Part II Particle and Nuclear Physics Outline solutions - High Energy ...

Part II Particle and Nuclear Physics
Outline solutions
D.R.Ward and C.G.Lester
Lent/Easter Terms 2011
1. Natural Units; practice in using natural units, and converting between natural and
SI units
(a)
λ = 1 m = 7.15 GeV−1 = 1.41 fm
using ¯hc = 0.197 GeV fm to change units.
(b) We have α ≈ 1/137 and √ s =91.2 GeV.
⇒ σ = 2.68 · 10 −8 GeV −2 = 2.68 · 10 −8 × (0.197 fm) 2 = 1.04 · 10 −11 barns = 0.0104 nb
(c) Using standard dimensional analysis to insert the missing factors of ¯h and c we find:
λ = ¯h and σ = 4πα2¯h 2 c 2
mc
3s
(noting that α is dimensionless, so equal to 1/137 in any system of units).
2. Semi-Empirical Mass Formula
(a) Bookwork.
(b) Treat nucleus as uniformly charged sphere of radius R = R 0 A3, 1 charge Ze. Find selfenergy
by bringing spheres of charge from infinity. Charge of sphere of radius r is
Ze(r/R) 3 , and charge of spherical shell between r and r + dr is Ze(4πr 2 dr)/( 4 3 πR3 ).
So, the self-energy is
∫ R
Hence,
0
Ze 4πr2 dr Ze(r/R) 3
4
3 πR3 4πǫ 0 r
a C =
= 3Z2 e 2
20πǫ 0 R
3e2
20πǫ 0 R 0
= 0.72 MeV
(c) The B/A vs A curve peaks around A ∼ 56, so for light nuclei fusion can be energetically
favourable, while for very heavy nuclei fission is possible.
(d) At fixed A the most stable nucleus will be that with lowest mass (not necessarily the
greatest binding energy). So we require ∂M )
= 0
∂Z
⇒ (m p − m n ) + 2a CZ
A 1 3
− 4a A(A − 2Z)
A
Inserting numerical values for A = 300 gives Z = 114 etc.
1
A
= 0 ⇒ Z = (m n − m p ) + 4a A
2a C A −1 3 + 8a A /A

(e) By analogy with (b), and neglecting the proton-neutron mass difference, we have the
gravitational p.e. given by:
− 3GM2
5R = −3Gm2 nA 2
5R 0 A 1 3
= −a G A 5 3 with a G = 3Gm2 n
5R 0
If neutrons only, N = A and Z = 0, so the binding energy reduces to
B = a V N − a S N 2 3 − aA N + a G N 5 3
= 5.8 · 10 −37 MeV
which must be > 0 if the ”nucleus” is bound. The a S term is negligible, so we find
N 2 3 > (a A − a V )/a G = 5 · 10 55 . So M > 8 · 10 28 kg which is 0.07 solar masses.
3. The Fermi Gas model
Density of states:
Fermi energy is defined by
g(ǫ) = BAǫ 1 2 where B = 4√ 2m 3 2R 3 0
3π¯h 3
N =
∫ ǫF
0
g(ǫ)dǫ = 2 3 BAǫ3 2
F ⇒ ǫ F =
( ) 3N
2 3
2BA
If N = Z = 1 A, neutrons and protons have the same Fermi energy
2
ǫ F =
( ) 3A
2 ( )2
3 3 3
= = 33.2 MeV
4BA 4B
(n.b. non-relativistic) and the corresponding Fermi momentum = √ 2mǫ F = 250 MeV/c.
Total KE carried by the neutrons is
∫ ǫF
0
Adding in the protons similarly, total KE is
ǫg(ǫ)dǫ = 2 5 BAǫ5 2
F = 2 ( ) 3N
5 ( )2
5 BA 3 3 3 3 5
= N 3
2BA 5 2BA
( )2
3 3 3 5 5
(N 3 + Z 3 )
5 2BA
Writing N = 1A(1 + α) and Z = 1 A(1 − α) where α = (N − Z)/A, we can expand to second
2 2
order in α
( )
(N 5 5 A 5 [
3
3 + Z 3 ) ≈
2
So the asymmetry term in the energy is
(
3 3
)2
5 2BA
3 ( A
2
2 + 10
9
(N − Z) 2 ]
A 2
) 5 3 10 (N − Z) 2 (N − Z) 2
≡ a
9 A 2 A
A
from which we find, after a little algebra, a A = 1 3 ǫ F = 11 MeV. This is about half the fitted
value for a A .
Pairing energy ∼ 1/g(ǫ F ) in this model.
g(ǫ F ) = BAǫ 1 2
F = 3A
4ǫ F
2

