-14, which is a radioactive isotope. After a tree dies, the proportion of carbon-14 in it decreases due to radioactive decay.

(a) (i) The half-life of carbon-14 is 5740 years.Calculate the radioactive decay constant in yr–1 of carbon-14.

decay constant ..................................... yr–1

(1)

(ii) A piece of wood taken from an axe handle found on an archaeological site has 0.375 times as many carbon-14 atoms as an equal mass of living wood.Calculate the age of the axe handle in years.

age ......................................... yr(3)

(b) Suggest why the method of carbon dating is likely to be unreliable if a sample is:

(i) less than 200 years old,


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(ii) more than 60 000 years old.

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(Total 6 marks)

Q5. The isotope of uranium, , decays into a stable isotope of lead, , by means of a series of α and β– decays.

(a) In this series of decays, α decay occurs 8 times and β– decay occurs n times.Calculate n.

answer = ...........................................(1)

(b) (i) Explain what is meant by the binding energy of a nucleus.

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(ii) Figure 1 shows the binding energy per nucleon for some stable nuclides.

Figure 1


Use Figure 1 to estimate the binding energy, in MeV, of the nucleus.

answer = ................................. MeV(1)

(c) The half-life of is 4.5 × 109 years, which is much larger than all the other half-lives of the decays in the series.

A rock sample when formed originally contained 3.0 × 1022 atoms of and no

atoms.

At any given time most of the atoms are either or with a negligible number of atoms in other forms in the decay series.

(i) Sketch on Figure 2 graphs to show how the number of atoms and the

number of atoms in the rock sample vary over a period of 1.0 × 1010 years from its formation.Label your graphs U and Pb.

Figure 2


(2)

(ii) A certain time, t, after its formation the sample contained twice as many

atoms as atoms.

Show that the number of atoms in the rock sample at time t was 2.0 × 1022.

(1)

(ii) Calculate t in years.


answer = ................................. years(3)

(Total 10 marks)

Q6. Complete the following equation showing the β+ decay of carbon-11.

(Total 3 marks)

Q7.The decay of a radioactive substance can be represented by the equation

A = A0e–λt

where A = the activity of the sample at time t A0 = the initial activity at time t = 0 λ = the decay constant

The half life, T½ of the radioactive substance is given by

T½ =

An experiment was performed to determine the half-life of a radioactive substance which was a beta emitter. The radioactive source was placed close to a detector. The total count for exactly 5 minutes was recorded. This was repeated at 20 minute intervals. The results are shown in the table below.


time, t /minutes

total count, C,recorded in5 minutes

count rate, R /counts minute–1

corrected count

rate, RC /counts minute–1

ln (RC / minute–1)

0 1016 203 183 5.21

20 892 178 158 5.06

40 774 155 135 4.90

60 665 133 113 4.73

80 608 122 102 4.62

100 546 109 89 4.49

(a) A correction has been made to the count rate, R, to give the corrected count rate, RC.Explain why this correction has been made and deduce its value from the table.

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(b) Draw an appropriate straight line through the plotted points.(1)

(c) Determine the gradient G of your graph.

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(d) Use your graph to determine the half-life in minutes of the radioactive substance used in this experiment.

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half-life, T½ .......................................... minutes(2)

(e) Due to the nature of a radioactive decay there will be an uncertainty in the total count recorded. What type of error is this called?

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(f) (i) It can be shown that the error in the total count C, is given by

uncertainty in total count C = ± √C

Using data from the table, calculate the uncertainty in the smallest total count, C.

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(ii) Hence calculate the percentage uncertainty in the smallest total count, C.

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(iii) Another student performed the same experiment with identical equipment but took total counts over a 1 minute period rather than a 5-minute period. The total count, C, at 140 minutes was equal to 84 counts. Estimate the percentage uncertainty in this total count, and hence explain the advantage of using a larger time.

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(Total 13 marks)

Q8. A nuclide of manganese undergoes beta+ decay to form a nuclide of chromium (Cr).

(a) Complete the equation for this decay process.

Mn Cr + β+ + (2)

(b) State the name of the exchange particle involved in this beta+ decay.

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(Total 3 marks)


Q9. Radioisotope thermoelectric generators (RTGs) are electrical generators powered by radioactive decay. As a radioisotope decays some of the energy released is converted into electricity by means of devices called thermocouples. In this way RTGs have been used as power sources in satellites, space probes and heart pacemakers.

The Cassini space probe was launched in 1997. It carried three RTGs each containing 11 kg of a nuclear fuel, plutonium oxide (a compound having two oxygen atoms combined with every plutonium-238 atom). In 1997, when the probe was launched, the power released from one gram of plutonium oxide was 500 mW.

Plutonium-238 is an alpha emitter, decaying into uranium(U).The half-life of the decay is 87.7 years.

mass of one mol of plutonium-238 = 238 gmass of one mol of oxygen atoms = 16 g

(a) State and explain why environmentalists might have been concerned by the use of such a large quantity of plutonium-238.

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(b) State and explain whether the activity of a given number of atoms of plutonium is affected when they are in the form of plutonium oxide.

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(c) (i) Calculate the decay constant, in s–1, for plutonium-238.


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decay constant ...................................... s–1

(2)

(ii) Calculate the number of plutonium-238 atoms in the total mass of the plutonium oxide in the Cassini probe at the beginning of its mission.

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number of plutonium-238 atoms ....................................(5)

(iii) Calculate the initial activity of the plutonium-238 in the Cassini probe.Give a suitable unit for your answer.

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initial activity of plutonium-238 ........................ unit .......................(3)

(d) (i) Write a nuclear equation for the decay.


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(ii) Assume the power released by the RTGs’ fuel originated as the kinetic energy

of the alpha particles emitted in the decay of .

Calculate the maximum kinetic energy of each alpha particle.

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kinetic energy of alpha particle ...................................... J(4)

(Total 20 marks)

Q10.The age of an ancient boat may be determined by comparing the radioactive decay of


from living wood with that of wood taken from the ancient boat.A sample of 3.00 × l023 atoms of carbon is removed for investigation from a block of living

wood. In living wood one in 1012 of the carbon atoms is of the radioactive isotope , which has a decay constant of 3.84 × 10–12 s–1.

(a) What is meant by the decay constant?

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(b) Calculate the half-life of in years, giving your answer to an appropriate number of significant figures.

1 year = 3.15 × 107 s

answer = ..................................... years(3)

(c) Show that the rate of decay of the atoms in the living wood sample is 1.15 Bq.

(2)

(d) A sample of 3.00 × 1023 atoms of carbon is removed from a piece of wood taken

from the ancient boat. The rate of decay due to the atoms in this sample is 0.65 Bq.Calculate the age of the ancient boat in years.


answer = ............................ years(3)

(e) Give two reasons why it is difficult to obtain a reliable age of the ancient boat from the carbon dating described.

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(Total 11 marks)

Q11. (a) Bi can decay into Pb by a β– followed by an α decay, or by an α followed by a β– decay. One or more of the following elements is involved in these decays:

Write out decay equations showing each stage in both of these decays.

First decay path Second decay path


(6)

(b) (i) Describe how you would perform an experiment that demonstrates that gamma radiation obeys an inverse square law.

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(ii) Explain why gamma radiation obeys an inverse square law but alpha and beta radiation do not.

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(Total 15 marks)


Q12. (a) Calculate the radius of the U nucleus.

r0 = 1.3 × 10–15 m

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(b) At a distance of 30 mm from a point source of rays the corrected count rate is C.Calculate the distance from the source at which the corrected count rate is 0.10 C, assuming that there is no absorption.

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(c) The activity of a source of particles falls to 85% of its initial value in 52 s. Calculate the decay constant of the source.

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(d) Explain why the isotope of technetium, 99Tc m, is often chosen as a suitable source of radiation for use in medical diagnosis.

You may be awarded additional marks to those shown in brackets for the quality of written communication in your answer.

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(Total 10 marks)

Q13. (a) Sketch, using the axes provided, a graph of neutron number, N, against proton number, Z, for stable nuclei over the range Z = 0 to Z = 80. Show suitable numerical values on the N axis.


(2)

(b) On the graph indicate, for each of the following, a possible position of a nuclide that may decay by

(i) α emission, labelling the position with W,

(ii) β– emission, labelling the position with X,

(iii) β+ emission, labelling the position with Y.(3)

(c) The isotope decays sequentially by emitting α particles and β– particles,

eventually forming the isotope . Four α particles are emitted in the sequence.

Calculate the number of β– particles in the sequence.


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(d) A particular nuclide is described as proton-rich. Discuss two ways in which the nuclide may decay. You may be awarded marks for the quality of written communication in your answer.

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(Total 10 marks)

Q14. (a) Suggest two reasons why an α particle causes more ionisation than a β particle of the same initial kinetic energy.

You may be awarded marks for the quality of written communication in your answer.

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(b) A radioactive source has an activity of 3.2 × 109 Bq and emits α particles, each with kinetic energy of 5.2 Me V. The source is enclosed in a small aluminium container of mass 2.0 × 10–4 kg which absorbs the radiation completely.

(i) Calculate the energy, in J, absorbed from the source each second by the aluminium container.

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(ii) Estimate the temperature rise of the aluminium container in 1 minute, assuming no energy is lost from the aluminium.

specific heat capacity of aluminium = 900 J kg–1 K–1

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(Total 7 marks)

Q15.The sodium isotope Na is a radioactive isotope that can be produced by bombarding the aluminium isotope Al with neutrons. Which line, A to D, in the table correctly represents the production of Na from the aluminium isotope Al and its subsequent decay?

production decay

A


B

C

D

(Total 1 mark)

Q16.(a) A solar panel of area 2.5 m2 is fitted to a satellite in orbit above the Earth. The panel produces a current of 2.4 A at a potential difference of 20 V when solar radiation is incident normally on it.

(i) Calculate the electrical power output of the panel.

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(ii) Solar radiation on the satellite has an intensity of 1.4 kW m–2. Calculate the efficiency of the panel.

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(b) The back-up power system in the satellite is provided by a radioactive isotope enclosed in a sealed container which absorbs the radiation from the isotope. Energy from the radiation is converted to electrical energy by means of a thermoelectric module.

(i) The isotope has an activity of 1.1 × 1014 Bq and produces α particles of energy 5.1 MeV. Show that the container absorbs energy from the α particles at a rate of 90 J s–1.


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(ii) The isotope has a half-life of 90 years. Calculate the decay constant λ of this isotope.

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(iii) The mass number of the isotope is 239. Calculate the mass of isotope needed for an activity of 1.1 × 1014 Bq.

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(Total 11 marks)

Q17.(a) When an α particle is emitted from a nucleus of the isotope , a nucleus of thallium, Tl, is formed. Complete the equation below.

→ α + Tl(2)


(b) The α particle in part (a) is emitted with 6.1 MeV of kinetic energy.

(i) The mass of the α particle is 4.0 u. Show that the speed of the α particle immediately after it has been emitted is 1.7 × 10–7 m s–1. Ignore relativistic effects.

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(ii) Calculate the speed of recoil of the daughter nucleus immediately after the α particle has been emitted. Assume the parent nucleus is initially at rest.

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(Total 8 marks)

Q18.(a) The nuclide can decay by electron capture to become an isotope of lead as shown in the following equation,

(i) Explain what is meant by electron capture.


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(ii) Give one reason why electromagnetic radiation is emitted following this process.

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(iii) Give the equation for another process in which is converted into an isotope of lead.

(5)

(b) The nuclide is also an α particle emitter. An initial measurement of the α particle activity of a sample of this isotope gives a corrected count rate of 1200 counts s–1. After an interval of 24 hours the corrected rate falls to 290 counts s–1. Assume that corrections have been made for the radiation both from daughter products and background radiation.

(i) Show that the decay constant of is about 1.6 × 10–5 s–1.

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(ii) Calculate the half-life of this sample.

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(iii) Calculate the number of nuclei in the sample when the corrected count rate was 1200 counts s–1.

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(Total 10 marks)

Q19.A radioactive nuclide decays by emitting α particles. The graph shows how the rate of decay At of the source changes with time t.


(a) Determine

(i) the half-life of the nuclide,

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(ii) the decay constant,

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(iii) the initial number of undecayed nuclei present at time t = 0.

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(b) Each decay releases 1.0 × 10–12 J. For the time interval between t = 30 s and t = 80 s, calculate

(i) the number of nuclei which decay,

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(ii) the energy released.

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(Total 9 marks)


Q20.(a) An α particle source of half-life 3420 years has a rate of decay of 450 kBq. Calculate

(i) the decay constant, in s–1,

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(ii) the number of radioactive atoms in the source.

