B4: Uno
cial Past Paper Solutions (2011-2016)
Toby Adkins
August 8, 2017
Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at
2011
Question 1
dσ Particles per solid angle seen at detector |j scattered |
= = r2
dΩ Incoming
ux |j incoming |
where j scattered and j incoming are the probability current associated with the scattered and
incoming particles respectively.
Suppose that in the absence of the scattering potential, the system satis
es:
h0 |φi = E |φi
Then, in the presence of the potential, it will satisfy:
H |ψi = (H0 + V ) |ψi = E |ψi
such that
(H0 − E) |ψi = −V |ψi
We can write the solution to this as
V
|ψi = |φi + |ψi
E − H0 ± iε
where we have added ±iε to the denominator of the fraction to avoid the divergence
associated with the singular operator 1/(E − H0 ). This also allows us to evaluate the
expression using contour integration in the complex plane. Thus,
V E
|ψi = |φi + ψ (±)
E − H0 ± iε
1
,Toby Adkins B4 Past Papers
The ψ (−) corresponds to solutions that are propagating towards the scattering centre,
which we are not interested in (and are not as physically signi
cant as the outwards prop-
agating solutions).
Under the Born approximation, we assume that the scattering potential is so weak that it
does not cause a large change in the incoming wave. This means that to
rst order, we
approximate that |ψi ≈ |φi. Now, assume that |φi is a plane wave with wavevector k.
Then,
0
0 eik·x
x φ =
(2π)3/2
such that
Z
m 0 0
f (1)
=− d3 x0 e−i(k−k )·x V (x0 )
2π~2
When the potential is spherically symmetric, we have that r = |x|, q = |k − k0 |.
Z 2π Z −1 Z ∞
m
f (1)
=− dφ d(cos θ) dr r2 V (r)e−iqr cos θ
2π~2 0 1 0
Z ∞ Z −1
m
=− 2π dr r 2
V (r) d(cos θ) e−iqr cos θ
2π~2
Z0 ∞ 1
m 1
=− 4π 2
dr r V (r) (eiqr − e−iqr )
2π~2 0 2iqr
2m ∞
Z
=− 2 dr rV (r) sin qr
~ q 0
as required. The scattering amplitude is related to the scattering cross-section by
dσ
= |f (1) |2
dΩ
Assume that
(
−V0 r<R
V (r) =
0 otherwise
so
Z R
2mV0
f (1) = dr r sin qr
~2 q 0
Over the region of integration, qr
1. This means that we can approximate the integral
of
2mV0 R
Z
(1) 1 3
f = 2 dr r qr − (qr) + . . .
~ q 0 6
3
2mV0 1 qR 1 3 4
= − q R + ...
~2 q 3 30
3 2
2mV0 R q R
= 1− + ...
~2 3 10
Then,
dσ (1) 2 2mV0 6 R 2
= |f | ≈ R 1 − q + ...
dΩ 9~2 5
2
,Toby Adkins B4 Past Papers
We note that
8mE
q 2 = 4k 2 sin2 θ/2 = sin2 θ/2
~2
Thus, we arrive at the result that
dσ 2mV0 6 8mE 2
= R 1 − sin θ/2 + . . .
dΩ 9~2 5~2
Question 2
Nuclei are most stable when their binding energy per nucleon is maximised. Fission of
heavier nuclei can increase the B/A of the products and release energy, while fusion can
increase the B/A of the fusion products and also release energy. These processes will move
the system towards the maximum of the curve of binding energy per nucleon.
The energy released is given by the Q value for the reaction, de
ned by
X X
Q= Mi − Mf
i f
The reaction can only proceed/is energetically favourable if Q > 0.
A sketch of the potential barrier is as follows:
Let us model the two separating nuclei as charged spheres. As they are pulled apart, the
magnitude of the Coulomb repulsion term is reduced. However, the contribution from the
3
, Toby Adkins B4 Past Papers
surface term in the SEMF will increase. This will create the potential barrier that must
be overcome. To estimate the height of the barrier, consider:
Coulomb Energy ~cαEM (Z/2)2
Ef v v
Surface Energy 2r0 (A/2)1/3
where we have assumed that A
1, so N v Z v A/2. By energy conservation, this means
that
T = Q + Ef
is the energy that the neutron mas have to induce
ssion.