so pairing energy ∼ 4ǫ F /3A ∼ 0.44 MeV, taking A = 100 and ǫ F = 33 MeV. For comparison,
SEMF fit gives 33.5 MeV/A 3 4 ∼ 1 MeV. Again the Fermi gas model is smaller by about a
factor of two, because it only takes account of KE, and not of any PE contribution, i.e. the
inter-nucleon attraction.
4. Practice in using cross-sections
(a) Number of scattering centres per unit area = ρ/M A where ρ = 0.1 kg m −2 , and M A
is the mass per nucleus. Rate of scattering = 10 5 (ρ/M A )σ = 688 s −1 (since the total
cross-section σ = 270.01 b); transmitted intensity = 99312 s −1 .
(b) Rate of fissions = 688 × (200/270.01) = 510 s −1 .
(c) Total rate of elastic scattering = 688 × (0.01/270.01) = 0.0255 s −1 . Area of sphere at
radius 10 m is 400π m 2 , so flux of neutrons = 0.0255/400π = 2 10 −5 s −1 m −2 .
5. Nuclear sizes and Form Factors
Start from
∫
F(q 2 ) =
d 3 r e iq·r ρ(r)
For the purpose of integration, take q as the polar axis, hence:
F(q 2 ) =
=
∫
2πd(cosθ ′ ) r 2 dr e iqr cos θ′ ρ(r)
∫ [ ] e
2πr 2 iqr cos θ ′ π
drρ(r) = 4π ∫ ∞
r sin qrρ(r)dr
iqr q
0
0
Then, for qr ≪ 1, expand the sin keeping the first two terms:
F(q 2 ) = 4π q
∫ ∞
0
r
[
qr − (qr)3
3!
· · ·
]
ρ(r)dr = 1 − 1 6 q2 R 2 · · ·
For a 200 MeV electron (ultrarelativistic) E = p = k in natural units, so k = 1.02 × 10 15 m −1 .
q = 2k sin 1 2 θ = 1.95 × 1014 m −1 . So, given |F(q 2 )| 2 = 0.7 we obtain √ R 2 = 5 × 10 −15 m.
Oscillatory structure in F(q 2 ) suggests non-smoothness in ρ(r), i.e. more like a top-hat than a
Gaussian form (c.f. Fourier transforms of a Gaussian and a top-hat in 1D.) Best fit is a top-hat
with rounded corners (e.g. a Fermi function).
6. Mirror Nuclei
The pair of mirror nuclei have (N, Z) = ( 1(A ± 1), 1 (A ∓ 1)). They have different Coulomb
2 2
energies;
∆E C = a [ (A )
C + 1 2 ( ) A − 1 2
]
− = a CA
= 3αA
A 1 3 2 2 A 1 3 5R
using a C = 3e 2 /(20πǫ 0 R 0 ) = 3α/5R 0 and R = R 0 A 1 3. The only other terms which are differ in
the SEMF between the mirror nuclei are Z(m p +m e )+(A−Z)m n , so M(A, Z +1)−M(A, Z) =
2m e + E max = ∆E C + m p + m e − m n ⇒ ∆E C = E max + 1.8 MeV. Insert numbers to get the
answers on the problem sheet. Find R/A 1 3 is reasonably constant, but higher than the usual
value of R 0 .
3

7. Scattering in Nuclear Physics
Born approximation is based on first order perturbation theory, so it is suitable when the
potential is weak. In contrast, the partial wave method is general, but involves the calculation
of an infinite number of partial waves. However, if the projectile’s energy is small enough,
and/or the potential is sufficiently short range, so that kR ≪ 1, then only the l = 0 partial
wave is significant in the neighbourhood of the potential, and the method is quite easy to apply.
Of course, if the potential is also weak, then either approach is viable.
The question doesn’t imply that V (r) is weak, but we are asked about the low energy limit, so
the partial wave approach is the obvious choice. Wave functions are:
√ √
2m(V0 − E) 2mV0
r < b ψ = A sinh qr where q =
¯h 2 ≈
¯h 2
√
2mE
r > b ψ = B sin(kr + δ 0 ) where k =
¯h 2
Matching boundary conditions at r = b we have:
Dividing:
Hence:
A sinh qb = B sin(kb + δ 0 ) ; qA cosh qb = kB cos(kb + δ 0 )
tanh qb
q
−f(θ) ≡ δ 0
k
= tan(kb + δ 0)
k
=
tanh qb
q
≈ kb + δ 0
k
as k, δ 0 → 0
− b ⇒ dσ [ ] 2
tanhqb
dΩ = − b
q
8. Breit-Wigner Formula in Nuclear Scattering
We have Γ n = 0.0104 eV and Γ γ = 0.105 eV, so that Γ = Γ n + Γ γ = 0.1154 eV, assuming
that there are no other decay modes. Hence, σ n /σ γ = Γ n /Γ γ , and given σ γ = 75 nb, we obtain
σ n = 7.4 kb.
The c.m. energy of the neutron is approximately √equal to the lab. energy of 2.2 eV, since the
target is much more massive. We thus infer λ = h 2 /2mE = 1.92 × 10 −11 m. We then know
everything in the cross-section formula apart from g (note that E = E 0 at the resonance peak).
We find g=0.77≈ 3 , which corresponds to J = 1 for the spin of the compound nucleus.
4
9. Nuclear Shell Model
Sequence of energy levels is given in lecture notes. Each has degeneracy (2j + 1). Shell model
predicts that the spin and parity is that of the unpaired nucleon. Hence:
Z N Unpaired nucleon predicted J P c.f. experiment?
3
2 He 2 1 1s 1+
1/2
yes
2
9
3 −
4Be 4 5 1p 3/2 yes
2
7
3 −
3Li 3 4 1p 3/2 yes
2
12
6 C 6 6 None 0+ yes
13
7 C 6 7 1p 1 −
1/2
yes
2
15
7 N 7 8 1p 1 −
1/2
yes
2
17
8 O 8 9 1d 5+
5/2
yes
2
23
11 Na 11 12 1d 5+
5/2
no
2
131
54 Xe 77 54 1h 11 −
11/2
no
2
207
82 Pb 125 82 2f −
5/2
no
4
5
2