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(b) A narrow beam of α particles is directed at a thin gold foil target in an evacuated vessel. Only a very small proportion of the α particles scatter backwards at an angle greater than 90° to the direction from which they came

(i) Describe what happens to the majority of the α particles incident on the gold foil.

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(ii) Several deductions may be made about the structure of gold atoms from the results of α– particle scattering. Write down two of these deductions.

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(Total 7 marks)


Q21.(a) Sketch a graph to show how the number of neutrons, N, varies with the number of protons, Z, for stable nuclei over the range Z = 0 to Z = 80. Draw a scale on the N axis.

(2)

(b) On the same graph, enclosing each region by a line, indicate the region in which nuclides are likely to decay, by

(i) α emission, labelling the region A,

(ii) β– emission, labelling the region B,


(iii) β+ emission, labelling the region C.(3)

(c) Complete the table.

mode of decay change in proton number Z change in neutron number N

α emission −2

β– emission

β+ emission

e capture

p emission 0

n emission 0

(3)(Total 8 marks)

Q22.A student attempted to determine the half-life of a radioactive substance, which emits α particles, by placing it near a suitable counter. He recorded C, the number of counts in 30 s, at various times, t, after the start of the experiment.

The results given in the table were obtained.

t / minute 0 10 20 30 40 50 60

number of counts in 30 s, C

60 42 35 23 18 14 10

ln C

(a) Explain what is meant by half-life.

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(b) Complete the table.(1)

(c) On the grid below

(i) plot ln C against t,

(ii) draw the best straight line through your points,

(iii) determine the gradient of your graph.

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(5)

(d) (i) Show that the decay constant of the substance is equal to the magnitude of the gradient of your graph.

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(ii) Calculate the half-life of the substance.

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(e) This particular experiment is likely to lead to an inaccurate value for the half-life. Suggest two ways in which the accuracy of the experiment could be improved.

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(f) The age of a piece of bone recovered from an archaeological site may be estimated by 14C dating. All living organisms absorb 14C but there is no further intake after death. The proportion of 14C is constant in living organisms.

A 1 g sample of bone from an archaeological site has an average rate of decay of 5.2 Bq due to 14C. A 1 g sample of bone from a modern skeleton has a rate of decay of 6.5 Bq. The counts are corrected for background radiation.

Calculate the age, in years, of the archaeological samples of bone.

half-life of 14C = 5730 years

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(Total 16 marks)

Q23.(a) Ancient rocks can be dated by measuring the proportion of trapped argon gas to the


radioactive isotope potassium-40. Potassium-40 produces argon as a result of electron capture. The gas is trapped in the molten rock when the rock solidifies.

(i) Write down an equation to represent the process of electron capture by a potassium nucleus.

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(ii) The atomic masses of potassium-40 and argon-40 are 39.96401 u and 39.96238 u, respectively. Calculate the energy released, in MeV, when the process given in part (a)(i) occurs.

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(iii) An argon atom formed in this way subsequently releases an X-ray photon. Explain how this occurs.

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(b) Potassium-40 also decays by beta emission to form calcium-40.

(i) Write down an equation to represent this beta decay.

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(ii) This process is eight times more probable than electron capture. A rock sample is found to contain 1 atom of argon-40 for every 5 atoms of potassium-40. The half-life of potassium-40 is 1250 million years. Calculate the age of


this rock.

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(Total 9 marks)

Q24.(a) The diagram is copied from a photograph taken of a cloud chamber containing a small radioactive source.

(i) What type of radiation is emitted from the source?

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(ii) State and explain what can be deduced about the energy of the particles emitted by the source.

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(4)

(b) Plutonium-239 is a radioactive isotope that emits α particles of energy 5.1 MeV and decays to form a radioactive isotope of uranium. This isotope of uranium emits α particles of energy 4.5 MeV to form an isotope of thorium which is also radioactive.

(i) Write down an equation to represent the decay of plutonium-239.

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(ii) Write down an equation to represent the decay of the uranium isotope.

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(iii) Which of the two radioactive isotopes, plutonium-239 or the uranium isotope, has the longer half-life? Give a reason for your answer.

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(iv) Explain why thorium is likely to be a β– emitter.

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(Total 9 marks)

Q25.Nuclei of decay by the emission of an α particle to form a stable isotope of an element X. You may assume that no γ emission accompanies the decay.

(a) (i) State the proton number and the nucleon number of X.

proton number ......................................................................................

nucleon number ....................................................................................

(ii) Identify the element X.

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(b) Each decaying nucleus of Po releases 8.6 × 10–13 J of energy.

(i) State the form in which this energy initially appears.

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(ii) Using only the information provided in the question, calculate the difference in mass between the original atom and the combined mass of an atom of X and an α particle.

speed of light in vacuum = 3.0 × 108 m s–1

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(Total 5 marks)


Q26.(a) (i) What is meant by the random nature of radioactive decay?

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(ii) Explain what is meant by each of the following.

isotopes ................................................................................................

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radioactive half-life ................................................................................

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radioactive decay constant ....................................................................

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(b) The radioactive isotope of iodine 131I has a half-life of 8.04 days. Calculate

(i) the decay constant of 131I,

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(ii) the number of atoms of 131I necessary to produce a sample with an activity of 5.0 × 104 disintegrations s–1 (Bq),

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(iii) the time taken, in hours, for the activity of the same sample of 131I to fall from 5.4 × 104 disintegrations s–1 to 5.0 × 104 disintegrations s–1.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(6)

(Total 12 marks)

Q27. The table below gives the values for the activity of a radioactive isotope over a period of a few minutes.

time/s 0 60 120 180 240 300

activity/Bq 480 366 280 214 163 124

(a) Complete the graph below by plotting the remaining points and drawing an appropriate curve.


(3)

(b) Use the graph to determine the half-life of the isotope.

half-life ......................................(3)

(c) Initially there were 1.1 × 105 atoms of the isotope present. Calculate the decay probability of the isotope.


decay probability.......................................(2)

(Total 8 marks)

Q28. (a) Radioactive lead – 214 changes to lead – 206 by a series of decays involving alpha (α) and negative beta (β–) emissions. Explain clearly how many alpha and beta particles are emitted during this change.

(4)

(b) The half-life of lead – 214 is 26.8 minutes.

(i) Explain what is meant by half-life.

...............................................................................................................

...............................................................................................................

...............................................................................................................

(ii) Show that the decay constant of lead – 214 is approximately 0.026 minute–1.


(iii) Calculate the percentage of the original number of nuclei of lead – 214 left in a sample after a period of 90 minutes.

(7)

(Total 11 marks)

Q29. (a) Carbon-14 ( C) decays with the emission of a beta particle to form nitrogen (N).

Nitrogen has a proton number of 7. Write down the full nuclear equation that describes this decay.

(3)

(b) A sample of pure C contains 6.3 × 1019 carbon atoms each with a decay probability of 3.8 × 10–12 s–1.

(i) State the S.I. unit of activity.

...............................................................................................................

(ii) Calculate the initial activity of the sample.


initial activity .........................................(3)

(c) All nuclei of carbon-14 ( C) have the same decay probability. Explain how this statement accounts for the observation that the number, N, of radioactive carbon atoms in the sample varies with time, t, as shown in the figure below.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(4)

(Total 10 marks)


Q30. (a) (i) Alpha and beta emissions are known as ionising radiations. State and explain why such radiations can be described as ionising.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(ii) Explain why beta particles have a greater range in air than alpha particles.

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(b) The figure below shows the variation with time of the number of Radon (220Ra) atoms in a radioactive sample.

(i) Use the graph to show that the half-life of the decay is approximately 53 s. Show your reasoning clearly.

(3)

(ii) The probability of decay (decay constant) for 220Ra is 1.3 × 10–2 s–1. Use data from the graph to find the activity of the sample at a time t = 72 s.


activity .....................................................(3)

(Total 10 marks)

##

Americium-241 ( Am) is a common laboratory source of alpha radiation. It decays spontaneously to neptunium (Np) with a decay constant of 4.8 × 10–11 s–1.

A school laboratory source has an activity due to the presence of americium of 3.7 × 104 Bq when purchased.

Avogadro constant = 6.0 × 1023 mol–1

one year = 3.2 × 107 s

(a) (i) Calculate the half-life, in years, of americium-241.

(2)

(ii) Calculate the number of radioactive americium atoms in the laboratory source when it was purchased.

(2)


(iii) Calculate the activity of the americium in the laboratory source 50 years after being purchased.

(3)

(iv) Suggest why the actual activity of the sources is likely to be greater than your answer to part (iii).

.............................................................................................................

.............................................................................................................(1)

(b) (i) Use the following data to deduce the energy released in the decay of one americium-241 nucleus.

mass of americium-241 nucleus = 4.00171 × 10–25 kgmass of an alpha particle = 0.06644 × 10–25 kgmass of neptunium nucleus = 3.93517 × 10–25 kgspeed of electromagnetic radiation = 3.00 × 108 m s–1

in free space


(3)

(ii) Explain what is meant by decays spontaneously and how consideration of the masses of particles involved in a proposed decay helps in deciding whether the decay is possible.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(Total 13 marks)

Q32. A freshly prepared radioactive source that emits negatively charged beta particles (β–) has an activity of 120 Bq and a half-life of 12 h.

(a) (i) State the effect on the proton number Z and the nucleon number A when a β– particle is emitted.

...............................................................................................................

...............................................................................................................(2)

(ii) Sketch, on the axes below, a graph that shows how the activity varies during the two days after the source was prepared.


(3)

(b) (i) The total energy released in each decay is 5.5 × 10–13 J.Calculate the initial energy produced each second by the source.

initial energy ..................................... J(1)

(ii) Figure 1 shows the energy spectrum for the beta particles emitted in the decay.

It shows that different energy beta particles are possible.

Figure 1


Explain why all the beta particles that are emitted do not have 5.5 × 10–13 J of energy.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(3)

(c) The probability of one of the radioactive atoms decaying each second is1.6 × 10–5.

How many radioactive atoms are present when the activity is 120 Bq?

number of radioactive atoms ...................................(1)

(d) A scientist undertaking an investigation places the freshly prepared source close to a Geiger-Müller tube as shown in Figure 2 and records a count rate of 50 counts per second.

Figure 2

State and explain two reasons why the measured count rate is lower than the activity of the source.

........................................................................................................................

........................................................................................................................

........................................................................................................................


........................................................................................................................(2)

(Total 12 marks)

Q33. Iodine-123 is a radioisotope used medically as a tracer to monitor thyroid and kidney functions. The decay of an iodine-123 nucleus produces a gamma ray which, when emitted from inside the body of a patient, can be detected externally.

(a) Why are gamma rays the most suitable type of nuclear radiation for this application?

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(b) In a laboratory experiment on a sample of iodine-123 the following data were collected.

time/h 0 4 8 12 16 20 24 28 32

count-rate /counts s–1 512 410 338 279 217 191 143 119 91

Why was it unnecessary to correct these values for background radiation?

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(c) On the axes provided in the diagram below, complete the graph of count-rate against time.


(2)

(d) Use your graph to find an accurate value for the half-life of iodine-123.Show clearly the method you use.

Half-life ..............................(3)

(e) Give two reasons why radioisotopes with short half-lives are particularly suitable for use as a medical tracer.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................


........................................................................................................................(2)

(Total 11 marks)

Q34. A nucleus of plutonium ( ) decays to form uranium (U) and an alpha-particle (α).

(a) Complete the equation that describes this decay:

(2)

(b) (i) Show that about 1 pJ of energy is released when one nucleus decays.

mass of plutonium nucleus = 3.98626 × 10–25 kgmass of uranium nucleus = 3.91970 × 10–25 kgmass of alpha particle = 6.64251 × 10–27 kgspeed of electromagnetic radiation = 2.99792 × 108 m s–1

(3)

(ii) The plutonium isotope has a half-life of 2.1 × 1011 s. Show that the decay constant of the plutonium is about 3 × 10–12 s–1.

(2)


(iii) A radioactive source in a school laboratory contains 3.2 × 1021 atoms of plutonium. Calculate the energy that will be released in one second by the decay of the plutonium described in part (b)(i).

(3)

(iv) Comment on whether the energy release due to the plutonium decay is likely to change by more than 5% during 100 years. Support your answer with a calculation.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(4)

(Total 14 marks)

Q35. The graph below shows the number of radioactive nuclei remaining in a sample of material against time. The radioactive isotope decays to a non-radioactive element.


