Tutorial Answers - School of Physics - The University of New South ...

THE UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF PHYSICS
PHYS3050 NUCLEAR PHYSICS
TUTORIAL PROBLEMS
1. Using the r α A 1/3 observation, estimate the average mass density of a nucleus.
ρ
M
A×
u
A×
1.66×
10
3×
1.66 × 10
−27
−27
17 −3
= = =
2.3 10
4 3
1
=
= × kg m
4
3 3
−15
3
V π r
4 (1.2 10 )
3
π ( R A ) × ×
3 o
π
It is interesting to compare this value with the highest ‘atomic’ density found in the Periodic Table
which is ~ 22600 kg m -3 for the metals Osmium and Iridium, a factor of 10 billion times smaller !
2. [Ashby & Miller 13.3]. Estimate the rest energy of 1 Å 3 of nuclear matter.
2 2
17 −30
16
E = mc = ρ Vc = 2.3×
10 × 10 × 9 × 10 = 20. 7 kJ
3. [A&M 13.8]. Calculate the distance of closest approach of a 12 MeV deuteron to a Silver
nucleus.
E = 12 × 10
6
× 1.6×
10
−19
q qAg
9 × 10
J = =
4πε
r
o
× 1×
47 × (1.6 × 10
r
9
−19
2
d −15
4. [Krane 3.9]. Calculate the binding energy per nucleon of (a) 7 Li and (b) 56 Fe.
)
∴r
= 5.64 × 10
m
7
B(
Li)
=
= 39.246MeV
B(
56
Fe)
=
1
7 2
[ 3m(
H ) + 4mn
− m(
Li)
] c = [ 3(1.00782504u
) + 4(1.008665u
) − 7.016003u]
∴
B / A = 39.246 / 7 = 5.607MeV
1
56
[ 26m(
H ) + 30mn
− m(
Fe)
]
= 492.262MeV
∴ B / A = 492.262 /56 = 8.790MeV
c
2
(931.5 MeV / u)
5. [K 3.11]. Calculate the mass defect of 238 U.
∆ =
238
[ m(
U ) − 238]
= 47.306MeV
6. [K 3.13]. Calculate the neutron separation energy of 91 Zr.
S
n
=
90
91
[ m(
Zr)
− m(
Zr)
+ mn
]
= 7.195MeV
c
2
c
2
= (238.050785 − 238) × 931.5 MeV / u
7. [K 3.13]. Calculate the proton separation energy of 197 Au.
196
197
1
[ m(
Pt)
− m(
Au)
+ m(
H )]
S p =
= 5. 783MeV
= (89.904703u
− 90.905644u
+ 1.008665u)
× 931.5 MeV / u
c
2
= (195.964926u
−196.966543u
+ 1.007825u)
× 931.5 MeV / u
8. [A&M 14.7] A π o meson decays into two gamma-rays. If the π o is at rest, calculate the
energy of each gamma-ray.
E 1 2 1
m o c = × 134.9745 MeV = 67. 4872MeV
γ =
2 π
2