23
The last three disagree with experiment. 11 Na has a significant electric quadrupole moment,
which tells us that the nucleus is not spherical, unlike the assumption in the simple shell model.
In the case of 131 207
54 Xe and 82 Pb, levels 2d 3/2 and 3p 1/2 respectively are predicted to lie very close
to the naively predicted level, which would explain the observed J P values. It is often favourable
to fill a higher-j level at the expense of a lower-lying lower-j level, because the pairing energy
is greater for high j.
10. Magnetic moments of nuclei
Method – find the state of the unpaired nucleon, and use the Schmidt limit formulae from
notes: µ = g J µ N m J for m J = J where
g J = g l ± g s − g l
2l + 1
for the cases j = l ± 1 2
and g s = 5.58 g l = 1 for an unpaired proton and g s = −3.83 g l = 0 for an unpaired neutron.
Hence we have:
3
2 He unpaired n in 1s 1/2 gives µ = −1.91µ N
9
4 Be unpaired n in 1p 3/2 gives µ = −1.91µ N
7
3 Li unpaired p in 1p 3/2 gives µ = 3.79µ N
13
6 C unpaired n in 1p 1/2 gives µ = +0.64µ N
15
7 N unpaired p in 1p 1/2 gives µ = −0.26µ N
17
8 O unpaired n in 1d 5/2 gives µ = −1.91µ N
Some correlation with experiment, but by no means perfect agreement. Least successful are 9 4 Be
and 7 3Li which have a partially filled 1p 3/2 — in this case the simple shell model usually predicts
correctly the largest component of the wavefunction, but there are other ways of coupling the
angular momenta of the nucleons, which presumably contribute. The other four nuclides are
all closed shell ±1 nucleon, and the model is closer to reality.
11. Energy Levels; excited states of nuclei
Shell model predicts that the ground state of even-even nuclei should have J P = 0 + . Only 18
is not even-even, so it must be (d).
18
8 O and 18
10Ne are mirror nuclei, so one expects similar level schemes. They must be (b) and (e).
The difference between them is the Coulomb energy, which will be greater for 18
10Ne. To first
order this has been subtracted out by measuring the energies w.r.t. the ground state. However,
the Coulomb repulsion is slightly less for excited states than for the ground state, because
their wavefunctions will be spatially larger, and we have therefore overcompensated for their
Coulomb energy by aligning the ground states. The upshot is that the excited states for 18
10Ne
will lie lower w.r.t. the ground state. Hence, 18
18
10Ne is (b) and 8 O is (e). This subtle effect is
called the Thomas-Ehrman shift.
208
82 Pb is doubly magic, so we expect it to be very stable with a large first excitation energy —
it must be (a).
By elimination, 166
68 Er must be (c). Its low-lying excitations clearly form a series of rotational
states with energy ∝ J(J + 1). This is as expected for a heavy nuclide far from closed shells,
which is quite likely to be non-spherical.
12. Rate equations
9 F
5

(a) Denote the number of 197 Au nuclei by N 1 and the number of 198 Au nuclei by N 2 . We then
have:
Ṅ 1 = R − λN 1
Ṅ 2 = λN 1
where the decay rate is the reciprocal of the mean lifetime, λ = 2.89 × 10 −6
R = 10 10 s −1 . Integrating, we obtain
s −1 , and
N 1 = R λ
(
1 − e
−λt ) = 2.69 × 10 15 for t = 6 days
Further integration gives
N 2 = R
[
t + e−λt
λ − 1 ]
λ
= 2.50 × 10 15
The equilibrium number of 197 Au nuclei is given by the limit of N 1 for large t, which is
given by N 1 = R/λ = 3.46 × 10 15 .
(b) The rate equations are:
Ṅ Cs = −λ Cs N Cs ⇒ N Cs (t) = N Cs (0)e −λ Cst
Ṅ Ba
= λ Cs N Cs − λ Ba N Ba = λ Cs N Cs (0)e −λ Cst − λ Ba N Ba
This can be solved by various standard methods (e.g. integrating factor; complementary
function + particular integral), noting the initial condition N Ba (0) = 0 to yield
N Ba (t) = λ CsN Cs (0)
λ Ba − λ Cs
[
e
−λ Cs t − e −λ Bat ]
The Ba activity (i.e. its rate of decay, λ Ba N Ba (t)) is maximised when N Ba (t) is maximised.
i.e.
λ Cs e −λCst = λ Ba e −λ Bat
⇒ t = ln(λ Cs/λ Ba )
= 33.5 min
(λ Cs − λ Ba )
Activity of Ba relative to initial activity of Cs is
so the maximum activity is 87 µCi.
13. Alpha decay
λ Ba N Ba (t)
λ Cs N Cs (0) = λ Ba λ [ Cs e
−λ Cs t − e ] −λ Bat
= 0.087
λ Cs λ Ba − λ Cs
(a) The likely decay scheme is this (not to scale):
200 X
1 − 0 −
α 4.687
α 4.650
γ 0.305
196 Y
γ 0.266
6

Note the energies of the α and γ almost add up to the same total, but not quite. This is
because of the small recoil kinetic energy carried by the daughter nucleus, which is slightly
different in the two decay chains.
The α-particle has J P = 0 + (even-even nucleus). The decay 1 − → 0 − would require
orbital angular momentum l = 1 in order to conserve total angular momentum, but then
parity wouldn’t be conserved. The excited states of 196 Y could be 1 − , with l = 0 in the
α-decay, followed by M1 photon transition to the ground state, or 1 + , with l = 1 in the
α-decay, and E1 transition to the ground state, or 2 + , with l = 1 in the α-decay, and E2
transition to the ground state.
(b) Geiger-Nuttall tells us that (for fixed Z as in this case):
ln τ1
2
= a + bQ −1 2
For the Thorium isotopes given, a plot of lnτ1 against Q −1 2 shows good linear behaviour.
2
One can thus use the graph (or a fit to the graph) to interpolate and determine τ1 for
224
90
Th using the given value of Q.
14. Nuclear β-decay
Main ingredients: Fermi transitions have S(eν) = 0 and Gamow-Teller have S(eν) = 1. Allowed
have l = 0 and ”n th -forbidden” have l = n. Then apply usual QM rules for combination of
angular momenta. Odd l leans to a parity change, even l does not. The lowest permitted value
of l will dominate.
(i) 1+ → 1+ is consistent with either Fermi or G-T transitions — can couple either S(eν) = 0
2 2
or S(eν) = 1 to 1 and get 1 . Superallowed, because overlap between nucleon wavefunctions
2 2
must be maximal.
(ii) 0 + → 1 + is consistent with l = 0 and S(eν) = 1 (but not S(eν) = 0). So pure G-T and
superallowed (based on the ft value).
(iii) 0 + → 0 + is consistent with l = 0 and S(eν) = 0 (but not S(eν) = 1). So pure Fermi and
superallowed (based on the ft value).
(iv) 3+ → 3+ is consistent with either Fermi or G-T transitions — can couple either S(eν) = 0
2 2
or S(eν) = 1 to 3 and get 3 . Allowed (based on ft).
2 2
(v) 2 − → 0 + has a parity change, so l is odd. Consistent with l = 1 and S(eν) = 1 (but not
S(eν) = 0). So pure G-T and first forbidden.
(vi) 1 − → 0 + has a parity change, so l is odd. Consistent with l = 1 and S(eν) = 0 or
S(eν) = 1, so mixed Fermi and G-T, first forbidden.
(vii) 7+ → 3+ has no parity change, but l = 0 does not work. Hence l = 2 with either
2 2
S(eν) = 0 or S(eν) = 1, so mixed Fermi and G-T, second forbidden.
15. Nuclear β-decay
Here, as well as considering selection rules, as in the preceding question, we also have to consider
which decays are energetically possible. In terms of atomic masses, the relevant constraints
are:
β + : M(Z) − M(Z + 1) > 0
β − : M(Z) − M(Z + 1) − 2m e > 0
ElectronCapture : M(Z) − M(Z + 1) > 0
2
7