(a) Use the graph to show that, after a time of 500 s, about 6 × 104 nuclei are decaying every second.

(3)

(b) Calculate the decay probability (decay constant) of a nucleus of the radioactive isotope in the sample.

Decay probability ...............................................(3)

(Total 6 marks)

Q36. Th is a radioactive isotope of thorium.


(a) State for an atom of Th,

(i) the number of protons in the nucleus,

...............................................................................................................(1)

(ii) the number of neutrons in the nucleus.

...............................................................................................................(1)

(b) A sample of pure Th that contains 2.6 × 1021 atoms is observed to decay at an initial rate of 3.0 × 1013 Bq.

(i) State the unit of decay probability ..........................................................(1)

(ii) Calculate the probability of decay for an atom of Th.

Probability of decay .......................................(2)

(Total 5 marks)

Q37.A Geiger-Muller tube and counter were used to detect β– particles emitted by a source


during a radioactivity experiment. The diagram below shows the corrected count rate detected by the counter plotted against time since the beginning of the experiment.

(a) Draw on the graph the curve that best fits these data.

Use the graph to determine the half-life of the radioactive source as accurately as you can.

Half-life ...................................................(3)

(b) State why the count rate has to be corrected before it is plotted on the graph.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(c) Explain how the graph and your final answer for the half-life would change if the correction were not applied.

........................................................................................................................

........................................................................................................................


........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(d) Describe, giving reasons, how you would determine an accurate value for the correction.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(3)

(Total 10 marks)

Q38.Potassium-42 decays with a half-life of 12 hours. When potassium-42 decays it emits β– particles and gamma rays. One freshly prepared source has an activity of 3.0 × 107 Bq.

(a) To determine the dose received by a scientist working with the source the number of gamma ray photons incident on each cm2 of the body has to be known.

One in every five of the decaying nuclei produces a gamma ray photon. A scientist is initially working 1.50 m from the fresh source with no shielding. Show that at this time approximately 21 gamma ray photons per second are incident on each cm2 of the scientist's body.

(2)

(b) The scientist returns 6 hours later and works at the same distance from the source.

(i) Calculate the new number of gamma ray photons incident per second on each cm2 of the scientist's body.

(3)


(ii) At what distance from the source could the scientist now work and receive the original dose of 21 photons per second per cm2 .

(2)

(c) Explain why it is not necessary to consider the beta particle emission when determining the dose of radiation the scientist receives.

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(Total 9 marks)

Q39.(a) (i) Explain what is meant by background radiation.

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(ii) Name a source of background radiation.

...............................................................................................................(1)

(b) State the meaning of each of the terms in the equation A = λN

A .....................................................................................................................

........................................................................................................................

λ ......................................................................................................................

........................................................................................................................

N .....................................................................................................................

........................................................................................................................(3)

(Total 6 marks)


Q40.(a) Sketch a graph to show how the activity of a radioactive sample changes with time. You are not expected to give any numerical values.

(2)

(b) The activity of 5.0 × 1016 nuclei of a radioisotope is 1.1 × 1014 Bq. Calculate the probability of decay for a nucleus of the sample.

(3)(Total 5 marks)

Q41.Which of the following does not give a value in seconds?

A capacitance × resistance

B

C half-life

D

(Total 1 mark)


Q42.The graph below shows how the nucleon number A changes with proton number Z for the decay series that starts with uranium-238. The half-lives of each decay are also shown.

(a) How many alpha particles and beta particles are emitted when a uranium-238 nucleus decays to radon-222 (222Rn)?

Number of alpha particles ..............................................................................

Number of beta particles.................................................................................


(2)

(b) How many neutrons are there in a nucleus of polonium-210 (210Po)?

........................................................................................................................(1)

(c) Identify the stable isotope that results from this decay chain.

........................................................................................................................(1)

(d) 214 g of bismuth-214 (214 Bi) contains 6.0 × 1023 atoms. A sample containing only bismuth-214 has an initial mass of 0.60 g.

(i) After what period of time will the mass of bismuth-214 present in the sample be 0.15 g?

(2)

(ii) Determine the number of bismuth-214 atoms present after this time.(1)

(iii) Calculate the activity of the bismuth-214 in the sample after this time.(4)

(iv) Explain how the total activity of the sample will be different from the value calculated in (iii).

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(v) The bismuth-214 decays into polonium-214. Explain why you would find very


little polonium-214 if you were to analyse the sample.

...............................................................................................................

...............................................................................................................(2)

(Total 15 marks)

Q43.A detector and counter are used to measure the count rate from a gamma source.

(a) Complete the graph to show how the corrected count rate will vary with the distance, d, between the source and the detector. One point has been plotted. To complete the graph accurately, you should perform a suitable calculation to determine the position of one other point on the graph.

(2)

(b) (i) State what is meant by corrected count rate.

...............................................................................................................

...............................................................................................................(1)

(ii) State one means by which you would ensure that the measurement of count rate is accurate.

...............................................................................................................

...............................................................................................................(1)

(Total 5 marks)


Q44.An isotope of technetium is a gamma emitter used by doctors as a tracer in the human body. It is injected into the patient’s blood stream. Scanners outside the body measure the gamma activity, enabling the blood flow to be monitored.

(a) The graph shows the variation of activity with time, t, for a sample of the isotope.

(i) Use data from the graph to determine the half-life of the technetium isotope.(3)

(ii) The decay constant of the technetium isotope is 3.2 × 10−5 s−1. Use data from the graph and the equation A = λN to calculate the number of nuclei of the radioactive technetium isotope present at time t = 0.

(2)

(b) (i) State why an alpha emitter would not be suitable in this application.

...............................................................................................................

...............................................................................................................(1)

(ii) State why the half-life of the technetium isotope makes it suitable for this application.


...............................................................................................................

...............................................................................................................(1)

(c) State and explain how the presence of the technetium isotope may do some damage to the patient.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(Total 9 marks)

Q45. is an isotope of uranium which decays through a decay chain to the isotope of lead,

, which is stable. During the sequence of decays, both alpha particles and beta particles are emitted.

(a) (i) represents an alpha particle.

State the values of:

A ....................................................................

Z ....................................................................(1)

(ii) represents a beta particle.

State the values of:

A ....................................................................

Z ....................................................................(1)

(iii) State and explain how many alpha particles are emitted during the decay chain


between .

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(iv) State and explain how many beta particles are emitted during the decay chain

between and .

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(b) The decay of uranium can be used to measure the age of rocks. A rock, which

contained N0 atoms of when it was formed, will contain NU atoms of the isotope at a later time t.

(i) The relationship between N0, NU and t is:

NU = N0e–λt

State the meaning of the decay constant, λ.

...............................................................................................................

...............................................................................................................(1)

(ii) Because the half-life of is much longer than that of any other member of

the decay chain, it may be assumed that the number of atoms of

present at time t is equal to the number of atoms which have decayed. Show that the ratio of the number of lead atoms (NP) to the number of uranium atoms at time t is given by:

= eλt – 1(2)


(iii) The half-life of is 4.5 × 109 year. Calculate the age of a rock sample

which contains 4.2 parts per million of and 0.62 parts per million of .

(4)

(iv) The above analysis assumes that the rock contained no lead when it was formed. State and explain the effect on the estimated age if the rock had contained lead when it was formed.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(v) State reasons why the measurement of the amount of uranium in a rock sample would be inaccurate when performed by a radiation counting technique.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(Total 17 marks)

Q46.The diagram below shows an arrangement used to maintain a constant thickness of sheet paper or steel as it is being rolled. A radioactive source and detector are used to monitor the thickness.


(a) Explain briefly how this arrangement works.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(3)

(b) Alpha, beta or gamma sources could be selected for use in such an arrangement.

State which source should be selected in each case and explain briefly why the others would not be suitable.

Paper:

Source selected .............................................................................................

Reasons why the others are unsuitable ........................................................

........................................................................................................................

........................................................................................................................

Steel:

Source selected .............................................................................................

Reasons why the others are unsuitable ........................................................

........................................................................................................................

........................................................................................................................(4)

(c) Cobalt-60 is commonly used as a source in such applications. This has a half-life of 5.3 years. When fresh the source contains 5.0 × 1020 radioactive atoms.


For it to be useful the source has to have an activity of at least 1.5 × 1012 Bq.

(i) What is meant by an activity of 1 Bq?

...............................................................................................................

...............................................................................................................(1)

(ii) Draw a graph showing the number of radioactive atoms in the source over a period of 3 half-lives. Include suitable scales on the axes.

(2)

(iii) Determine the decay constant of cobalt-60 in s–1.(2)

(iv) After what time will it be necessary to replace the source?(3)

(Total 15 marks)

Q47.The radioisotope iodine-131 ( ) is used in medicine to treat over-active thyroid glands. It decays into an isotope of xenon (Xe) by β– emission with a half-life of 8.1 days. The xenon subsequently emits a γ ray.

(a) Explain what is meant by:

(i) isotope;

...............................................................................................................


...............................................................................................................(2)

(ii) half-life.

...............................................................................................................

...............................................................................................................(1)

(b) Write down the equation which represents the nuclear reaction.(3)

(c) Calculate the time (in days) for a sample of iodine to decay to 1% of its initial activity.

(4)

(d) State and explain which decay product can be detected outside the body during treatment.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(Total 12 marks)

Q48.The actinium series of radioactive decays starts with an isotope of uranium, nucleon (mass) number 235, proton (atomic) number 92.

Which line in the table shows the nucleon number and proton number of the isotope after the emission of 5 α particles and 2 β– particles?

Nucleon number proton number

A 213 82

B 215 80


C 215 84

D 227 87

(Total 1 mark)

Q49.A small portion of the hydrogen in air is the isotope tritium . This is continually being formed in the upper atmosphere by cosmic radiation so that the tritium content of air is constant. Tritium is a beta emitter with a half-life of 12.3 years.

(a) (i) Write down the symbols for the two isotopes of hydrogen, the atoms of which have lower masses than those of tritium.

...............................................................................................................(1)

(ii) Write down the nuclear equation that represents the decay of tritium using the symbol X for the daughter nucleus.

...............................................................................................................(2)

(iii) Calculate the decay constant for tritium in year–1.(1)

(b) When wine is sealed in a bottle no new tritium forms and the activity of the tritium content of the wine gradually decreases with time. At one time the activity of the tritium in an old bottle of wine is found to be 12% of that in a new bottle. Calculate the approximate age of the old wine.

(3)

(c)


mass of a tritium nucleusmass of a protonmass of a neutronatomic mass unit, uspeed of electromagnetic radiation in free space

= 3.016050 u= 1.007277 u= 1.008665 u= 1.660566 × 10–27 kg= 3.0 × 108 m s–1

Calculate:

(i) the mass change, in kg, when a tritium nucleus is formed from its component parts,

(2)

(ii) the binding energy, in J, of a tritium nucleus.(2)

(Total 11 marks)


M1.(a) (3.0 × 10-10/24) × 6.02 × 1023 seen✓

(7.52 × 1010)1

(b) Decay constant = (0.69 / 14.8 h-1) or 1.3 × 10-5 s-1✓

A = 1.30 × 10-5 × 7.5 × 1010 ✓

9.75 × 105 Bq ✓

Allow 2 or 3 sfAllow use of A = λN with an incorrectly calculated decay constant

3

(c) Activity 3.5 h later should be A = 9.8 × 105 e-0.0466 x 3.5 ✓

8.33 × 105 Bq✓

Volume of liquid = (8.33 × 105 / 3600 ) × 15 = 3470 cm3✓3

(d) Estimate gives 3700 compared with 3500✓

Flask has more mass than average / liquid is not water✓2

[9]

M2.D[1]

M3.D[1]


M4.(a) (i) λ ( = ln 2 / T1 / 2 = 0.693 / 5740 ) = 1.2 × 10−4 (yr−1) ✓ (1.21 × 10−4 yr−1)

only allow 3.83 × 10−12 s−1 if the unit has been changed working is not necessary for mark

1

(ii) (use of Nt = No e– λt and activity is proportional to N At = Ao e– λt ) 0.375 = exp - (1.21 × 10−4 × t) ✓

t = ✓ t = 8100 or 8200(yr) ✓

1st mark substitution, allow EC from (i) 2nd mark rearranging, allow EC from (i) Allow t / T1 / 2 = 2 n approach 3rd mark no EC (so it is not necessary to evaluate a CE) so max 2 for a CE full marks can be given for final answer alone. A minus in the final answer will lose the last mark