9. [K 4.1]. Calculate the minimum photon energy necessary to dissociate the deuteron i.e.
γ + d → p + n. Take the deuteron binding energy to be 2.224589 MeV and use a nonrelativistic
approach.
Conserve energy and linear momentum:
p
γ
→ E
= p
γ
p
+ p
= [ m
p
n
&
+ m ] c
n
E
2
γ
+ m
⎪⎧
⋅ ⎨1
−
⎪⎩
d
c
2
= m
2md
m + m
p
p
c
n
2
+ K
p
+ m c
+ K
⎪⎫
−1⎬
= 2.226219MeV
⎪⎭
10. [A&M 14.11]. On the basis of tabulated masses, which of the isobars, 17 N, 17 O and 17 F, would
you expect to be the most stable ? Atomic masses are:
17
17
17
M ( N)
= 17.008450 u,
M ( O)
= 16.999131u,
M ( F)
= 17.002095 u,
17 O has the lowest mass and therefore the highest binding energy.
11. [K 5.1] Give the expected shell-model spin and parity assignments for the ground states of (a)
7 Li; (b) 11 B; (c) 15 C; (d) 17 F; (e) 31 P and (f) 141 Pr. [(a) 3/2 and –1; (b) 3/2 and –1; (c) 5/2 and +1;
(d) 5/2 and +1; (e) 1/2 and +1; (f) 5/2 and +1]. See left-hand side of figure 5.6, p123.
n
2
n
( a)
( b)
( c)
( d)
( e)
( f )
Li,
Z = 3, N = 4. p in 1p
7
11
15
31
B,
Z = 5, N = 6. p in 1p
C,
Z = 6, N = 9.
17
F,
Z = 9, N = 8. p in 1d
P,
Z = 15, N = 16. p in 2s
141
n in 1d
3
2
3
2
5
2
5
2
1
2
Pr, Z = 59, N = 82. p in 2d
∴ I = 3/ 2, π = ( −1)
∴ I = 3/ 2, π = ( −1)
∴ I = 5/ 2, π = ( −1)
∴ I = 5/ 2, π = ( −1)
∴ I = 1/ 2, π = ( −1)
5
2
= ( −1)
= ( −1)
= ( −1)
= ( −1)
∴ I = 5/ 2, π = ( −1)
l
l
l
l
l
1
1
2
2
= ( −1)
l
= −1∴
I
= −1∴
I
= + 1∴
I
= + 1∴
I
0
= ( −1)
= + 1∴
I
2
π
π
π
π
=
=
=
π
=
3 −
2
3 −
2
5 +
2
= + 1∴
I
5 +
2
=
1 +
2
π
=
5 +
2
12. [K5.7] Calculate the shell-model quadrupole moment of 209 Bi (I = 9/2 – ). (Experimental value =
–0.37 b).
Q
=
Q
sp
2 j −1
= −
2( j + 1)
⎛
9
2 ⋅ −1⎞
2
3 2
= −⎜
⎟
× × 1.2 × 209
11
2
⎝ ⋅ 5
2 ⎠
r
2
2 j −1
3
= − ⋅ R
2( j + 1) 5
2
3
= −22.13
fm
2
2
2 j −1
3
= − ⋅ R
2( j + 1) 5
= −0.22b
13. [K5.2] The low-lying energy levels of 13 C are: Ground state (1/2–); 3.09 MeV (1/2+); 3.68
MeV (3/2–) and 3.85 MeV (5/2+). Interpret these states according to the shell-model.
13
C , Z = 6, N = 7.
Protons are paired; the energy states of 13 C are determined by the unpaired (7 th ) neutron.
2
0
A
2
3

2s (1/2)
1d (5/2)
1p (1/2)
1p (3/2)
1s (1/2)
1
Therefore, the energy states are (L to R):
2
−
1
,
2
+
3
,
2
−
5
,
2
+
14. [K10.7]. A certain decay process leads to final states in an even-Z, even-N nucleus and gives
only three γ rays of energies 100, 200 and 300 keV, whose multipolarities are E1, E2 and E3,
respectively. Construct two possible level schemes for this nucleus and label the states with their
most likely spin-parity assignments.
The parity of the En γ-ray is (-1) n so E1 and E3 involve a change in parity between the two
nuclear levels involved. E2 involves no such change. Furthermore, the ground state of the even-even
nucleus is 0 + . The ‘n’ in En gives the change in nuclear spin.
E1 E2 E3 E1 E2 E3
3 – 300 keV 3 – 300 keV
2 +
200 keV
1 –
100 keV
0 +
0 keV
0 +
0 keV

15. [Serway et al 15.17]. By considering the quark makeup of the various particles, deduce the
identity of the unknown particle in the reaction
Σ
0
+ p → Σ
+
+ γ + ?
Σ
u
d
0
: 3
= uds;
p = uud;
Σ
: 2
s :1 → 1
Mystery particle has quark makeup udd i.e. neutron.
16. [K9.9]. The complete processes are:
3 3
ν + He→
H + e
− 8 8
e + B→
Be + ν
40
K →ν
+ e
+
+
+
40
Ar
→ 2 + ? ∴u
→ 0∴
dd
6
He→
ν +
40
12
+
6
C→
= uus
Li + e
12
−
N + e
−
K →ν
+ e +
17. [K9.4]. The maximum kinetic energy of the positron spectrum emitted in the decay 11 C→
11 B
Q = K
∴ m(
11
is 1.983±0.003 MeV. Use this information and the known mass of 11 B to calculate the mass of
11 C.
max
β
= 1.983(3) MeV =
⎛ Q ⎞
C)
= ⎜ + m(
2
⎟
⎝ c ⎠
11
B)
+ 2m
11
6
C→
11
5
+
B + β + ν
11
11
[ m(
C)
− m(
B)
− 2me
]
e
c
2
e
+ ν
−
40
Ca
⎧ 1.983(3) MeV ⎫
= ⎨
⎬ + 11.009305u
+ 2(5.485803×
10
⎩931.502
MeV / u ⎭
−4
) = 11.012531(3) u
18. [Serway 45.59]. A by-product of some fission reactors is 239 Pu which is an α-emitter with a
half-life of 24,120 years. Consider 1 kg of 239 Pu at t=0. (a) What is the number of 239 Pu nuclei at
t=0 ? (b) What is the initial activity ? (c) For how long would you need to store the Plutonium
until it had decayed to a safe activity level of 0.1 Bq ?
( a)
1 nucleus = 239 × 1.66×
10
( b)
( c)
R
0
= λN
0
ln 2
= N
t 1
0.1 = 2.305×
10
2
12
e
0
−27
= 2.305×
10
= 3.97 × 10
12
Bq
∴t
= 1.07 × 10
−λt 6
yrs
−25
kg ∴ N
0
= 1/(3.97 × 10
−25
) = 2.519 × 10
24
19. [A&M 15.14]. Calculate the Q value for the β – decay of 108 Ag.
108
47
Ag→
108
48
−
Cd + β + ν
2
Q = ( M P − M D ) c = (107.905952 −107.904176)
u × 931.5MeV
/ u = 1. 65MeV
20. [A&M 15.19]. On the basis of Q values, determine if the 98 Tc nucleus can decay by (a) β –
(a)
98
43
decay, (b) β + decay, (c) electron capture.
Tc→
98
44
−
Ru + β + ν
e
2
Q = ( M − M ) c = (97.907215 − 97.905287) u × 931.5MeV
/ u = 1.796MeV
i.
e.
> 0∴Yes
P
D
e