Based on this, the Ce and Nd nuclides should be stable. The possible decays are:
• La→Ce by β − decay. 2 − → 0 + implies first forbidden G-T (c.f. Qu. 14(v)).
• Pr→Ce by EC. 2 − → 0 + implies first forbidden G-T.
• Pr→Nd by β − decay. 2 − → 0 + implies first forbidden G-T.
• Pm→Nd by EC or β + decay. 1 + → 0 + implies allowed G-T (c.f. Qu. 14(ii)).
• Sm→Pm by EC or β + decay. 0 + → 1 + implies allowed G-T.
The shortest lifetime will correspond to the allowed decay with the greatest energy release, i.e.
Pm. The longest lifetime will correspond to the most forbidden decay with the smallest energy
release, i.e. Pr. Agrees with experiment.
16. Gamma decay; selection rules
We are told the ground state is J P = 3 2+ . γ 3 is an E2 transition to the g.s., so ∆J = 2 and no
parity change. This is consistent with the 2.51 MeV state having J P = 1+ 3
, 2+ , 5 2+ or 7+ . But
2
2
the first three of these could also decay to the g.s. by M1 transition, not seen. So the 2.51 MeV
state has J P = 2+ 7 .
γ 1 is M1, so ∆J = 1 and no parity change. This is consistent with the 2.10 MeV state having
J P = 5+ , 7+ or 9+ . But if it was 5+ or 7+ it could decay to the g.s. by E2 (or M1 in the latter
2 2 2
2 2
case), not seen. So the 2.10 MeV state has J P = 9 2+ .
γ 2 is E1, so ∆J = 1 and a parity change. This is consistent with the 1.20 MeV state having
J P = 5 − , 7 − or 9 − . But if it was 5 − it could decay to the g.s. by E1, which is not seen. So the
2 2 2
2
1.20 MeV state has J P = 9 − or 7 − (which can decay to the g.s. by the higher order processes
2 2
E3 or M2/E3 respectively).
17. Nuclear Fission
(a) Suppose initial neutron (momentum p 0 , mass m) hits nucleus (mass Am where A = 12
in this case) head on. Final particles have momenta p 1 and p 2 in the same direction. We
then have:
p 0 = p 1 + p 2 (conservation of momentum)
2m = p2 1
2m + p2 2
2Am
Eliminate p 2 and solve for p 1 /p 0 we find
p 2 0
(conservation of energy)
p 1
= 1 − A
p 0 1 + A = −11 13
Hence the maximal fractional energy loss for the neutron in one collision is 1 −(11/13) 2 =
48
. The smallest number of collisions, n, to reduce the energy to 0.025 eV is therefore
169
given by
( ) 11 2n
= 0.025 ⇒ n = 55
13 2.5 × 10 6
This requires n consecutive head-on collisions, which is highly unlikely, so this is a considerable
underestimate.
8

(b) Call the numbers of each nuclide present N U , N C and N B in obvious notation. The fraction
of 235 U by weight is
235N U
f =
235N U + 12N C
where the contribution of B is negligible in f. We therefore have
N U
N C
=
12f
235(1 − f)
;
N B
= 12
N C 10 × 10−6 = 1.2 × 10 −6
The probability that a neutron is absorbed by a 235 U nucleus and generates fission is
therefore
P =
N U σ U;fiss.
N C σ C + N U σ U + N B σ B
=
N U
N C
× 580
0.04 + N U
N C
× 700 + 1.2 × 10 −6 × 3800
The multiplication factor is given by k = 2.5P < 1, which leads to N U
N C
which in turn implies f < 1.16 × 10 −3 .
< 5.94 × 10 −5
18. Nuclear Fusion The two oxygen nuclei each have radius R ∼ 1.2 fm × A 1 3 = 3.0 fm. The
Coulomb barrier to fusion is therefore the potential energy when the nuclei are 2R apart, i.e.
Z 2 e 2
4πǫ 0 × 2R = Z2 α
2R = 0.08 fm−1 = 15 MeV.
Setting this equal to k B T gives T = 1.7 × 10 8 K.
19. Relativistic Kinematics
(a) Conservation of energy:
Conservation of momentum:
m X = E a + E b
√ √
Ea 2 − m 2 a = Eb 2 − m2 b
Eliminating E b we have
E 2 b = m 2 X + E 2 a − 2m X E a = E 2 a − m 2 a + m 2 b
⇒ E a = m2 X + m2 a − m2 b
2m X
(b)
and the result for E b follows by interchanging a and b. If m a = m b both reduce to
E a = E b = 1 2 m X.
p a =
√
E 2 a − m2 a = √
m 4 X + m 4 a + m 4 b − 2m2 Xm 2 a − 2m 2 Xm 2 b − 2m2 am 2 b
2m X
If m a = m b this reduces to p a = p b = 1 2
(c) If m b = 0, p a = m2 X − m 2 a
2m X
.
√
m 2 X − 4m2 a
= p b
9