3

(b) (i) (it is difficult to measure accurately) the small drop / change in activity / count-rate the small change / drop in the ratio of C-14 to C-12 ✓

the activity would be very small / comparable to the background or the ratio of C-14 to C-12 is too small or there are too few C-14 atoms or there is very little decay or the level of C-14 (in the biosphere) is uncertain (this long ago) ✓

1st mark needs some reference to a change in count-rate or activity for the mark be lenient in 2nd mark in reading a script assume C-14 is the subject. Eg ‘there is little activity to work with’ scores mark. Also allow any reasonable suggestion. Eg carbon may have been removed by bonding to surrounding material Don't allow, ‘All the carbon has decayed’

2[6]

M5. (a)


β = 6 1

(b) (i) the energy required to split up the nucleus

into its individual neutrons and protons/nucleons

(or the energy released to form/hold the nucleus

from its individual neutrons and protons/nucleons )2

(ii) 7.88 × 206 = 1620 MeV (allow 1600-1640 MeV)1

(c) (i) U, a graph starting at 3 × 1022 showing exponential fall passing through

0.75 × 1022 near 9 × 109 years

Pb, inverted graph of the above so that the graphs cross at 1.5 × 1022 near4.5 × 109 years

2

(ii) (u represents the number of uranium atoms then)

u = 6 × 1022 – 2u

u = 2 × 1022 atoms1

(iii) (use of N = No e-λt)

2 × 1022 = 3 × 1022 × e-λt

t = ln 1.5 / λ

(use of λ = ln 2 / t1/2)

λ = ln 2 / 4.5 × 109 = 1.54 × 10-10

t = 2.6 × 109 years (or 2.7 × 109 years)3


[10]

M6. boron numbers correct: A = 11; Z = 5

B1

β+ correct: A = 0; Z = (+)1

B1

νe (not anti neutrino) with numbers correct: 0,0

B13

[3]

M7.(a) Background radiation ✔Background count = 20 count/minute unit required ✔

Ignore any –ve sign for background countMust be written to 2 sf

2

2

(b) Correct line of best fit ✔The line must be a straight line (as instructed), with approximatelyan equal number of points on either side of the line.

1

(c) Triangle drawn with smallest side at least 8 cm ✔(or 8 grid squares)correct values read from graph ✔gradient = – 0.00698 (± 0.00030) min–1 ✔must have –ve sign and must be to 2 or 3 sf ✔


Gradient must lie within limits stated. No ecf from incorrectly readvalues unless it falls within stated limits. No unit penalty.

3

(d) Recognises gradient = (–) λ orUses gradient for value of λ = 7.0 (± 0.30) × 10–3 minute–1 ✔T½ = 99 minutes to 2 or 3 sf ✔

For 1st mark accept evidence that value of gradient has been substituted into correct formula for half life. No penalty for missing –ve sign. Allow ect from incorrect gradient value. Unit penalty if half life has been calculated in different unit (to minutes stated in question)

2

(e) Random ✔1

(f) (i) Uncertainty = ( ± √429) = ± 21No sf penalty ✔

Details of calculation not required. Marks can be awarded for correct numerical answers. Also no penalty for quoting uncertainty or % uncertainty without ‘±’.

1

(ii) % uncertainty = ± 4.9% ✔No sf penalty. (Note that % uncertainty in total count is same as % uncertainty in corresponding count rate.)

Accept also 4.8% (number achieved keeping all sig figs in calculator)No penalty for omitting % or ‘±’.No sf penalty

1

(iii) % uncertainty for 84 counts is ± 10.9% ✔Taking data over larger time period / larger total count will have smaller percentage uncertainty. ✔

Accept ±11%No penalty for omission of ± sign. No sf penalty for estimated % uncertainties.

2[13]


M8. (a) correct numbers for beta+ (0, (+)1) and chromium (52)

B1

(electron) neutrino with correct numbers (0,0)

B12

(b) W+/W/(intermediate vector) boson (not Z boson)

B11

[3]

M9. (a) plutonium is toxic/large mass of plutonium

B1

harmful if released into atmosphere/explosion occurred

B1

alphas dangerous when ingested/during launch etc

B1max2

(b) unaffected

B1

chemical bonding involves electrons (atomic)radioactivity is nuclear (owtte)/same number of nucleipresent

B12


(c) (i) T½ = ln2/λ

C1

2.51 × 10–10

A12

(ii) molar mass calculated (0.270 kg)

C1

use of 33 kg

C1

number of moles in sample (122.2)

C1

multiplication of value by Avogadro’s number

C1

7.36 × 1025

A15

(iii) (c) (i) × (c) (ii)

C1

1.83 × 1016 cao

A1

Bq

B13

(d) (i) uranium correct (234,92)

B1

alpha correct (4,2) – accept He or α symbol

B12


(ii) use of 1 g generating 500 mW

C1

16500 W total

C1

recognition that activity × energy of one alpha = power

C1

9.00 × 10–13 (J)

A14

[20]

M10.(a) probability of decay per unit time/given time period

or fraction of atoms decaying per second

or the rate of radioactive decay is proportional to the number of (unstable)nuclei

and nuclear decay constant is the constant of proportionality (1)1

(b) use of =

= ln2/3.84 × 10–12 s (1) (1.805 × 1011 s)= (1.805 × 1011/3.15 × 107) = 5730 y (1)

answer given to 3 sf (1)3

(c) number of nuclei = N = 3.00 × 1023 × 1/1012 (1)


(= 3.00 × 1011 nuclei)

(using )

rate of decay = 3.84 × 10–12 × 3.00 × 1011 (1)

(= 1.15 Bq)2

(d) (N = N0e–λt and activity is proportional to the number of nuclei A N use ofA = A0e–λt)

0.65 = 1.15 × (1)

t = 4720 y (1)3

(e) the boat may have been made with the wood some time after the tree wascut down

the background activity is high compared to the observed count rates

the count rates are low or sample size/mass is small or there is statisticalvariation in the recorded results

possible contamination

uncertainty in the ratio of carbon-14 in carbon thousands of years agoany two (1)(1)

2[11]

M11. (a) number correct for alpha (1)

number correct for beta (1)

alpha decay first goes via Tl (1)

numbers correct for Tl (208, 81) (1)

beta decay first goes via Po (1)


numbers correct for Po (212, 84) (1)6

(b) (i) use of GM tube + counter/rate-meter (1)measurement of count rate (1)at range of distances + suitable ruler or tape measure (1)specifies suitable range (1)determines background & corrects (1)safety precaution given (1)graph of count rate or corrected count rate against 1/d2 (1)

max 6

(ii) gamma not absorbed (1)spreads uniformly from a pointsource/spherically symmetrically (1)area over which it spreads is proportionalto radius squared (1)alpha and beta are absorbed in addition to spreading out (1)

max 3[15]

M12. (a) R (= r0A1/3) = 1.3 × 10–5 × (238)1/3 (1)= 8.0(6) × 10–15m (1)

2

(b) (use of inverse square law e.g. gives)

10 = (1)

x = 0.095 m (1)(0.0949 m)

2

(c) (use of A = Aoexp(–λt gives) 0.85 = 1.0 exp (–λ52) (1)

(1)


= 3.1(3) × 10–3s–1 (1)3

(d) it only emits γ rays (1)relevant properties of γ radiation e.g. may be detected outsidethe body/weak ioniser and causes little damage (1)it has a short enough half-life and will not remain activein the body after use (1)it has a long enough half-life to remain active during diagnosis (1)the substance has a toxicity that can be tolerated by the body (1)it may be prepared on site (1)

any three (1)(1)(1)3

[10]

M13. (a) graph passes through N = 10/11 when Z = 10 and N increases as Zincreases (1)N = 115 → 125 when Z = 80 and graph must bend upwards (1)

2

(b) (i) W at Z > 60 just (within one diagonal of a square) below line (1)

(ii) X just (within one diagonal of a square) above line (1)

(iii) Y just (within one diagonal of a square) below line (1)3

(c) working showing the change due to emission of four α particles (1)four β– particles (1)

1

(d) Any two from the following list of processes:

β+

describe the changes to N (up by 1) and Z (down by 1)[or allow p change to n]

α


move closer to line of stability[or state the proton to neutron ratio is reduced]

ponly if nuclide is very proton rich[or electrostatic repulsion has to overcome the strong nuclear force][or highly unstable][or rare process]

e– capturedescribe the changes to N (up by 1) and Z (down by 1)allow p changes to n

marking: listing two processes (1)discussing each of the two processes (1) (1)

3QWC 1

[10]

M14. (a) reasons:α particle has much more mass/momentum than β particleα particle has twice as much charge as a β particleα particle travels much slower than a β particle any two (1) (1)

2QWC 1

(b) (i) energy absorbed per sec (= energy released per sec)= 3.2 × 109 × 5.2 × 106 ×1.6 ×10–19 (1)= 2.7 ×10–3 (J) (1) (2.66 × 10–3 (J))

(ii) temperature rise in 1 minute

= (for numerator) (1) (for denominator) (1)

= 0.90 K (or °C) (1)

(allow C.E. for incorrect value in (i))5

[7]


M15.B[1]

M16.(a) (i) (use of P = VI gives) P (= 2.4 × 20) = 48 W (1)

(ii) incident (solar) power (= 1.4 × 2.5) = 3.5 (kW) (1)

efficiency = (1)

= 0.014 (1) (or 1.4%)

(allow C.E. for incorrect values of input and output power)

4

(b) (i) in 1 s source emits 1.1 × 1014 particles (1) energy emitted in 1 s = 1.1 × 1014 × 5.1 × 1.6 × 10–13 (J) (1) (= 90 J)

(ii) T1/2 = + correct use or λ = (1)

= 2.44 × 10–10 s–1 (1)

[or λ = = 7.7 × 10–3 yr–1]


(iii) (allow C.E. for incorrect value of λ in (ii))

mass of isotope = (1)

= 0.18 kg (1)(allow C.E. for incorrect no. of nuclei)

7[11]

M17.(a)

either (1) (for both atomic mass numbers, 4 and 208) and (1) (for both atomic numbers, 2 and 81) [or (1) for Tl and incorrect α]

2

(b) (i) Ek = (½mv2) = 6.1 × 106 × 1.6 × 10-19(J) (1) substitution for m = 4.0 × 1.66 × 10-27(kg) (1)

(1) (= 1.7 × 107m s-1)

(ii) correct use of conservation of momentum mTl vrecoil = mα v (1) substitution of mTl = 208u (1) (allow C.E. for mass = 208)

vrecoil = = 3.3 × 105 m s-1 (1)

(allow C.E. for value of v)6

[8]

M18.(a) (i) (inner) orbiting electron [or electron surrounding the nucleus] (1) captured by a proton (in the nucleus) (1) converted into a neutron (1)


The Quality of Written Communication marks are awarded for the quality of answers to this question.

(ii) daughter nuclide / nucleus / atom might be excited and energy given up as electromagnetic radiation [or orbiting electrons drop down to fill space (left by captured electron)] (1)

The Quality of Written Communication marks are awarded for the quality of answers to this question.

