Post on 03-Jun-2018
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Nuclear Physics
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Recap
The famous Geiger-Marsden Alpha scattering
experiment (under Rutherfords guidance)
In 1909, Geiger and Marsden were
studying how alpha particles are scattered
by a thin gold foil.
Alpha
source
Thin gold foil
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Geiger-Marsden
As expected, most alpha particles were
detected at very small scattering angles
Alpha particles
Thin gold foil Small-angle
scattering
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Geiger-Marsden
To their great surprise, they found that
some alpha particles (1 in 20 000) had
very large scattering angles
Alpha particles
Thin gold foil Small-angle
scattering
Large-angle
scattering
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Explaining Geiger and Marsdens results
The results suggested that the positive (repulsive) charge must beconcentrated at the centre of the atom. Most alpha particles do not pass
close to this so pass undisturbed, only alpha particles passing very close to
this small nucleus get repelled backwards (the nucleus must also be very
massive for this to happen).
Remember on this scale, if the
nucleus is 2 cm wide, the atom
would be 200 m wide!
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Rutherford did the calculations!
Rutherford calculated theoretically the
number of alpha particles that should be
scattered at different angles (using
Coulombs law). He found agreement withthe experimental results if he assumed the
atomic nucleus was confined to a diameter
of about 10-15metres.
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Closest approach (new)
Using the idea of energy conservation, Rutherford found
it is possible to calculate the closest an alpha particle
can get to the nucleus during a head-on collision.
Alpha particle
nucleus
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Closest approach
Initially, the alpha particle has kinetic
energy = mu2
K.E. = mu2
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Closest approach
At the point of closest approach, the
particle reaches a distance rfrom the
nucleus and comes momentarily to rest.
r
K.E. = 0
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Closest approach
All the initial kinetic energyhas been
transformedto electrical potential energy.
K.E. = 0
r
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Closest approach
Using the formula for electrical potential energywhich is derived from Coulombs law
Kinetic energy lost = Electrical potential
mu2= 1 q1q2
4o r
K.E. = 0
r
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Closest approach
Rearranging we get;
r = 1 q1q2
4o mu2
In this case k is the Coulomb constant
K.E. = 0
r
umkr qq
2
21
21
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Closest approach
For an alpha particle, m = 6.7 x 10-27kg, q1= 2 x (1.6 x
10-19C) and u is around 2 x 107m.s-1. If the foil is made
of gold, q2is 79 x (1.6 x 10-19C).
r = 1 q1q2
4o mu2
r = 1 x (2 x 1.6 x 10-19C) x (79 x 1.6 x 10-19C)
4x 8.85 x 10-12Fm-1 x 6.7 x 10-27kg x (2 x 107m.s-1)2
r = 2.7 x 10-14m
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Rutherfords findings From the equation Rutherford found that the alpha particles
approached nuclei to within 3.2 x 10-14 m when the foil was made of
gold. Hence the radius of the gold nucleus must be less than this
value.
For silver atoms the closest distance of approach was found to be
2 x 10-14 m.
From these results, Rutherford concluded that the positive charge is
concentrated in a small sphere, that he called the nucleus. Whose
radii is no greater than about 10-14 m.
A new length was suggested for these small values called the
femtometer (fm) where 1 fm 10
-15
m. Since the time of Rutherfords scattering experiments, a multitude of
other experiments have shown that most nuclei are roughly
spherical and have an average radius given by;
A is the mass number and r0is a constant
equal to 1.2 x 10-15 m.
31
0Arr
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Video
Closest approach, and mass spectrometer
http://www.youtube.com/watch?v=PiTtRaqqJAshttp://www.youtube.com/watch?v=PiTtRaqqJAs8/12/2019 4 - Radius of Nucleus & Mass Spectrometer
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The mass spectrometer
A VERY useful machine for measuring the
masses of atoms (ions) and their relative
abundances.