20. Relativistic Kinematics
(a) In vacuo we have:
θ
e − ; (E 1 ,p 1 )
e + ; (E 2 ,p 2 )
The initial state has E 2 − p 2 =0, so equating this to the invariant for the final state:
0 = (E 1 + E 2 ) 2 − (p 1 + p 2 ) 2 = 2m 2 e + 2E 1E 2 − 2p 1 p 2 cos θ
and the right hand side is manifestly greater than zero. Hence the decay is forbidden. In
matter, we can have the following:
e − ; (E 1 ,p 1 )
e + ; (E 2 ,p 2 )
Z
The internal electron line is virtual, so no problem to conserve (E, p) at each vertex. The
spectator (labelled ”Z”) could be an electron, or more likely a heavy nucleus which would
have a greater coupling.
(b) Energies of π − and Ξ 0 are 313.8 and 2316 MeV respectively. So,
m 2 Ω = m2 π + m2 Ξ + 2E πE Ξ − 2p π p Ξ cosθ ⇒
Hence E Ω = 2316 + 313.8 = 2630 MeV and p Ω =
m Ω = 1689 MeV
√
E 2 Ω − p 2 Ω = 2016 MeV.
(c) In the lab, t = L/β, and the proper lifetime τ = t/γ. Noting that β = p/E and γ = E/m
we have τ = mL/p = 7 10 −11 s (all formulae in natural units).
21. Partial width, decay rate and branching fraction
Total width of decay Γ(K + ) = ¯h/τ = 5.47 10 −8 eV. Hence branching fraction for the decay
K + →π + π 0 is given by Γ(K + →π + π 0 )/Γ(K + ) = 1.2/5.47 = 0.219.
22. Breit-Wigner Formula
In this case, we have Γ = 2.5 GeV; Γ i = Γ f = 0.0337Γ = 0.08425 GeV; E = 90 GeV,
E 0 = 91.2 GeV. Since the Z 0 has J = 1, the spin degeneracy factor g = 3 4 and λ =
2π/(45 GeV)=0.0275 fm (remembering that this is the wavelength of the colliding particles
in the c.m. frame). Inserting all this into the Breit-Wigner formula we obtain σ = 1.07 nb.
23. Feynman Diagrams and QED
Draw the lowest order Feynman diagram(s) for each of the following processes:
(a) γ → e + e − (in matter)
10

e −
e +
Z
(b) e − + e − → e − + e −
e −
or
e −
e −
e −
(c) e + + e − → e + + e − e +
e +
e − e −
e +
e −
e +
e −
or
(d) e + + e − → µ + + µ − µ +
e +
e −
µ −
(e) e + + e − → γ + γ
e +
e −
γ
γ
or
e +
e −
γ
γ
(f) γ + γ → γ + γ
γ
e −
γ
γ
γ
(other charged particles could appear in the loop, and the final photons could be crossed).
24. Feynman Diagrams and QED
(a) The relevant Feynman diagrams are:
u
γ
u e +
e −
u
γ
u
γ
u e + u
e +
u
e − u
e −
11
e +
e −

Also can have dd initial state, but less likely owing to charge. Compare each decay with
the predominant γγ mode, counting powers of α:
• e + e − γ — one extra vertex means matrix element is multiplied by a factor 2 √ α, where
the factor 2 asises because either photon could convert to e + e − , and hence the rate
∝ 4α ∼ 2.9%;
• e + e − e + e − — two extra vertices, so rate ∝ α 2 ∼ 5 10 −5 ;
• e + e − — again two extra vertices, so rate ∝ α 2 ∼ 5 10 −5 again. Presumably the extra
electron propagator accounts for the additional suppression compared to e + e − e + e − .
Note that the quarks can’t just decay into a single virtual photon, since this would
not conserve angular momentum (the photon has J = 1).
(b) Unlike the π 0 , the ρ 0 has the same J P as the photon, so this Feynman diagram is allowed:
u
e +
u
e −
The partial width is given by Γ e + e − = B(e+ e − )¯h/τ. Inserting numbers, we get Γ e + e − =
6.0 keV for the ρ 0 and Γ e + e − = 1.6 µeV for the π0 ; a ratio of 4 × 10 9 . This is because of
the extra vertex factors and propagators in the π 0 case, and also the greater phase space
in ρ 0 decay.
25. Drell-Yan process
Typical Feynman diagram:
q
q
q
qq
µ +
µ −
The cross section will be proportional to ∑ i e 2 i , where the sum runs over all qq pairs which can
annihilate, and e i is the charge of the i th quark. Hence, in π + (ud) interactions with a proton
(uud) the cross-section ∝ 1 32 , in π + (ud) interactions with a neutron (udd) the cross-section
∝ 2 × 12 , in π − (du) interactions with a proton (uud) the cross-section ∝ 2 × 22 and in π − (du)
3
3
interactions with a neutron (udd) the cross-section ∝ 2 32 . Hence the ratios are 1 : 2 : 8 : 4.
In pp scattering, there are no antiquarks, so the cross-section is zero to leading order. (However,
in higher orders in QCD, the gluons in the proton can create virtual qq pairs, so the rate is
non-zero.) In ¯pp collisions the rate should be ∝ ( 4 × 22 + 1 2 ) = 17, i.e. 17 times that for 3 3 9 π+ p.
26. Spin and Parity
The reaction is this:
π − d }{{}
l=0
−→ nn }{{}
L
We are given that the deuteron has spin-parity J P = 1 + and the pion has spin 0. Hence,
the initial state has total J = 1 and parity equal to that of the pion, P π . J P must both be
conserved. Combining the angular momenta in the final state, we see that there are three
ways of achieving J = 1, namely 3 S 1 , 3 P 1 and 1 P 1 in spectroscopic notation. However, the two
neutrons are identical fermions, and their overall wavefunction must be antisymmetric under
12