(iii) max 5

(b) (i) (use of N = Noe-λt and N ∝ activity gives) 290 = 1200 exp(-λ × 24 × 60 × 60) (1)

λ = (1) (= 1.64 × 10–5 s–1)

(ii) (use of T½ = ln 2 / λ gives) T½ = (1)

= 4.2(3) × 104s (1) (= 11.(7) hr)

(use of λ = 1.6 × 10-5s-1 gives T½ = 4.3 × 104s or 12 hr)

(iii) (use of = λN gives) (-)1200 = (-)1.64 × 10-5 N (1)

N = 7.3(2) × 107 (nuclei) (1) (use of λ = 1.6 × 10-5s-1 gives N = 7.5 × 107 (nuclei))

max 5[10]

M19.(a) (i) T1/2 = 50 s (1) (from graph)

(ii) λ = (1) = 0.014 s–1 (1)

(iii) (5)


(b) (i) elapsed time = 50 s = 1 half-life (1)N30 = N0e–30λ (1) = 1.71 × 107 e–30×0.014 = 1.12 × 107 (1)

∴ no. decayed from t = 30 s to t = 80 s is = 0·56 × 107 (1)

[alternative (b)(i)N30 = N0e–30λ and N80 = N0e–80λ (1)give 1.12 × 107 (1) and 0.56 × 107 (1)number decayed (= 1.12 × 107 – 0.56 × 107) = 0.56 × 107 (1)]

(ii) energy released = 0.56 × 107 × 1.0 × 10–12 = 5.6 × 10–6 J (1)(max 4)

[9]

M20.(a) (i) (1)

= 6.43 × 10–12 (s–1) (1)

(ii)

= 7.0 × 1016 (1)(4)

(b) (i) pass through with no [or very small] deflection (1)


(ii) volume of nucleus << volume of atom (*)[or nucleus small and atom mostly empty space] (*)most of mass in nucleus (*)nucleus has positive charge (*)size of nucleus << separation (*)(*) any two (1) (1)

(3)[7]

M21.(a)

straight line between (Z = 0, N = 0) to (Z = 20, N = 20) (1)curving upwards to Z = 80; N = 110 – 130 (1)

(2)

(b) (i) A = any region below the line of stabilitybut N > 80 and Z > 60

(ii) B = any region above and close to the line of stability (1)

(iii) C = any region below and close to the line of stability (1)(3)


(c)

mode of decay change in proton

number, Zchange in neutron

number, N

α emission –2 –2

β– emission +1 –1

β+ emission –1 +1

e capture –1 +1

p emission –1 0

n emission 0 –1

(1)(1) (1) – lose one mark for each row in error(3)

[8]

M22.(a) time for half of (active) nuclei (of radioactive substance) to decay (1)(1)

(b)

t / minute 0 10 20 30 40 50 60

number of counts in 30s, C 60 42 35 23 18 14 10

ln C 4.094 3.738 3.555 3.135 2.890 2.639 2.303

Correct values of ln C above (1)(1)

(c) (i) seven points correctly plotted (1) (1) [six points correct (1)]


(ii) best straight line through points (1)

sensible scale (1)

(iii) from sensible triangle on graph (1)

gradient = – (1) 0.030 [0.294] (1) (min–1)(max 5)

(d) (i) C = C0e–λt, ln C – ln C0 = –λthence using y = mx + c, λ = (–)gradient (1)

(ii) half-life = (1) = 23 min (1)(3)

(e) count over longer period than half minute [or repeat experiment] (1)

use stronger source (1)

use background count correctly(1)

Colonel Frank Seely School(max 2)

(f) for 14C, λ = = 1.21 × 104 (year) (1)

t = = 1840 (year) (1)(4)

[16]

M23.(a) (i)

(ii) Δm = 39.96401 – 39.96238 = 0.00163 u (1)

Q (= 0.00163 × 931) = 1.5 (MeV) (1)

(iii) orbital electron vacancy due to electron capture (1)

outer electron fills vacancy and emits X-ray photon (1)(max 5)

(b) (i)

(ii) use of beta emission 8 times more probable than electron capture (1)

use of number of potassium atoms equals 5 times number of

argon atoms (1)

leading to evaluation of 36% (1)


use of the decay equation (1)

correct answer (1)

for example:

14 atoms of K-40 originally

becomes 8 atoms of Ca-40 + 1 atom of Ar-40 + 5 atoms K-40 unchanged

36% of K-40 not decayed

use N = N0e–λt to give t = = 1860 × 106 yr(max 4)

[9]

M24.(a) (i) alpha (1)

(ii) two different track lengths (1)

short range particles have lower energy than long range particles (1)

particles in each range have same energy (1)(4)

(b) (i)

(ii)

(iii) U-235 (1)

because of the inverse relationship

between half-life and alpha particle energy (1)


(iv) because the Th-90 nucleus is neutron-rich compared

with U-235 [or Pu-239] (1)(5)

[9]

M25.(a) (i) proton number 82 and nucleon number 214 (1)

(ii) Pb (1)2

(b) (i) kinetic energy [or electrostatic potential energy] (1)

(ii) Δm = (1)

= = 9.6 × 10–30 kg (1)3

[5]

M26.(a) (i) which atom decays (1)at what time is chance (1)

(ii) isotopes are (different forms) of same element (1)same proton number, Z, different nucleon number, A[or with same number of protons but different number of neutrons] (1)

half-life is time for number of nuclei to halve [or to halve activity] (1)for a particular isotope (1)

= -λN(1)

λ is constant of proportionality [or probability of decay] (1)[or λ is probability of decay (1) in unit time (1)]

max 6


(b) (i) λ = (1) = 1.0 × 10–6 s–1 (1)

(ii) N = (1) = 5.0 × 1010

(iii) ln = λt (1)

t = (1) = 7.7 × 104 s (1)

= = 21(.4) (hour) (1)max 6

[12]

M27. (a) all plots correct to ½ small squarededuct 1 mark for one incorrect, 2 marks for 2+ incorrect

B2

line appropriate

B13

(b) one correct determination from correct numbers

B1

154 ± 10 s

B1

two correct determinations and average

B13

(c) (use of A = λN) 480 = λ × 1.1 × 10–5

[allow λ = ln 2/t ½ ]


C1

4.4 × 10–3 s–1 [4.36]

A12

[8]

M28. (a) alpha decay Z-2 and A-4

B1

beta decay Z+1 and A const

B1

2α’s

B1

2α’s and 4β’s

B14

(b) (i) time taken for half number of nuclei to decaytime for activity to halve not sample/particles etc

B1

(ii) T ½ = 0.69/λ

C1

0.0257 ( minute–1)

A1

(iii) N = N0e−λt

C1

candidate’s substituted values ( or 3.36 half-lives)

C1

correct use of ratio N/N0 (or 0.5 3.36)


C1

9.6% → 9.9%

A1

C1 for little longer than 3 half-lives7

[11]

M29. (a) 147N on rhs

B1

0-1e/0-1β on rhs

B1

antineutrino shown

B13

(b) (i) Becquerel/Bq

B1

(ii) 6.3 × 1019 × 3.8 × 10-12

C13

= 2.4 × 10 + [no unit required]

A1

(c) route A

activity is proportional to the number of atoms


B1

activity is the change in the number of atoms per second

B1

max 2 from:

graph shows:N decreases as activity decreases

B1Max 4

exponential

B1

constant half life

B1

gradient decreases as N decreases

B1

or

route B

defines half life in words or by example

B1

t1/2 related to λ

B1Max 4

same λ implies constant t1/2

B1

graph shows constant half-life behaviour

B1[10]

M30. (a) (i) remove electrons from atoms


B1

by colliding with them/knocking into them

B12

(ii) fewer collisions per metre

B1

because beta particles are smaller/less ionising(accept smaller charge) not less mass

B1

so lose energy over a larger distance

B1max 2

(b) (i) one correct construction shown on graph (52 s – 53 s)

M1

at least one other determination visible on graph

M1

averaging process shown

A13

(ii) N = 1.08 to 1.1 × 1021 (condone errors in powers of 10)

B1

A = λN

C1

1.4(3) × 1019 Bq (allow decays/second) allow ecffor minor error in N

A1

orattempt to find gradient at 72 scorrect extraction from a tangent of graph1.3 to 1.5 × 1019 Bq (allow decays / second)

Colonel Frank Seely School3

[10]

M31. (a) (i) half life = 1.44 × 1010 s or half life = 0.69/λ or ln2/λ

C1

450 (449) y (or 460(456)y if they use 3.15 × 107 s = 1y)

A12

(ii) A = λN

C1

7.7 × 1014

A12

(iii) A = Ao e−λt

C1

correct substitution (A = 3.7 ×104e )

C1

3.4 × 104 Bq

A13

(iv) the decay products/neptunium may also be radioactive

B11


(b) (i) attempt to determine mass change or quotes E = mc2

C1

mass change = 1.00 × 10−29kg

C1

9.0 × 10−13 J or 5.63 MeV(condone correct answer in J followed by incorrectattempt to convert to MeV)

A13

(ii) decays without the need for any external stimulus(owtte)

B1

mass of products must be less than original mass(of nucleus for possible spontaneous decay)

B12

[13]

M32. (a) (i) Z increases by 1

B1

A remains the same

B12

(ii) Correct curvature starting at 120 Bq

B1

60 (or 0.5 × their start value) at 12 h days later

B1

30 (or half their value at 12 h)and continuing to fall thereafter approximatelyexponentially


B13

(b) (i) 6.6 × 10–11 J (s–1) (120 × 5.5 × 10–13)

B11

(ii) another particle is emitted in each decay (not gammaradiation)orthe nucleus recoils

B1

anti-neutrino emitted (this would get first and secondmark 2 marks)

B1

the other particle/neutrino/antineutrino/nucleus takessome/varying amounts of the energy

B13

(c) 7.5 × 106

B11

(d) Particles are emitted in all directions/particles do notall go to detector

B1

Detector only detects some of the particles that enterit/mention of dead time or recovery time(not detector does not detect all the particles - this addsnothing)

B1

Some particles are absorbed by the window

B1max 2

[12]


M33. (a) Gamma rays are very penetrating/alpha/beta rays wouldnot be detected

B1

(outside body)Gamma rays are less ionising/less hazardous(to patients)/ alpha/beta are more ionising/more hazardous

B12

(b) Background radiation/count is much smaller/negligible

B1

Random fluctuations in the readings greater than background

B12

(c) Accurate plotting check all four points ( ½ square)

M1

reasonably smooth curve with even point scatter

A12

(d) two or more half-lives averaged

B1

Half-life calculated from best fit line

C1

Half-life = 13 1 hour

A13

allow ecf from inaccurate plotting, but straight line = P.E.

(e) High activity (so only a small sample needed)


B1

Decays quickly

B1

Less risk to patient/other people

B1

(Short half-life ok because) medical test doesn‘t last long

B1

Any two from four2

[11]

M34. (a) 236/92/U

B1

4/2/α [4/2/He]

B1

(b) (i) Equation correct or Evaluates mass difference

B1

(1.349 × 10–29 kg)Uses E = mc2

B1

to yield energy (1.21 pJ)

B1

(ii) uses t½ = [loge 2/λ] = 0.69/2.1 × 1011

M1

to yield λ = 3.29 × 10–12 s–1

C1

(iii) uses A = λN [= 1.05 × 1010] or N1 = N0e–λt


C1

uses A × 1.21 × 10–12 or (N0 – N1) × 1.21 × 10–12

C1

= 12.7 mJ

A1

(iv) A = A0e–λt

C1

0.95 = e–3.29 × 10–12 t [or log expression]

C1

t = 1.56 × 1010 s = 495 years

C1

correct deduction from candidate answer

B1

or100 y = 3.19 × 109 s

C1

A = A0e–λt = 1.056 × 1010 e–0.0104 [ecf from first mark]

C1

A = 1.046 × 1010 [ecf from first mark]

C1

Change is 1 part in 105 OWTTE so no significantchange

B1

orHalf life calc/fractional change/2n/99% left so no sigchangeorfurther alternative

[14]


M35. (a) Appropriate method

B1

sensible and correct readoffs

B1

correct evaluation from readoffs

B13

(b) correct readoff on y-axis

B1

use of λ = A/N

C1

correct evaluation from readoff [condone use of 6.0 here]

or determines T1/2 /uses T1/2 =0.69/λ/λ = 0.69/725

A13

[6]

M36. (a) (i) 90

B11

(ii) [228-90 =] 138

B11


(b) (i) second–1

B11

(ii) λ = 3.0 × 1013 /2.6 × 1021

M1

= 1.2 × 10–8 [s–1]

A12

[5]

M37.(a) Draw curve in on graph (at least one point missed)B1

Clear use of graph to determine half-lifeB1

[repeat and average required] Answer in range 650 ± 50 s [600–700]

B1

(b) Mention of background radiationB1

because this increases the count rateB1

(c) Curve displaced upwardsB1

leading to half-life that is too longB1

(d) Take count with source absentB1

Over long time (5+ min) or Average several times (3

or more) and…

Colonel Frank Seely SchoolB1

Because background is small or subtract correction Or correction is negative

B1[10]

M38.(a) number of gamma ray photons per sec = (= 6.0 × 106)B1

correct use of 4πr2; substitution of data

= 21.2

NB they may determine number per m2 and divide by 10 000B1

(b) (i) decay constant = 0.69 / 12 = 0.0575 h–1 or 1.6 × 10–5

(or time =.5 half life)Cl

dose = 21e–(6×0.0575)

dose = 21 / 20.5Cl

or new (gamma) activity = 6 × 106 e–(6×0.0575)

or new (total) activity = 3 × 107 e–(6×0.0575)

Cl

15 (gamma rays per cm2 per second) Condone 14.8 – 14.9(no up)

Al

(ii) clear attempt to apply inverse square lawCl

1.3 (1.26) mAl

(c) beta particles are more heavily ionising than gamma radiationorloses energy rapidly by ionising the air / matter

Colonel Frank Seely SchoolBl

beta particle range / penetration (in air) is loworbeta particle rangeis about 30 cmoris less than 1.5 moris much lower than gamma radiationNB: mention of not able to penetrate skin or clothing is talk out