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The Mass Spectrometer
ions
produced
velocity selector
ionsaccelerated
Region of
magnetic field
detector
ion beam
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The Mass Spectrometer
Electrons are produced at a
hot cathode and accelerated
by an electric field. Collisionwith gas particles produces
ions.
Substance to be tested
is turned into a gas byheating.
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The Mass Spectrometer
Ions acclerated by
an electric fieldNarrow slit through
which ions pass
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The Mass Spectrometer
Velocity selector In many experiments involving the motion of charged particles it is
important to have particles moving with the same velocity. This is
achieved by using a combination of an electric field and a magnetic
field, as shown below. The orientation is known as crossed fields.
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Mass Spec.Velocity selector
A uniform electric field vertically downward is provided by a pair ofcharged parallel plates, while a uniform magnetic field is applied
perpendicular to the page. For q positive, the magnetic force qvB is
upward and the electric force qE is downward.
If the fields are chosen such that the magnetic force balances the
electric force, the particle moves in a straight horizontal line throughthe region of the fields.
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Mass Spec.Velocity selector
If we adjust the size of the two fields so that qvB = qE, we get
Only those particles having this speed pass undeflected through the
perpendicular electric and magnetic fields. The magnetic force
acting on particles with speeds greater than this is stronger than the
electric force, and these particles are deflected upward. Thosehaving speeds less than this are deflected downward.
B
Ev
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The Mass Spectrometer
Ions enter the velocity selectorwhich
contains an electric field (down the
page) and magnetic field (into the
page) at right anglesto each other.
By choosing a suitable value for the
magnetic field the ions continue in a
straight line. i.e. The force produced
by the electric filed (qE) is equal to
the force produced by the magnetic
field (Bqv). qE = Bqv
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The Mass Spectrometer
qE = Bqv
orv = E/B
This means that only ions with a
specific velocity pass through this
region. (Hence velocity selector)
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The Mass Spectrometer
The selected ions all with the
same velocity (but different
masses of course) enter the
second region of magnetic
field (also into the page). They
are deflected in a circular
path.
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The Mass Spectrometer
Heavier ions continue forward
and hit the sides, as do ions
that are too light. Only ions of
one particluar mass reach thedetector.
The radius of the circle is
given by R = mv/qB so the
mass can be calculated from
m = RqB/v
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The Mass Spectrometer
The magnetic field can be varied
so that ions of different mass can
be detected (higher B would
mean that ions of larger mass
could be directed at the detector).
In an experiment B is gradually
varied from low to high so that all
ions pass through and their
relative abundance measured.
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The Mass Spectrometer
The detector can measure the
numbers of ions detected,hence giving an idea of
relative abundance of different
ions.
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The Mass Spectrometer
The mass spectrometer is
particulary useful foridentifying isotopesof the
same element.
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Nuclear energy levels
We have seen previously that electrons exist inspecific energy levels around the atom.
There is evidence that energy levels existinside the nucleus too.
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Nuclear energy levels
When a nucleus decays by
emitting an alpha particle or
a gamma ray, the particles
or photons emitted are onlyat specific energies (there is
not a complete range of
energies emitted, only
certain specific levels).
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Nuclear energy levels
An alpha particle or photon thus has an
energy equal to the difference between
energy levels of the nucleus.
energy
levels in235U
(MeV)
51.57
0.051
0.013
0.000
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Nuclear energy levels
In the alpha decay of 239Pu to 235U, the plutonium
nucleus with an energy of 51.57 MeV can decay into
Uranium at 3 different energy levels.
energy
levels in235U
(MeV)
51.57
0.051
0.013
0.000
Plutonium-239
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Nuclear energy levels
If the 239Pu (51.57 MeV) decays to the ground stateof235U (0 MeV), an alpha particle of energy 51.57 MeV is
emitted.