particle exchange. The only state which satisfies this is 3 P 1 . Hence L = 1 and the parity of
the final state is (−1) L = −1, and consequently P π = −1.
27. Hadron masses in the quark model
(a) The quark model prediction for the mass of a meson with spin J and l = 0 is
M = m 1 + m 2 +
A (J(J + 1) − 3 )
2m 1 m 2 2
i.e. m 1 + m 2 + A/4m 1 m 2 if J = 1 and m 1 + m 2 − 3A/4m 1 m 2 if J = 0.
Most of the mesons should be straightforward, given their quark compositions and spins.
Only two cases may cause some difficulty:
The η has quark wavefunction 1(uu + dd − 2ss), i.e. a probability 2 of being in the ss
6 3
configuration etc. This gives the mass as 2 × 768 + 1 × 140=559 MeV.
3 3
The η ′ has quark wavefunction 1 (uu + dd + ss), i.e. a probablity 1 of being in the ss
3 3
configuration etc. This gives the mass as 1 × 768 + 2 × 140=349 MeV – not a successful
3 3
prediction!
(b) Each pair of quarks in a J P = 3 2+ baryon must have S = 1. Hence the mass formula gives:
M = ∑ i
m i + A 4
∑
i

(c)
µ u =
2
3 e¯h
2m u
; µ d = −1 3 e¯h
2m u
= − 1 2 µ u
(d) Using the wavefunctions just derived, p has probability 2 of being in the |u↑u↑d↓〉 state
3
etc., so:
µ p = 2(2µ 3 u − µ d ) + 1µ 3 d = 4µ 3 u − 1µ 3 d = 3µ 2 u
µ n = 2 3 (2µ d − µ u ) + 1 3 µ u = 4 3 µ d − 1 3 µ u = −µ u
and the ratio µ n /µ p = 2 3 , compared to data: 0.684. Not too bad. Requires m u ≈ 0.33 GeV,
which is around one third of the proton’s mass – not stupid.
29. Feynman diagrams in QCD; conservation laws
(a) ρ 0 → π 0 γ
u
u
γ
Electromagnetic decay - needs to conserve J P , which is fine, so long as the photon is
emitted in a 1 + state, i.e. it is a magnetic dipole (M1) transition.
(b) ρ 0 → π + π −
u
u
d
d
OK, by the strong interaction. J P is conserved, so long as the two pions are formed in an
l = 1 orbital state.
(c) ρ 0 → π 0 π 0 . Same Feynman diagram as for π + π − can be drawn, just replacing d by u.
But in this case the two π 0 s are identical bosons, so they are forbidden to be in an l = 1
(anitsymmetric) state. Hence angular momentum cannot be conserved, and the decay is
absolutely forbidden.
Expect π + π − to have the highest rate (strong) followed by π 0 γ while π 0 π 0 is simply forbidden.
30. α s in charmonium decay
c
c
g
g
g
By analogy with QED we expect the rate for charmonium decay to be
Γ(ggg) = 2(π2 − 9)
9π
14
( 4 α ) 6
3 s mc .

Taking the charm quark mass to be m c ≈ 1 2 M J//ψ ≈ 1.55 GeV, and assuming Γ(J/ψ → hadrons) = Γ(ggg),
we obtain α s = 0.23.
Similarly for the Υ, replacing the charm quark mass by m b ≈ 1 2 M Υ ≈ 4.73 GeV, we obtain
α s = 0.18. We note that α s runs (decreases) with energy scale as expected.
31. Breit-Wigner for J/ψ production; Zweig’s rule
(a) In this case, J = 1, and Γ i = Γ f = BΓ. Integrate the Breit-Wigner formula using the
substitution E − E 0 = 1 2 Γ tanθ:
∫
σ ′ = 3λ2
∞
16π B2 Γ 2 1
0 [(E − E 0 ) 2 + 1 4 Γ2 ] dE
= 3λ2
16π B2 Γ 2 ∫ π/2
−π/2
1
Γ 2 sec2 θ dθ
1
4 Γ2 sec 2 θ
= 3λ2
16π B2 Γ 22π Γ = 3 8 λ2 B 2 Γ
Note we have implicitly assumed that Γ ≪ E 0 , allowing us to set the lower limit of
integration to − 1 2 π.
(b) The measured cross-section is given by a convolution:
σ meas (E) =
∫ ∞
0
σ true (E ′ )f(E ′ − E)dE ′ .
Now, the spectrum f is a probability distribution, so
Hence,
∫ ∞
0
f(E ′ − E)dE =
∫ ∞
0
f(E ′ − E)dE ′ = 1 .
∫ ∞
0
σ meas (E)dE =
∫ ∞ ∫ ∞
by performing the integration over E ′ .
(c) The fraction is given by
(since dΩ ∝ d cosθ).
0
0
σ true (E ′ )f(E ′ − E)dEdE ′ =
[
∫ 0.6
−0.6 (1 + cos2 θ)d cosθ cos θ +
1
∫ 1
−1 (1 + cos2 θ)d cosθ = 3 cos3 θ ] 0.6
−0.6
[
cosθ +
1
3 cos3 θ ] 1
−1
∫ ∞
0
= 0.504
σ true (E)dE ,
(d) Only possible to make rough estimates — these are mine.... Taking e + e − for example,
peak cross-section above background ∼ 80 nb, FWHM=Γ ∼ 0.003 GeV, so that
σ ′ ∼ 80 × 3
0.504
= 460 nb.MeV
(with a large uncertainty). Peak cross-section above background for µ + µ − ∼ 90 nb and
for hadrons ∼ 2200 nb. So B ∼ 80/(80 + 90 + 2200) ∼ 0.034. The wavelength λ =
(2π/1550) MeV −1 = 0.8 fm. Hence Γ ∼ 0.16 MeV and Γ e + e − ∼ 5 keV.
15