Bl[9]

M39.(a) (i) mention of radioactivity / decay / nuclear radiationB1

ever present / independent of source being in proximity / always there / cannot be eliminated

B1(2)

(ii) radon / rocks / cosmic rays / nuclear fallout / medicine / space / sunB1

(1)

(b) A – activity / rate of decayB1

λ – decay constant / probability of decayB1

N – number of nuclei (radioactive atoms) presentnot number of isotopes / atoms / particles

B1[6]

M40.(a) curve that has intercept on activity axis at t = 0 (not 0,0)B1

tending to but not intersecting t axis; exponential decay shape single line

B1(2)


(b) A = λN or A = – λN or substituted values 1.1 × 1014 / 5.0 × 1016

C1

2.2 × 10–3

A1

s–1 or / s or Bq or “each second” not %B1

(3)[5]

M41.D[1]

M42.(a) number of alpha particles = 4B1

number of beta particles = 2B1

(2)

(b) 126B1

(1)

(c) Pb-206B1

(1)

(d) (i) number of half lives = 2

or half life = 20 minutesC1

40 minutesA1

(2)

(ii) 4.2 × 1020

Colonel Frank Seely SchoolB1

(1)

(iii) decay constant = 0.69 / half life (allow e.c.f. from (i))

or = N0e–λt1 / 2

C1

5.75 × 10–4 s–1 or 5.78 × 10–4 s–1 or 0.0345 min–1 (allow if calculation is done in (ii))

C1

A = λNC1

2.4 (2.42) × 1017 Bq (or decays per s)

or 1.5 (1.45) × 1019 decays per minuteA1

(4)

(iv) the (daughter) products are also decaying [or are radioactive]M1

activity will be greaterA1

(2)

(v) any 2 of:

polonium-214 has a half life of 1.6 × 10–4 sB1

decays almost as soon as it is formed or decays very quicklyB1

only some of the bismuth-214 decays via polonium-214B1

max 2[15]

M43.(a) curve drawn of approximately correct general shapeC1

curve shows inverse square law e.g.includes the point (3.0 , 12.5) or (1.5, 50) or (2.0, 28.1)

A1(2)


(b) (i) measure and deduct background count (rate)B1

(1)

(ii) count for large periods (to ensure large N)or repeat counts and average

B1(1)

[4]

M44.(a) (i) correct construction method seen on graphor quotes appropriate values from the graph

C12.0 × 104 s to 2.4 × 104 s allow 1 or2 s.f.

A1repeats and averages

B1(3)

(ii) A = 3.7 or 3.8 × 1016 (Bq) at t = 0C1

1.2 × 1021

A1(2)

(b) (i) alpha will not penetrate the body or risk to patient from ionisationsB1

(1)

(ii) long enough half life to make measurements / short enough half life so does not remain long in body

B1(1)

(c) causes ionisations / damageB1

to cell / cell nucleus / body tissue / DNAB1

(2)


[9]

M45.(a) (i) A = 4; Z = 2B1

(1)

(ii) A = 0; Z = –1B1

(1)

(iii) 8 alphas c.a.o.B1

explanation consistent with their (i) / correct equations including values of A and Z

B1(2)

(iv) 6 betas e.c.f.B1

explanation consistent with their (i) and (ii) / correct equations including values of A and Z

B1(2)

(b) (i) probability of a nucleus decaying per unit time orratio of activity to number of nuclei (of that type) present or

with terms defined or

with terms definedB1

(1)

(ii) NP = N0 – N0e–λt / Nu = (Nu + Np) e–λt / Np = No – Nu

B1

correct manipulationB1

(2)


(iii) NP / NU = 0.62 / 4.2C1

λ = (ln2) / 4.5 × 109

or0.69 / 4.5 × 109

C1correct manipulation of logs

C19.0 × 108 y / 8.96 × 108 y

A1look for alternative approach based on No = 4.82 and thenuse of N = Noe–λt

(4)

(iv) age will be overestimatedB1

not all of the Pb is the result of uranium decay or words to that effectB1

(2)

(v) change of count rate for U will be small as half life is largeB1

comment about background count (rate) or background radiationB1

absorption in sampleB1

presence of other radioactive nuclides / daughter nucleiB1

max 2[17]

M46.(a) thicker material absorbs more particlesB1

count rate (number detected) falls if material is thickerB1

fall in count rate produces change to adjust process to produce thinner material / restore to original thickness

allow 1 mark for ‘change in thickness changes count rate and rollers adjust to compensate’

B1(3)

(b) use a beta sourceM0


alphas would be absorbed by paperA1

gammas would not be affectedA1

use a gamma sourceM0

beta would be absorbed completelyA1

alphas would be absorbed completely

allow beta if candidate includes statement about the steel sheet being thinA1

(4)

(c) (i) 1 disintegration / decay / particle emitted per second (per unit time) not one count per second

B1(1)

(ii) correct curvature starting at 5 × 1020; time scale inserted up to 15 (unit not necessary) or labelled T1/2, 2T1/2, 3T1/2

M1

sensible scales (not multiples of 3); correct number of atoms at each half-life; reasonable curve and unit for time

A1(2)

(iii) half-life = 0.69 / decay constantC1

4.1 – 4.2 × 10–9 (s–1)A1

(2)

(iv) A = (–)λNC1

number of R / A atoms when activity is 1.5 × 1012 Bq = 3.6 × 1020

C1

correct time read from graphA1


(2.5 y / 920 days / 8.0 × 107s)

or determines original activity or final number of atoms 2.1 × 1012 Bq or 3.6 × 1020 allow ecf from (iii)

C1

N = N0e–λt or A = A0e–λt

C1

940 d or 2.6 y (answer depends on where rounding off has been done)A1

(3)[15]

M47.(a) (i) nuclides or atoms of same element or same proton / atomic numbernot just ‘same element’different mass / nucleon number or numbers of neutrons

B1B1

(2)

(ii) time taken for half the sample or half the number of nuclei(not just ‘atoms’) to decay or activity to fall to half initial value

B1(1)

(b) B1B1B1

(3)

(c) A / A0 = 0.01 or A = 1 and A0 = 100C1

λ = 0.69 / 8.1 or or 0.086 (0.0856)

attempt to take logsC1

54 days (53.8)

Colonel Frank Seely SchoolA1

(for between 6 and 7 half-lives, 50% → 25% → 12.5%)C1

(4)

(d) γ radiation detectedM1

β– cannot penetrate tissue or γ can penetrate tissueA1

(2)[12]

M48.C[1]

M49.(a) (i) and or and condone etcB1

(1)

(ii) → X + β

M0

Z correct throughoutA1

A correct throughoutA1

(2)

(iii) 0.056(4) (no unit penalty)B1

(1)

(b) A = A0e–λt

C1


12 = 100e–λt or other progress toward answerC1

38 (40) y allow e.c.f. from (iii)A1

or evidence of working in half-livesC1

statement that age is 3 × TC1

age ≈ 37 yA1

or A = N0e–λt (using incorrect formula from sheet)C1

λN = N0e–λt and = 0.12 or = 0.12C1

t = 89 yA1

(3)

(c) (i) (mass of proton + 2 × mass of neutron – mass of tritium) = 0.0086 uB1

multiplies any mass in u by 1.660566 × 10−27

A1

1.42 × 10–29 kg(no significant figure penalty) (no unit penalty) (condone –ve answer)

(2)

(ii) E = mc2

C1

E = 1.28 × 10–12 J(no unit penalty) (allow e.c.f. from (i) for m in kg) (condone –ve answer)

A1

or using recall of u = 931 MeV energy change in MeV = 8.557 × 931 = 7.97 MeV

C1

energy change in J = 7.97 × 1.6 × 10–19 = 1.27 × 10–12 JA1

(2)[11]


E4.The introductory part (a)(i) was done extremely well with over 90% gaining the mark. A few got the equation incorrect and fewer still lost their mark by calculating the half-life in s−1 units.

Less able candidates failed at the first step in part (a)(ii) by not using the exponential decay equation. In other cases it was almost universal that mistakes were made at the substitution stage. Some common errors were to swap round the abundance of C-12 atoms initially and at a later time in the equation. Others sometimes added 1 to the proportion 0.375 and used N0 = 1.375 and Nt = .375. Many candidates understood how to process the mathematics and there were a high percentage of full marks.

Although students appeared to have a good idea of the difficulties in carbon dating young and old objects in part (b)(i + ii) many could not express their ideas well enough to score marks. It was common to see, 'In less than 200 years none of the C-14 would have decayed'. Very few referred to the small difference in activity when comparing a new sample with the 200 year old sample. Candidates were much more successful in the second part when referring to the much older sample. Most knew that the activity would be very small and it would be difficult to get statistically significant data. Several did not gain this last mark because of lack of care in what they put down on paper. It was common to see variations of, 'There has been so much decay there is no C-14 atoms left'.

E5. Even though part (a) needed a little thought almost all students obtained the correct answer. By contrast part (b)(i) was simply a factual recall question, which was answered poorly by a significant minority. The main error was for students not to state the energy needs to be given out or is required, when a nucleus was formed or broken up. It was common to see written, ‘The energy to keep the nucleus together’. In part (b)(ii) a majority of students simply read the value from the graph and gave an answer near 7.88 MeV without appreciating the ‘per nucleon’ on the y-axis of the graph. Part (c)(i) was done well by most students. Some students missed marks due to a lack of care in choosing specific coordinates for the graphs to pass through. Most students made a good attempt at part (c)(ii). Part (c)(iii) was more difficult and only the better student could correctly combine the two equations required to answer the question. A common mistake made by a few students who looked as if they were going to get the correct answer was for them to confuse the time units they were using. These students obtained the correct answer but then multiplied it by 60×60×24×365.

E6. The majority of students got most of the data correct. A few had no idea of the nature of the reaction involved. Misidentification of the electron neutrino was quite common from


those who made minor errors.

E7.For part (a), most candidates identified background radiation and were able to calculate the value of 20 counts/minute. Unfortunately, a large proportion of candidates lost a mark for omitting the unit.

In part (b) a proportion of candidates were unable to draw a satisfactory line of best fit.

In answer to part (c), most candidates drew an appropriate triangle (or used data points sufficiently far apart), and were able to correctly read off and process the data values. The gradient value must have a negative sign, be within the specified limits and quoted to two or three significant figures.

Part (d) was answered well by a large proportion of candidates who correctly calculate the decay constant from the gradient. The half life must have been quoted in minutes.

Part (e) provided an easy question, correctly answered by almost all candidates.

For parts (f)(i) & (ii) most candidates were able to correctly calculate the uncertainty and percentage uncertainty from the data provided.

Although many students were able to calculate the appropriate percentage uncertainty in part (g)(iii), a much small proportion appreciated the significance in terms of larger time/larger total count having a lower percentage uncertainty.

E8. Surprisingly, few candidates achieved full marks in part (a). The most common error was to include the (electron) antineutrino instead of the (electron) neutrino.

Part (b) was poorly answered; many candidates had no idea of the exchange particle involved in β+ decay. Common errors included gluon, neutron, pion and weak force.


E9. For part (a), few candidates stated more than that the plutonium is very radioactive. Credit was given for implied d of explosions during launch or passing through the atmosphere.

In part (b), more candidates believed that being in a compound would affect the activity of the plutonium. Of those recognising that it would be unaffected, few mentioned the nuclear nature of the decay.

Part (c) (i) was almost invariably correct.

Part (c) (ii) discriminated well. Few gained full marks, but many performed different aspects of the calculation to gain some credit.

With error carried forwards several candidates gained full marks in part (c) (iii).

Nearly all candidates gave the correct nucleon and proton numbers for the uranium isotope and the alpha particle in answer to part (d) (i).

Very few candidates gained any credit for part (d) (ii), with several simply not attempting it.

E10.Less than half the candidates could explain the meaning of the decay constant. By contrast almost all candidates could find the half-life in part (b) and a majority could answer part (c). Some candidates did not gain credit because they conveniently removed 1012 in their calculation without showing the division. So lines like, 1.15 × 1012 Bq = 1.15 Bq, were seen. Most candidates who tackled part (d) using the exponential decay of the activity equation got full marks. Only a few candidates could not rearrange the equation. By contrast almost all candidates who tried to use the exponential decay in the number of nuclei got confused. Most had numbers of nuclei on one side of the equation but activity on the other.