Suppose that a thermal neutron is incident on a nucleus with neutron number N . If N =
even, this will lead to an odd N nucleus. If N = odd, then this will lead to an even N nu-
cleus, which will liberate the pairing energy term δ in the SEMF. This will release enough
energy to induce
ssion, even with thermal neutrons. This is why 235 U has a signi
cant
cross-section even for low energy reactions, while 238 U requires quite high energy neutrons
in order to induce
ssion.
Typical products of 235 U
ssion are A v 140, 95. These are neutron rich, and so will decay
towards the the valley of stability by β − decay.
We can control
ssion reactions on the timescale of seconds due to the delays neutrons;
these are neutrons emitted by the decay products after the
ssion decay on the timescale
of seconds. These have lower kinetic energy, and so can be morea easily controlled.
Assume that the probability of a process occurring is proportional to the collision cross-
section. Then:
σnf cσ 235 + (1 − c)σnf
238
p= = nf235 + (1 − c)σ 238
σtot cσtot tot
At E = 2 MeV,
235
σnf = 1.1 bn, 238
σnf = 0.7 bn
235
σel = 1.2 bn, 238
σel = 1.2 bn
235
σtot = 7 bn, 238
σtot = 10 bn
Using these values, p ≈ 0.07.
In the second case, we de
ne the absorption cross-section as σabs = σtot − σel . Then,
235 + (c − 1)σ 238
cσnf
σnf nf
q= = 235 − σ 235 + σ 235 + (c − 1) σ 238 − σ 238 + σ 238
≈ 0.12
σabs + σnf c σtot el nf tot el nf
at E = 2 MeV. With thermal neutrons,
3
E = kB T = 0.07 eV
2
p ≈ 0.17
q ≈ 0.79
4
cial Past Paper Solutions (2011-2016)
Toby Adkins
August 8, 2017
Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at
2011
Question 1
dσ Particles per solid angle seen at detector |j scattered |
= = r2
dΩ Incoming
ux |j incoming |
where j scattered and j incoming are the probability current associated with the scattered and
incoming particles respectively.
Suppose that in the absence of the scattering potential, the system satis
es:
h0 |φi = E |φi
Then, in the presence of the potential, it will satisfy:
H |ψi = (H0 + V ) |ψi = E |ψi
such that
(H0 − E) |ψi = −V |ψi
We can write the solution to this as
V
|ψi = |φi + |ψi
E − H0 ± iε
where we have added ±iε to the denominator of the fraction to avoid the divergence
associated with the singular operator 1/(E − H0 ). This also allows us to evaluate the
expression using contour integration in the complex plane. Thus,
V E
|ψi = |φi + ψ (±)
E − H0 ± iε
1
,Toby Adkins B4 Past Papers
The ψ (−) corresponds to solutions that are propagating towards the scattering centre,
which we are not interested in (and are not as physically signi
cant as the outwards prop-
agating solutions).
Under the Born approximation, we assume that the scattering potential is so weak that it
does not cause a large change in the incoming wave. This means that to
rst order, we
approximate that |ψi ≈ |φi. Now, assume that |φi is a plane wave with wavevector k.
Then,
0
0 eik·x
x φ =
(2π)3/2
such that
Z
m 0 0
f (1)
=− d3 x0 e−i(k−k )·x V (x0 )
2π~2
When the potential is spherically symmetric, we have that r = |x|, q = |k − k0 |.