energy
levels in235U
(MeV)
51.57
0.051
0.013
0.000
Plutonium-239
Alpha emission (51.57
MeV)
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Nuclear energy levels
If the 239Pu (51.57 MeV) decays to the 2nd excited state
of 235U (0.051 MeV), an alpha particle of energy 51.57-
0.051 = 51.52 MeV is emitted. The uranium nucleus is
now in an excited state so can decay further by gamma
emission to the ground state.energy
levels in235U
(MeV)
51.57
0.051
0.013
0.000
Plutonium-239
Alpha emission
(51.52 MeV)
Gamma
emission (0.051
MeV)
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Nuclear energy levels
In fact the nucleus could decay first to the 0.013 level,
and then the ground state, thus emitting two gamma
photons.
energy
levels in235U
(MeV)
51.57
0.051
0.013
0.000
Plutonium-239
Alpha emission
(51.52 MeV)
Gammaemission (0.038
MeV)
Gamma
emission
(0.013MeV)
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Nuclear energy levels
Notice that the energy of the emitted alpha
particles can only have certain discrete
values and that the gamma rays are
observed with a discrete spectra.
Remember gamma rays are part of the
EMS.
This is evidence that the nucleus like theatom is a quantum system and so has
discrete energy levels.
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Radioactive decay
Beta () and positron decay (+)
decay
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Radioactive decay
In beta decay, a neutron in the nucleus
decays into a proton, an electron and an
antineutrino.
n p + e + ve1
0
1
1
0
-1
0
0
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Radioactive decay
In positron decay, a proton in the nucleus
decays into a neutron, a positron (the
antiparticle of the electron) and a neutrino.
p n + e + ve
http://www.youtube.com/watch?v=lAAmAbJvvJg
1
1
1
0
0
+1
0
0
http://www.youtube.com/watch?v=lAAmAbJvvJghttp://www.youtube.com/watch?v=lAAmAbJvvJg8/12/2019 4 - Radius of Nucleus & Mass Spectrometer
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The antineutrino
The antineutrinoin beta decaywas not
detected until 1953, although its presence
had been predicted theoretically.
n p + e + ve1
0
1
1
0
-1
0
0
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The antineutrino
The mass of the neutron is bigger than
that of the proton and electron together.
n p + e + ve
1.008665u(1.007276 + 0.0005486)u = 0.00084u
1
0
1
1
0
-1
0
0
1.008665 u 1.007276 u 0.0005486 u
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The antineutrino
This corresponds (using E = mc2) to an
energy of 0.783 MeV.
n p + e + ve1
0
1
1
0
-1
0
0
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The antineutrino
This extra energy should show up as kinetic energy ofthe products (proton and electron). Since the electronshould carry most of the kinetic energy away, so weshould observe electrons with an energy of about 0.783
MeV.
n p + e + ve
In fact we observe electrons with a rangeof energiesfrom zero up to 0.783 MeV.
1
01
1
0
-1
0
0
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The antineutrino
Where is the missing energy? In 1933 Wolfgang Pauli and
Enrico Fermi hypothesized the existence of a third very light
particle produced during the decay. Enrico Fermi coined the
term neutrino for the little neutral one
n p + e + ve1
0
1
1
0
-1
0
0
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Radioactive decay
The rate of decay (number of nuclei that will
decay per second) is proportional to the
number of undecayed nuclei
dN = -N
dt
= decay constant = the likelihood that an
individual particle will decay in a second
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The radioactive decay law
If the number of nuclei present in a sample at t =
0 is N0, the number Nstill present at time tlater
is given by
N = Noe-t
where is the decay constant (the probabilitythat a nucleus will decay in unit time)
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Half-life
After one half-life, the number of original
nuclei present is equal to N0/2. Putting this
into the radioactive decay law;
N0/2 = N0e(-t)where tis the half-life
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Activity
A =N = N = Noe-t
t
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Measuring half-life
For short half-lives, the half life can usually
be measured directly.
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Measuring half-life
For longer half lifes, values of activitycan
be measured and the decay law can be
used to calculate and thus t.
Measure the activity A and chemicallyfindthe number of atoms of the isotope.
UseA = N and then t= ln2