Part (e) did discriminate but only between scoring zero marks or one mark. Very few candidates attempted two reasons. Most acceptable answers to this question were difficult for the candidate to express. For example, in question (d) it states that the decay rate due to carbon-14 is 0.65 Bq, indicating it is a corrected count rate. So an answer to part (e) like, ‘the background can effect the result’, is not acceptable. This is not the same as saying it is difficult to obtain the results for the sample activity because the background activity is high in comparison. This example is also ambiguous in that it suggests the surroundings can influence the rate of decay. Another answer that was not acceptable was, ‘radioactive decay is random so it’s bound to give false values’. To gain a mark following this line of thought it was necessary to refer to its effect on the statistics. The most common answers that candidates found easy to express included the following; the tree died well before the boat was made; or the boat was repaired later in its life with fresh


wood; or that carbon based microbes died in the wood when the boat was rotting at the end of its useful life.

E13. In scripts where candidates showed no numerical values on the N axis, it was not possible to judge the value of answers in part (a), so no marks could be awarded. The graph ought to start in an approximately linear fashion. To satisfy the requirements of the question, it should continue in an upward curve to meet Z = 80 at an N value around 120. The initial linearity was generally known, but the features of the upper part of the graph were less familiar to candidates. Some had no idea about how to respond, their answers bearing more resemblance to NIZ graphs showing α and β– decay chains.

Attempts at part (b) were usually more successful. Whatever mistakes had been made on the N/Z curve, examiners simply wanted candidates to show points that were respectively slightly below (at Z > 60), slightly above, and slightly below whatever line was drawn.

Part (c) attracted many highly successful solutions, with convincing (and often very different) explanations about why the loss of four a particles had to be accompanied by the emission of four β– particles in order to form the isotope 206Pb.

Candidates’ responses to part (d) were somewhat variable. It was necessary to identify two acceptable processes for the first mark; this was then followed by one mark each for the two discussions. Candidates who referred to β+ emission and to electron capture found it easier to score the discussion marks than those who chose a emission or proton emission. In the cases of β+ emission and electron capture, the conversion of a proton into a neutron produces an immediate and obvious change in the p/n ratio, a emission requires rather more careful explanation to indicate the effect on this ratio, whilst proton emission is very rare and occurs only to nuclides with sufficient instability for electrostatic repulsion to overcome the strong force.

E14. Many answers in part (a) lacked sufficient detail to gain more than one or two marks. Although candidates knew that the mass of an α particle was greater than that of a β particle, few stated that its mass or momentum was much greater. Again, candidates knew that an α particle had more charge than a β particle but few stated it had twice as much charge. Few candidates realised that an α particle travels much slower than a β particle with the same kinetic energy; some even claimed that it travelled faster because it had more momentum. In addition, candidates often wrote at length about the relative penetrating powers of the two types of radiation.

In part (b) (i), many candidates gave a correct calculation although some candidates


made arithmetical errors, often in the conversion of MeV to J, or in rounding the answer incorrectly. In part (ii), the underlying principles behind the calculation were known but some candidates failed to realise that the increase in temperature in one minute was required.

E16.Part (a) produced good results and many candidates scored full marks, although some were not aware of the expression for efficiency. Other candidates forgot to take account of the area in part (ii) and thereby lost a mark.

Many candidates provided a clearly explained calculation in part (b)(i). Candidates who failed to score both marks usually made an arithmetical error in the conversion from MeV to J. The most common error in part (ii) was the failure to include the appropriate unit of s−1

or year−1 for the decay constant. Weaker candidates made poor progress in part (iii), often being unaware of how to proceed or making a pointless attempt to use the radioactive decay equation. Incorrect answers by the better candidates arose because the decay constant was not converted to s−1 for use in part (iii) or through failure to use the mass number correctly or converting mass, given in atomic mass units, into kg.

E17.Most candidates were able to complete the equation in part (a) correctly, although a few lost marks by adding an extra particle to the right hand side of the equation.

In part (b)(i) the majority of candidates were aware of the correct conversions from MeV to J and from u to kg although some candidates used the proton / neutron rest mass of 1.67 x 10-27 kg instead of 1.66 x 10-27 kg to convert u to kg. The expression for kinetic energy was generally correctly used. In part (b)(ii) most candidates gave a complete and correct calculation. The conservation of momentum was generally applied correctly although a significant number of candidates incorrectly used 210 u as the mass of the recoil nucleus. Other candidates attempted to calculate the mass of the nucleus from the combined mass of its protons, neutrons and electrons. Some candidates unnecessarily converted the correct masses to kilograms.

E18.The question on nuclear instability realised some good responses but part (a)(i) was not always answered as well as one would have hoped. Candidates were aware that an


electron was captured by a proton which then became a neutron, but the answers were not specific enough as to where the electron came from. Statements such as ‘an electron is captured’ were not sufficient to gain the mark. It needed to be made clear that it was an orbiting electron. Answers to part (a)(ii) were sometimes vague, with only about 50 % of the candidates realising that electron capture excites the daughter nuclide. The equation in part (iii) usually gained some marks and it was good to see that only rarely was an anti positron or anti electron inserted in the equation.

Part (b) realised very good marks and a significant number of candidates were awarded the maximum of five. The conversion from hours into seconds was usually performed correctly in part (i) to give the required result. Part (ii) likewise gave good results although it was here that problems with significant figures became apparent. There was no real difficulty with part (iii), candidates using ΔN / Δt correctly.

E19.Nearly all candidates determined the half-life of the nuclide successfully from the graph in part (a)(i). Many candidates went on to calculate the decay constant correctly in part (a)(ii), but a significant number omitted to include a unit (s–1) Although many candidates could correctly relate activity and number of nuclei, few could determine the number of undecayed nuclei present initially.

Only a small minority of candidates appreciated that part (b)(i) required the calculation of the number of nuclei present at t = 30 s; too many attempted to calculate the number present at t = 50 s. Those candidates who calculated a nuclei number in part (b)(i) which was not ridiculously small could usually earn a mark for their subsequent attempt to calculate the energy released. Disappointingly, several candidates calculated the answer to part (b)(i) to be less than one decay and then carried on into part (b)(ii) to compute the energy from less than one decay.

E20.Many candidates scored full marks on this generally high scoring question. Most of those candidates who knew the meaning of decay constant could complete the calculation in part (a). There were some arithmetic errors, a few candidates did not change years into seconds, and in part (a)(ii) some mistakenly introduced N = N0 e−λt.

In part (b) marks were often lost through careless wording. Most candidates realised that the majority of alpha particles travelled straight through the foil, but others said that they were deflected through small angles, sometimes up to 90°. A statement that the volume of the nucleus was small compared with the volume of the atom was expected, but most candidates simply referred to a small nucleus. This was accepted if it was accompanied


by the statement that most of the atom was empty space, which seemed almost a reflex response. Other examples of lack of precision were to omit the word nucleus completely, or to confuse nuclei with atoms, or to say that the nucleus was highly charged, or a concentration of charge instead of positively charged. Describing the nucleus as massive in isolation was not accepted as a substitute for a statement that most of the mass of the atom is concentrated in the nucleus.

E21.This was very much a ‘hit or miss’ question. Graphs in part (a) varied considerably and few candidates appeared not to know how the values of N corresponded to various Z values, or took the care to relate them. Regions where different emitters are positioned were vaguely marked, often far from the line of stability. It was also not uncommon to see vertical lines used to separate the regions.

Only the best candidates scored full marks in part (c). The most common mistake was in the electron capture row. There was also a large number of candidates who failed to read the last column heading avid treated it as mass or nucleon number.

E22.The usual wrong definitions for half life were seen frequently in part (a) - especially those involving mass.

The inclusion of the results box for In C in the table avoided most confusion in part (b) and allowed even weak candidates to score on the graph work in part (c). Most candidates completed the table accurately, but a significant number incorrectly restricted their In C values to one decimal place. The most common error on the graph was to use too small a scale - candidates should not be afraid to suppress their zero if this is required in order to make full use of the graph paper.

Although producing and understanding ln C − ln C0 = λt was beyond several candidates, there were many good attempts at part (d)(i), some of these involved correct substitution of figures which was accepted. Part (d)(ii) was generally done correctly, but several candidates lost a mark by incorrectly giving s, s−1 or mm−1 as their unit.


Part (e) was expected to be straightforward, but candidates did not score well because their statements were too vague. For example, credit was not given for bald, unexplained statements such as “consider background radiation”.

Although part (f) produced a wide variety of incorrect equations, many candidates answered it correctly and scored full marks. Several candidates made things unnecessarily difficult for themselves by converting to seconds for λ only to have to convert back again at the end.

E23.Most scripts omitted the neutrino in the electron capture equation but far fewer left out the antineutrino in the beta decay equation. It was only the weaker candidates who failed to score the first mark in each equation. In part (a)(ii) the electron masses were not thought to balance by most candidates and the answer 2.0 MeV was given by even some of the strongest candidates. Part (a)(iii) turned out to be a good discriminator – the X-ray was vaguely explained away in terms of an excited nucleus by weaker candidates.

Only the best candidates obtained the correct answer to the calculation in part (b)(ii). Half the candidates attempted to use the decay equation on some data and, as a result, only lost one mark.

E24.Part (a) was answered well but the last mark, which related path length to the energy of the ionising particle, was often missed. It was curious to note that candidates who gave the answer “beta particle” in part (a) (i) then, contrary to plain facts, often stated in part (a)(ii) “. . because there is a continuous range of path lengths... “. In part (b)(i) the equations showing alpha decay were easy marks for most candidates and the problem relating alpha particle energy to half-life only caused difficulties for the weaker candidates.

Most candidates failed to score in part (b)(iv) because it was necessary to state a little more than “because its neutron rich” to gain the mark. This was because it could be said that U-235 has a high neutron to proton ratio.


E25.Part (a) was answered well by the great majority of candidates, although a significant number reversed the values of proton number and nucleon number.

Surprisingly, part (b)(i) caused many candidates difficulty, with too many believing that heat was the form of the initial energy. Those candidates who answered potential energy without making it clear that it was electrostatic were not allowed the mark. There were very many correct answers to part (b)(ii).

The usual error was failure to use E = mc2 or, when doing so, failure to square c. The unit of m was often wrong.

E26.All parts of this question had a spread of good and poor answers. The greatest difficulty for candidates was the use of precise terms to express ideas about which they had only partial knowledge. In part (a)(i) many candidates made little attempt to explain what was meant by “random” in this context. Many candidates simply repeated the question. Few candidates made it clear in part (a)(ii) that isotopes are different forms of the same element. Furthermore, in explaining the meaning of half-life, far too many candidates made the usual basic error of believing that it was the time for the mass of the sample to halve. The use of the word radioactivity was not allowed as a substitute for activity. Far too many candidates said that the decay constant was a measure of the rate of decay.

Several candidates scored full marks in part (b). As long as the correct unit was given, correct answers in s– or days–1 were accepted in part (b)(i); the usual error was to omit the unit of days–1. The biggest cause of disaster in parts (b)(ii) and (iii) was the use of the wrong equations.

E27. (a) Graph drawing was poor in this question. Data points were misplaced, curves


were inappropriate or, more commonly, simply careless with kinks and double lines that avoided being truly best-fit.

(b) Many candidates failed to obtain full credit because they only made one determination of the half-life. Two or more with a clear calculation of the mean value were required for full credit. Most values were within the tolerance of ± 10 s. A common error (treated as a lost determination) was to assume that the t = 0 s count-rate was 500 s–1 not the 480 that was both clearly stated in the table and also shown on the graph.

(c) Although most could calculate the decay constant with some facility, a correct unit (indeed, any unit at all) was rare. Candidates evidently do not recognise the physics here with any clarity even though they can jump through the arithmetic hoop.

E28. (a) A large number of candidates recognised that two alpha particles and four beta particles were emitted; many however, did not explain their answers clearly enough to be awarded all four marks; often candidates simply described what alpha/beta were.

(b) (i) Explanations of half-life were too often imprecise and involved words such as atoms, particles, element or mass. Reference should have been made to nuclei or activity decaying.

(ii) Many candidates, having written the relationship between half-life and decay constant, failed to do a precise calculation and were penalised for having equated the equation to the 0.026 minute–1 given.

(iii) Most candidates made a valiant stab at this part, but there was a penalty of two marks when the nucleon number was used for number of nuclei.

E29. (a) Many candidates are now aware of beta decay and it was rare to see the


omission of clear notation to indicate the true nature of the anti-neutrino.

(b) (i) The unit of activity was known by many but correctly spelt by few.

(ii) The calculation of initial activity was also done well, with almost all scoring the first mark. However the calculation itself (a multiplication involving powers of ten) led to many powers of ten errors.