(e) The J/ψ and φ mesons have similar leptonic widths Γ ee because both decays involve the
same Feynman diagram — annihilation into a single virtual photon. The total widths
Γ are dominated by the hadronic decay modes. The φ can decay into a pair of strange
K-mesons
s
s
whilst the J/ψ is below the threshold for decay into charmed mesons, so the quarks must
annihilate into three gluons
q
q
c
c
g
g
g
This more complicated decay is ”Zweig-suppressed” by the extra coupling constants and
propagators needed.
32. Fermi theory of β-decay
Bookwork gives:
dΓ
dp = G2 F
2π 3 (E 0 − E) 2 p 2
(omitting the suffices “e” for brevity). This is the momentum spectrum of the electrons, which
can be regarded as the probability density function for electron momentum, once normalised.
If the electron is highly relativistic we approximate p = E, so the mean energy is.
∫ E0
E · (E 0 − E) 2 E 2 dE
0
∫ E0
= 1 E
(E 0 − E) 2 E 2 2 0
dE
0
If the electron is non-relativistic, we have E = p 2 /2m, and the upper limit of the integral is
√ 2mE0 , so the mean kinetic energy is
∫ √ 2mE 0
0
∫ √ (
2mE 0
0
p 2 ( ) 2
2m · E 0 − p2
p 2 dp
2m
) 2
E 0 − p2
p 2 dp
2m
= 1 3 E 0
In the relativistic case, the total rate is
33. Weak decays of the τ-lepton
The relevant Feynman diagrams will be:
∫
Γ = G2 E0
F
(E
2π 3 0 − E) 2 E 2 dE = G2 F E5 0
.
0
60π 3
ν τ
;
ν τ
µ − τ −
;
τ −
e −
d
ν τ
τ − ν µ
ν e
u
16

The matrix elements for τ − → e − ν τ ¯ν e and τ − → µ − ν τ ¯ν µ should be equal, and the small
difference in the rates simply reflects phase space. The decay τ − → du should be multiplied
by a factor 3 for colour, and by cos 2 θ c where θ c is the Cabbibo angle. However, there is also
the possibility of the decay τ − → su with rate ∝ 3 sin 2 θ c , so the total rate for hadronic final
states is enhanced simply by the colour factor of 3. Energy conservation does not permit the τ
to decay to heavier quarks sich as charm. The observed ratio of 3.5 is accounted for by higher
order QCD corrections, such as:
ν τ
τ −
d
g
u
We have Γ(τ → e) = B(τ → e)/τ τ and Γ(µ → e) = 1/τ µ . Then , from Sargent’s rule, since
E 0 ∝ m l −,
( ) 5
Γ(τ → e)
Γ(µ → e) = mτ
m µ
and hence
τ τ = B(τ → e) · τ µ ·
( mµ
m τ
) 5
= 0.298 ps
34. Weak interactions of the ν τ ; kinematics
In a beam of antineutrinos, it is proposed to search for ¯ν τ via their interactions on nucleons in
a stationary target to produce τ-leptons.
(a) Simplest process is ν τ p −→ τ + n:
ν τ τ +
u
u
d
d
u
d
(b) Equating the E 2 −p 2 invariant before and after the collision, we have at threshold for this
reaction:
s = (E ν + m p ) 2 − E 2 ν = 2E νm p + m 2 p = (m n + m τ ) 2
in obvious notation, and hence, rearranging,
E ν = (m n + m τ ) 2 − m 2 p
2m p
= 3.47 GeV
(c) The γ-factor of the c.m. system is (E ν + m p )/(m n + m τ ) = 1.62, and hence in the lab,
E τ = γm τ = 2.88 GeV. (Many other ways to do this)
(d) Velocity of c.m. system is β = 0.787; lab lifetime t τ = γτ τ , and so average lab distance
travelled is βγτ τ in natural units, = 1.1 × 10 −4 m.
(e) See preceding question.
35. Decays of the Ω −
Ω − → Ξ 0 π − :
17

s
s
s
W −
d
u
u
s
s
Ξ 0 → Λ 0 π 0 :
s
u
s
W −
u
u
d
u
s
Λ 0 → pπ − :
s
d
u
W −
d
u
u
d
u
Strong decay Ω − → Ξ − K 0
s
s
s
g
s
d
d
s
s
Looks plausible, but the decay is not observed because energy is not conserved — the daughters
are more massive than the parent, so absolutely forbidden.
Several possibilities for Ω − → Λ 0 π − , for example:
s
s
s
W −
u
d
u
d
u
s
However, all possibilities involve two W-bosons, because we have to turn two strange quarks
into non-strange. So this is second-order weak, and highly suppressed relative to the permitted
first-order decays to Ξπ or Λ 0 K − .
36. Helicity in the weak interaction
(a) In the centre-of-mass frame of the pion:
ν l
⇒
π + l
⇐
+
18

The double arrows indicate the helicities of the emitted particles. The neutrino always
has negative helicity, so angular momentum conservation demands that the l + also have
negative helicity (the disfavoured state).
(b) Calling the c.m. momentum p ∗ , energy conservation gives
which, after rearrangement yields:
m π = p ∗ +
√
p ∗2 + m 2 l
p ∗ = m2 π − m 2 l
2m π
.
The velocity is given by v/c = p ∗ /E ∗ , so plugging in numbers gives v/c = 0.99997 for the
decay to electron, and 0.271 for the decay to muon.
(c) This is the Feynman diagram:
W +
π + l +
(d) The probability for producing the l + in the ”wrong” negative helicity state is proportional
to (1 − v c ), =3×10−5 for the electron case and 0.73 for the muon: a ratio of 4 ×10 −5 . This
accounts for the large part of the observation, with the difference in phase space also
contributing.
37. The Z boson
(a) From a graph we can make rough estimates: M Z 0 ∼91.5 GeV and Γ Z ∼2.8 GeV. The peak
cross-section (setting E = M Z 0 and Γ e = Γ τ ) is
ν l
σ 0 (Z 0 −→ τ + τ − ) = 3π
4p 2 4Γ e Γ τ
Γ 2 Z
1
where p is the c.m. momentum of the beams at the peak, i.e. M 2 Z0 = 45.75 GeV. From
the graph σ 0 is ∼ 1.5 nb, so inserting our estimates we get Γ τ /Γ=0.0146 and hence
Γ τ ≈82 MeV.
The main reason why the measured resonance curve is asymmetric is initial-state radiation,
i.e. emission of one or more photons from the electron and positron before they form the
Z 0 . This reduces the collision energy at the point when the resonance is formed. This
will tend to decrease the cross-section at or below the peak, and increase the cross-section
when we are significantly above the peak. The distorting effect needs to be calculated
carefully from QED before extracting a reliable value for the Z mass and width. There
are also small additional effects, like tides and trains to be considered, but initial-state
radiation is much the largest effect.
(b) The total width is obtained by summing all contributions:
Γ Z = Γ µ + µ −(3 + R) + Γ νν · 3
where R is measured as 20.7, and the factors of 3 reflect lepton universality, assuming
that only three generations of leptons contribute. This gives Γ = 2473.7 MeV, and τ Z =
1/Γ = 2.65 × 10 −25 s.
19

