Physics 1201 Final Exam information – Spring 2013

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Physics 1201 Final Exam information – Spring 2013 1) Format: 35 multiple-choice questions, 5 points each 175 points total 2) Closed book and closed notes 3) Equations and constants are provided on the final exam 4) Cumulative – it covers all of the material covered in the course 5) Approximate weightings of material: ~10 questions from material covered for midterm 1 ~10 questions from material covered for midterm 2 ~15 questions on material covered since midterm 2 6) See the Assignment Sheet for the date and time of your final exam

Transcript of Physics 1201 Final Exam information – Spring 2013

Page 1: Physics 1201 Final Exam information – Spring 2013

Physics 1201 Final Exam information – Spring 2013

1)  Format: 35 multiple-choice questions, 5 points each à 175 points total

2)  Closed book and closed notes

3)  Equations and constants are provided on the final exam

4)  Cumulative – it covers all of the material covered in the course

5)  Approximate weightings of material: ~10 questions from material covered for midterm 1 ~10 questions from material covered for midterm 2 ~15 questions on material covered since midterm 2 6) See the Assignment Sheet for the date and time of your final exam

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PRELIMINARY

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PRELIMINARY

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Chapter 30

Nuclear Physics and Radioactivity

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Radioactivity

A magnetic field separates three types of particles emitted by radioactive nuclei. The decay takes the general form:

parent→ daughter + x

Q ≡mparentc2 − mdaughter +mx( )c2 ⇒Q− value→ KE released

Q > 0⇒ exothermic Q < 0⇒ endothermic

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Radioactivity α DECAY

ZAP → Z−2

A−4 D + 24 He

Q =mPc2 − mD +mα( )c2

= 238.0508 u− 234.0436 u+ 4.0026 u( )"# $% 931.5MeVu

&

'(

)

*+

= 4.3MeV

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Radioactivity

A smoke detector

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Radioactivity

β DECAY

e D P 011 −+ +→ A

ZAZ

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Radioactivity

γ DECAY

γ P P +→∗ AZ

AZ

excited energy state lower energy

state

88226Ra*→ 88

226Ra+γ Eγ = 0.186 MeV

a photon that originates from the energy-level transition of the nucleus

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The Neutrino

During beta decay, energy is released. However, it is found that most beta particles do not have enough kinetic energy to account for all of the energy released. The additional energy is carried away by a neutrino à

90234 Th → 91

234 Pa + −10 e + ν

mass ≈ 0, q = 0

A free neutron beta decays in about 12 minutes,

01n → 1

1p + −10 e + ν

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Radioactive Decay and Activity

The half-life of a radioactive decay is the time in which ½ of the radioactive nuclei disintegrate.

with calculus⇒ N = Noe−λ t

λ2ln

21 =T

ΔN = −λNΔt, λ = decay rate s−1( )∴ΔNΔt

= −λN ⇒ #decayed / time⇒ activity

The number of nuclei decaying, ΔN, in a time Δt in a population of N nuclei is given by,

initial population

SI for activity = becquerel = Bq = 1 decay/s 1 Curie = 1 Ci = 3.70 x 1010 Bq

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Radioactive Decay and Activity

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Example: Activity per gram of in a living organism. a)  Determine the number of atoms present for every gram of in a living organism. b) Find the decay constant and c) the activity of this sample. Assume that in a living organism there is one atom of for every 8.3 x 1011 atoms of

614C614C

612C

614C

612C

a) #atoms 614C

1 gram 612C

=112mol

!

"#

$

%& 6.02×1023

atomsmol

!

"#

$

%&

18.3×1011!

"#

$

%&= 6.0×1010atoms

b)T1 2 = 5730 yr =1.81×1011s

λ =0.693T1 2

=0.693

1.81×1011s= 3.83×10−12 s−1

c) ΔNΔt

= λN = 3.83×10−12 s−1( ) 6.0×1010atoms( ) = 0.23 Bq

When an organism dies, this activity begins to decay away.

Activity of for every gram of in a living organism

614C

612C

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Radioactive Dating dating of the “Ice Man.” 6

14C

The “Ice Man” was found in the Italian Alps in 1991. If the body has a activity of 0.121 Bq per gram of carbon, find the age of the Ice Man’s remains.

614C

NN0

= e−λt ⇒ ln NN0

#

$%

&

'(= −λt

t = − 1λ

#

$%

&

'(ln

NN0

#

$%

&

'(

activity ≡ A = ΔNΔt

= λN ⇒NN0

=λNλN0

=AA0

∴t = − 1λ

&

'(

)

*+ln

NN0

&

'(

)

*+= −

T1 20.693&

'(

)

*+ln

AA0

&

'(

)

*+= −

5730 yr0.693

&

'(

)

*+ln

0.1210.23

&

'(

)

*+

= 5300 yr

Know λ Need to find

NN0

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Radioactive Decay Series

The sequential decay of one nucleus after another is called a radioactive decay series.

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Radioactive Decay Series

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Chapter 31

Nuclear Energy, Effects and uses

of radiation

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Induced Nuclear Reactions

A nuclear reaction is said to occur whenever the incident nucleus, particle, or photon causes a change to occur in the target nucleus.

Charge and nucleon number are conserved in a nuclear reaction.

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Induced Nuclear Reactions

Example: An Induced Nuclear Transmutation An alpha particle strikes an aluminum nucleus. As a result, and unknown nucleus and a neutron are produced.

P3015

24He + 13

27Al→ ZAX + 0

1n

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Induced Nuclear Reactions

Nuclear reactions can produce elements not found in nature……. ……an induced nuclear reaction in which uranium is transmuted into the transuranium elements neptunium and finally plutonium.

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Nuclear Fission

A slowly moving neutron causes the uranium nucleus to fission into barium, krypton, and three neutrons.

Average energy released in the fission reaction

n+ 92235U⇒≈ 200 MeV

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Nuclear Fission

A chain reaction

….assuming each fission creates two neutrons which can cause addition fissions in other nearby nuclei….. This results in an uncontrolled chain reaction à e.g. atom bomb

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Nuclear Fission

In a controlled chain reaction, only one neutron, on average, causes another neutron to fission. This is the reaction mode used in nuclear reactors.

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Nuclear Reactors

A nuclear reactor consists of fuel elements, control rods, and a moderator.

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Nuclear Reactors

The moderator slows neutrons and the control rods absorb neutrons.

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Nuclear Fusion

Two nuclei of very low mass can combine to generate energy. This process is called nuclear fusion.

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Nuclear Fusion

12H + 1

3H→ 24He+ 0

1n

energy released = Δm( )c2

= m 12H( )+m 1

3H( )−m 24He( )−m 0

1n( )$%

&'c

2

= 2.0141u+3.0161u− 4.0026 u−1.0087 u[ ] 931.5 MeVu

(

)*

+

,-

= 0.0189 u[ ] 931.5 MeVu

(

)*

+

,-=17.6 MeV