Z 2π Z −1 Z ∞
m
f (1)
=− dφ d(cos θ) dr r2 V (r)e−iqr cos θ
2π~2 0 1 0
Z ∞ Z −1
m
=− 2π dr r 2
V (r) d(cos θ) e−iqr cos θ
2π~2
Z0 ∞ 1
m 1
=− 4π 2
dr r V (r) (eiqr − e−iqr )
2π~2 0 2iqr
2m ∞
Z
=− 2 dr rV (r) sin qr
~ q 0
as required. The scattering amplitude is related to the scattering cross-section by
dσ
= |f (1) |2
dΩ
Assume that
(
−V0 r<R
V (r) =
0 otherwise
so
Z R
2mV0
f (1) = dr r sin qr
~2 q 0
Over the region of integration, qr
1. This means that we can approximate the integral
of
2mV0 R
Z
(1) 1 3
f = 2 dr r qr − (qr) + . . .
~ q 0 6
3
2mV0 1 qR 1 3 4
= − q R + ...
~2 q 3 30
3 2
2mV0 R q R
= 1− + ...
~2 3 10
Then,
dσ (1) 2 2mV0 6 R 2
= |f | ≈ R 1 − q + ...
dΩ 9~2 5
2
,Toby Adkins B4 Past Papers
We note that
8mE
q 2 = 4k 2 sin2 θ/2 = sin2 θ/2
~2
Thus, we arrive at the result that
dσ 2mV0 6 8mE 2
= R 1 − sin θ/2 + . . .
dΩ 9~2 5~2
Question 2
Nuclei are most stable when their binding energy per nucleon is maximised. Fission of
heavier nuclei can increase the B/A of the products and release energy, while fusion can
increase the B/A of the fusion products and also release energy. These processes will move
the system towards the maximum of the curve of binding energy per nucleon.
The energy released is given by the Q value for the reaction, de
ned by
X X
Q= Mi − Mf
i f
The reaction can only proceed/is energetically favourable if Q > 0.
A sketch of the potential barrier is as follows:
Let us model the two separating nuclei as charged spheres. As they are pulled apart, the
magnitude of the Coulomb repulsion term is reduced. However, the contribution from the
3
, Toby Adkins B4 Past Papers
surface term in the SEMF will increase. This will create the potential barrier that must
be overcome. To estimate the height of the barrier, consider:
Coulomb Energy ~cαEM (Z/2)2
Ef v v
Surface Energy 2r0 (A/2)1/3
where we have assumed that A
1, so N v Z v A/2. By energy conservation, this means
that
T = Q + Ef
is the energy that the neutron mas have to induce
ssion.
Suppose that a thermal neutron is incident on a nucleus with neutron number N . If N =
even, this will lead to an odd N nucleus. If N = odd, then this will lead to an even N nu-
cleus, which will liberate the pairing energy term δ in the SEMF. This will release enough
energy to induce
ssion, even with thermal neutrons. This is why 235 U has a signi
cant
cross-section even for low energy reactions, while 238 U requires quite high energy neutrons
in order to induce
ssion.
Typical products of 235 U
ssion are A v 140, 95. These are neutron rich, and so will decay
towards the the valley of stability by β − decay.
We can control
ssion reactions on the timescale of seconds due to the delays neutrons;
these are neutrons emitted by the decay products after the
ssion decay on the timescale
of seconds. These have lower kinetic energy, and so can be morea easily controlled.
Assume that the probability of a process occurring is proportional to the collision cross-
section. Then:
σnf cσ 235 + (1 − c)σnf
238
p= = nf235 + (1 − c)σ 238
σtot cσtot tot
At E = 2 MeV,
235
σnf = 1.1 bn, 238
σnf = 0.7 bn
235
σel = 1.2 bn, 238
σel = 1.2 bn
235
σtot = 7 bn, 238
σtot = 10 bn
Using these values, p ≈ 0.07.
In the second case, we de
ne the absorption cross-section as σabs = σtot − σel . Then,
235 + (c − 1)σ 238
cσnf
σnf nf
q= = 235 − σ 235 + σ 235 + (c − 1) σ 238 − σ 238 + σ 238
≈ 0.12
σabs + σnf c σtot el nf tot el nf
at E = 2 MeV. With thermal neutrons,
3
E = kB T = 0.07 eV
2
p ≈ 0.17
q ≈ 0.79
4