38. Decays of the W boson
Plugging in numbers,
Γ(W − → e −¯ν e ) = G F
√
2
M 3 W
6π
= 0.23 GeV .
The total width is the sum of the leptonic and hadronic contributions:
Γ W = n ν Γ(W − → e −¯ν e ) + 2 × 3 × Γ(W − → e −¯ν e ) = (6 + n ν )Γ(W − → e −¯ν e )
where the first term reflects the number of lepton generations accessible, n ν , and the second
term the hadronic decays. Hadronic decays to the third generation involving the top quark
are kinematically forbidden, so we have a factor 2 for generations and 3 for colour. Given
Γ W = 2.1 GeV, we deduce n ν = 3.1, i.e. 3 since it must be an integer. More correctly there
should be a QCD correction factor (of ∼1.04) applied to the hadronic term, which gives a closer
estimate.
39. Neutrino scattering
(a) ν e e − → ν e e − W − e −
ν e ν e
Z 0
e − ν e e − e −
ν e
(b) ¯ν e e − → ¯ν e e − W − e −
ν e
Z 0 e −
ν e
ν e
ν e
e −
e −
(c) ν µ e − → ν µ e − Z 0 e −
ν µ
ν µ
e −
(d) ¯ν µ e − → ¯ν µ e − Z 0 e −
ν µ
ν µ
e −
(e) ν e n → e − p
ν e
n
20
p
e −
W −

40. Weak decays and conservation laws
(a) π 0 → γγ OK, electromagnetic decay. This is the Feynman diagram:
u
γ
u
γ
π 0 → π − e + ν e – Feynman diagram is OK (see below), but kinematically impossible (π − is
heavier than the π 0 .
d
d
W +
ν e + e
d u
In the centre-of-mass frame of the pion the decay looks like this:
ν
⇒
π 0
⇒
ν
We see that π 0 → ν¯ν is absolutely forbidden by angular momentum conservation since neutrinos
(antineutrinos) always have the helicity states shown, and the π 0 has J = 0.
(b) e + e − → τ + τ − is an allowed electroweak process.
e +
γ/Z 0
τ +
e −
τ −
¯ν µ + τ − → τ − + ¯ν µ is an allowed weak process.
ν µ
ν µ
Z 0
τ −
τ −
ν τ + p → τ + + n does not conserve tau-lepton number. There is no valid Feynman diagram.
(c) B 0 (bd) → D − (cd)π + is a weak (flavour changing) decay. Rate ∝ V bc cosθ C .
b
d
W +
d c u
d
B 0 → π + π − is a weak (flavour changing) decay. Rate ∝ V bu cos θ C . Much suppressed compared
to Dπ + because of the small CKM matrix element V bu .
21

d
W +
d u u
d
B 0 → J/ψK 0 is a weak (flavour changing) decay. Rate ∝ V bc cosθ C . Same vertices as Dπ + but
suppressed a bit by phase space.
b
d
W +
d s c c
(d) D 0 (cu) → K − π + is an allowed weak decay; rate ∝ cos 2 θ C .
c
u
W +
u
d
s
u
D 0 → π + π − is an allowed weak decay; rate ∝ cosθ C sin θ C .
c
u
W +
d
u
u
d
D 0 → K + π − is an allowed weak decay; rate ∝ sin 2 θ C .
c
u
W +
d
u
u
s
41. Neutrino Oscillations
At t = 0 the state is
ψ(0) = |ν µ 〉 = ν 2 cosθ + ν 3 sin θ ,
so inserting the usual time dependence for the mass eigenstates:
ψ(t) = ν 2 e −iE 2t/¯h cosθ + ν 3 e −iE 3t/¯h sin θ .
Hence the probability of being in the ν µ state at time t is given by
|〈ν µ |ψ(t)〉| 2 =
∣
∣e −iE2t/¯h cos 2 θ + e −iE3t/¯h sin 2 θ∣ 2
= cos 4 θ + sin 4 θ + 2 sin 2 θ cos 2 θ cos(E 3 − E 2 )t/¯h
= 1 − sin 2 2θ sin 2 1 2 (E 3 − E 2 )t/¯h
22

which gives the required answer, noting that t ≈ L/c for neutrinos.
If m 2 and m 3 are very much less than the neutrino momentum, p,
and hence
( )
E 2 = (p 2 + m 2 2 )1 2 ≈ p 1 + m2 2
2p 2
(E 3 − E 2 ) = (m 2 3 − m2 2 )/2p
which yields the required answer; now in natural units.
Inserting the given numerical values into the formula, we have
ν µ (L)
2 (
= 1 − 0.9 sin 2 1.27 × 2.5 × )
10−3 × 730
∣ ν µ (0) ∣
p
Evaluate at some typical momenta and sketch:
p /GeV ∣ νµ(L) ∣
ν µ(0)
1 0.51
1.5 0.10
2 0.24
3 0.56
4 0.73
5 0.80
From the result of qu.34, the threshold for production of a τ − is 3.5 GeV, so observation is just
about possible, though the oscillation probabllity is not high at and above this energy.
∣ 2
23