(c) This was a difficult question and was intended to discriminate between the best candidates. Few were able to score full credit, but two and three marks out of four were very common. However, it did highlight for the examiners some of the common misconceptions to which students at this level are prone. It is, for example, commonly held that the decay probability somehow changes with time and that the probability itself relates to the ensemble of atoms and not to an individual nucleus. Candidates often found it hard to explain the concept of activity and whilst they could describe how it changes in terms of time, they had not yet understood the idea in terms of its relationship to the overall number of atoms that remain undecayed.

E30. (a) (i) This part was done reasonably well. Some candidates distracted themselves with ideas about positive and negative ions or with the idea that the alpha particles were themselves ions.

(ii) Most of the candidates knew that beta particles were less ionising or that they suffered fewer collisions. They found it more difficult to express ideas relating to the number of collisions per millimetre of travel or the consequent notion that alpha particles lost all of their kinetic energy over a smaller distance than beta particles

(b) (i) Many candidates could use the graph to find the half life. Some made mistakes over the interpretation of the scales and others did not show their constructions clearly enough for credit. Fewer candidates thought it worth finding more than one determination from the graph and then averaging their answers.

(ii) This was done quite well but significant figure errors and unit penalties were common and a surprising number of candidates made powers often mistakes.


E31. (a) (i) The most common error was not converting the half-life to years.

(ii) This was usually completed successfully.

(iii) This was also usually completed successfully but again failure to convert the time to seconds was a common error.

(iv) Most candidates appreciated the possible activity of neptunium or other decay products.

(b) (i) This was usually done well although many incurred a significant figure penalty. A common error was to quote the equation E = mc2 correctly but to fail to square (3.00 × 108) in the calculation.

(ii) This was usually done well and most candidates were able to gain one or both marks.

E32. (a) (i) In this straight-forward question it was surprising how many candidates thought that the proton number would decrease. Most appreciated that the nucleon number remained the same.

(ii) This was usually done well. Common errors were to show the starting activity of 120 Bq at 12 h or to show an activity of 140 Bq at time = 0. Failure to complete the graph for the 48 h period and very poor line drawing also cost some the third mark.

(b) (i) That the energy per second would be the activity x the energy per decay was understood by relatively few candidates. The number of seconds in an hour or a day appeared in the working of many candidates.


(ii) Although there were many excellent answers the fact that another particle (antineutrino) was the reason for the different energies was not as well known as one would have hoped. Many concentrated only on the recoil of the nucleus. Others wrote about the energy of emission becoming weaker as the source decayed, the randomness of radioactive decay or the source becoming charged.

(c) This was usually completed successfully.

(d) There were many good answers. Many candidates appreciated that particles were emitted in all directions so that only a proportion of those emitted would reach the detector. A misconception in some answers was that if the source had been shielded then all the particles would have gone in the same direction (being reflected by the shielding). Other points about the effect of absorption by the window to the tube and the limitations of the tube in detecting particles that arrive at the same time, or within a certain time, were also appreciated by many.

E33. This question gave nearly every candidate the opportunity to gain some marks and a pleasing number scored very well indeed.

Part (a) returned at least one mark to most candidates, but part (b) was answered poorly. The examiners were hoping to be told that a typical background count-rate was much smaller than the figures given in the table and negligible compared to the random fluctuations apparent in these data.

A surprising number of candidates were unable to plot all four points accurately in part (c), and some did not even attempt to do so. The better candidates averaged two or more half-life readings taken from their line of best fit in part (d), although some then lost a mark for an incorrect unit.

Most candidates provided at least one appropriate reason in answer to part (e).


E34. (a) It was disappointing that a sizeable minority could not write this simple equation out correctly. Some do not know the nuclear disposition of the alpha particle; some attributed it to the element Hydrogen. Although not a failure point, many feel that neutrinos are emitted during this nuclear reaction.

(b) (i) A large number were able to carry this mass defect calculation very well, obtaining full marks. There was no one source of error, some forgot to square the speed of light in the calculation, others could not calculate the mass defect correctly.

(ii) There were few errors in this part, a piece of work from PHB2.

(iii) Again, many could calculate the energy released per second by the plutonium with real fluency from A= λ N. Those trying to work from an exponential equation (a possible approach) however, found that their calculators suggested that there was no decay in the first second (the number is a small fraction of the total number of atoms present), took this at face value, said so without thinking, and therefore lost significant marks in this part.

(iv) Although there were many correct pieces of working and explanation in this part, discussion of the answer was often poor and garbled. Again, this was a calculation with a number of parts to negotiate and candidates did not find it easy to control their working and set it out well. Unchecked calculations with simple slips were rife.

E35. (a) There was a widespread inability to carry through this part in any sensible way. Many obviously failed to understand the consequences and implications of this graph. Only rarely did examiners see a clear tangent drawn on the graph and then a serious attempt to evaluate its value, this from candidates who in PHB1 will happily and accurately evaluate the gradient of a distance–time graph to calculate a speed.

(b) This part was better but very frequently marred by misreads from the graph and errors in expressing the unit of decay constant.

E36. (a) This was a two-part calculation of the numbers of protons and neutrons in a nucleus and it was answered correctly by the large majority of candidates.

(b) (i) The correct answer of s–1 was common, but so was the answer of ‘becquerel’.

(ii) Many were able to manipulate A = λN well but even these often failed at the


final hurdle and rounded the answer incorrectly or displayed an inappropriate number of significant figures. Failures usually centred on the evaluation of 3.0 × 1013 × 2.6 × 1021 indicating that the candidate either did not understand the meaning of the symbols in the equation or could not manipulate the equation correctly.

E37.(a) Curves were very often poorly drawn and only a small number of candidates took an average of two or more determinations of the half-life. The generous range of values allowed in the mark scheme was still not wide enough for many.

(b) Although a large number recognised that background radiation is the reason for the correction, only a few indicated that it would increase the count rate. Answers for this second mark were much too vague or simply factually incorrect.

(c) Candidates who discussed the curve usually recognised that the uncorrected line lies vertically above the corrected line: vague statements such as ‘It is higher’ are far too ambiguous at this level for credit. Many realised and stated clearly that the half-life would increase, or it would vary with time.

(d) In order to gain full credit candidates needed to mention the absence of the source during the background measurement, a process of averaging (either by time or repeat) and either the way in which the correction would be applied to the uncorrected values or a reason for the treatment they proposed. A common mark was 2 out of 3 with many forgetting to mention the need to remove the source from the vicinity during background measurement.

E38.Only the better candidates scored well on this and many of these obtained full or near full marks. However, it was disappointing that this was generally a low scoring question.

(a) The need to identify the approach and then to extract the appropriate data from both the stem and the question proved to be too demanding for most. Candidates needed


to appreciate that the gamma activity is 1 / 5 of the total activity and that the gamma ray photons pass through an area of 4πr2. Many tried using the decay formula in this part.

(b) (i) Candidates are required to be able to use the decay equations and here the decay formula could be applied here using the total activity or the gamma activity or the value 21 given in (a), the last being the easiest approach. Many thought that if the activity halves in 12 hours it will be reduced by a quarter in 6 hours. Errors made by those who used the correct approach usually involved an incorrect decay constant or inability to do the arithmetic.

(ii) Candidates could gain credit if it was clear that they were applying the inverse square law but this was far from evident in many scripts.

(c) The majority of candidates stated that beta particles have a short range but few stated why this was the case. A significant proportion demonstrated unclear thinking by saying that the beta particles would not pass through clothes or skin hence talking themselves out of the mark. A worrying number thought that beta particles were harmless since the skin prevents them entering the body. Of those who mentioned ionisation a number thought that the beta particles themselves are ionised.

E39.(a) (i) Although many candidates were able to suggest a sensible source for background radiation few gave a full enough explanation, of what it is, to gain full credit

(ii) It was unclear in many cases whether or not candidates understood that the source of background radiation was radioactive decay.

(b) Most candidates were able to define the symbols in the equation but too often, having given a correct quantity, the candidate would then give an alternative meaning for the symbol that was contradictory – inevitably this was penalised. Unsurprisingly, many weak candidates took λ to mean wavelength.

E40.(a) This was well answered by the majority of candidates. Weaker candidates were penalised for showing no activity value at zero time or the curve intersecting the time axis or curving upwards.

(b) Nearly all candidates were able to write down the equation for the probability of decay but many confused the activity with the number of nuclei and calculated the reciprocal of the correct answer. Few candidates gave the correct unit for probability of decay (s−1 ) and many quoted their answer as a percentage.


E46.(a) There were many well explained answers to this part. However, the question exposed many misunderstandings. Many candidates wrote that either the source or radioactive particles passed through the material. Some thought that the material itself was radioactive. Although many referred to less radiation reaching the detector, it was disappointing how few referred to the radiation being ‘absorbed’. (Did they think it was reflected?) Many candidates referred to radiation being detected or not detected as thickness changed rather than that there being a variation in count rate.

(b) (i) To gain credit in either part the correct source first had to be identified. There were many who stated that alpha sources should be used for paper although the fact that alpha particles are absorbed by paper and travel only a short distance in air should be well-known.

(ii) A beta source was stated by many to be suitable. This was allowed only if they also stated that the steel would be thin. Many stated or implied in one or other of the two parts that gamma radiation could pass through anything without any change in intensity.

(c) (i) There was a surprising number of incorrect answers to this. Statements such as ’1 Bq means that one radioactive atom is radiated from the source per second’ or simply that ‘it is the activity of a source’ were not uncommon. Many associate the value with the count rate of a detector rather than a property of the source.

(ii) Poor graph drawing skills cost many candidates a mark here. To gain the first mark the correct value at t = 0 had to be plotted and indicated at 5 × 1020 and the curvature had to be correct though not accurate. For the second mark the scale should have been sensible (e.g. not 5.3, 10.6 etc at the 2 cm grid markings), and the values were expected to be reasonably accurate at times equal to 1, 2 and 3 half lives.


(iii) Most candidates did this correctly. Common faults were giving the answer as 0.13 year−1 or as 0.13 s−1.

(iv) Whilst there were many correct answers, many were confused. A common response was 1.5 × 10−2 = 5 × 1020 e−λt. These candidates did not appreciate that they needed to find either the original activity or the final number of radioactive atoms. There were also many instances where units were mixed. Calculation of the number of atoms remaining when the activity is 1.2 × 1012 Bq and reading the time from the graph was an expected, easy route to the answer but this approach was rarely used.

E47.(a) This was rather disappointingly answered with a number of inconsistencies. Isotope was rarely used as a comparative term and candidates frequently referred to an isotope as being ‘an element with the same proton number but different neutron numbers’. Half-life was often referred to as being ‘the time for the number of atoms / nuclides / molecules to be reduced to half’. None of these responses was given credit. Candidates should be encouraged to talk in terms of decay of nuclei or the activity of a sample of material.

(b) This part was done relatively well, although a minority of candidates was very confused by the nomenclature used in the equation representing this nuclear reaction.

(c) This part was well answered with the majority of candidates being able to calculate a value for the decay constant and the majority of those candidates being able to use their value in order to calculate a sensible time for the reduction of the activity. Weaker candidates were awarded a compensation mark for estimating that the time would be between six and seven half-lives.

(d) Although the majority of candidates recognised that the γ radiation would be detected outside the body, few supported their assertion in an unambiguous manner. Weaker candidates suggested numerous different ‘decay products’ ranging from α and β particles to iodine and xenon.


E49.(a) (i) Most candidates completed this successfully.

(ii) There was a good proportion of correct equations but there were many who were unable to write the symbolic equation correctly using X and β

(iii) Most candidates did this correctly. A minority gave the answer in s−1.

(b) The majority of the candidates used the route of determining the approximate number of half lives and arrived at the correct answer. Although a more difficult route, most others used the decay equation correctly. A common fault by those using the equation was to assume that the activity dropped by 12% rather than to 12%. A very small number quoted the incorrectly printed formula on the formula sheet.

Most of these immediately realised the error and proceeded correctly. A few combined the incorrect formula with A = λN and proceeded correctly to obtain an answer of 89 years. Candidates who proceeded logically in this way gained full credit.

(c) (i) Most candidates gained credit for undertaking the conversion from u to kg correctly. Many failed thereafter because they assumed a tritium nucleus to have either one proton and one neutron or two protons and one neutron. Arithmetic, presumably done using a calculator, presented a problem for some who knew the correct constituents.

(ii) Having obtained the mass change most proceeded correctly in this part. The usual error was to determine the energy equivalent of a tritium nucleus.

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