(b)
98
43
Tc→
98
42
+
Mo + β + ν
e
Q = ( M
P
− M
D
i.
e.
> 0∴Yes
− 2m
) c
e
2
= (97.907215 − 97.905407 − 2 × 5.485803×
10
−4
) u × 931.5MeV
/ u = 0.662MeV
(c)
e
+
Tc→
− 98 98
43 42
Mo + ν
e
2
Q = ( M − M ) c = (97.907215 − 97.905407) u × 931.5MeV
/ u = 1.684MeV
i.
e.
> 0∴Yes
P
D
21. [S 46.31]. A 2 MeV neutron is emitted in a fission reactor. If it loses half of its kinetic
energy in each collision with a moderator atom, how many collisions must it undergo to achieve
thermal energy (0.039 eV) ?
2
n
6
2 × 10
7
= = 5.13×
10 ∴ n = 25.6 → 26
0.039
22. [S 46.37]. Assume a deuteron and a triton are at rest when they fuse according to
d + t → α + n . This reaction has a Q value of 17.6 MeV. Determine the kinetic energies
acquired by the α and the n.
p
2 2
pα
pn
pn & + = 17.6MeV
= Kα
+ K n → K = 3.5MeV
& K n = 14. 1MeV
2m
2m
α = α
α n
23. [Rohlf 11.15]. If the activity of a substance drops by a factor of 32 in 5 seconds, what is its
half-life ? 32 5
= 2 ∴5s = 5×
t 1 ∴t
1 = 1s
2 2
24. [R 11.26]. Show that the decay n → p + e cannot conserve angular momentum.
All particles have spin 1/2. Two spin-1/2s cannot be combined to yield s=1/2.
25. [R 17.16]. Which of the following strong interactions are allowed ? If a process is
forbidden, state the reason.
0
p + p → π
−
π + p → K
+
p + p → π
0
p + p → π
K
−
+ n
0
+ p → π + Λ
0
+ n
+ n + n
+
+ π
−
+ π
0
−
Electric Charge Baryon # Strangeness
Yes No Yes
Yes Yes No
No Yes Yes
Yes Yes Yes
Yes Yes Yes

26. [K 18.6]. Analyse the following decays or reactions for possible violations of the basic
conservation laws. In each case, state which conservation laws, if any, are violated and through
which interaction the process will most likely proceed (if at all):
( a)
+
π + p → p + p + n
( e)
K
+
→ π
+
+ π
+
0
+ π + π
−
( b)
( c)
Σ
K
+
+
→ n + e
→ π
+
+
+ e
+ ν
+
e
+ e
−
( f )
( g)
K
+
0
→ π
+
+ e
+
Λ + p → Σ
+
+ µ
+ n
−
( d)
−
π + p → Λ + Σ
0
0
( h)
Λ
0
→ p + K
−
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
conserves B, L e , L µ , S, Q, T & T 3 ; proceeds via the strong interaction
violates S; proceeds via the weak interaction
violates S; proceeds via the weak interaction
violates B, S, T & T 3 ; forbidden due to B-violation
violates conservation of energy so forbidden
violates L e & L µ and is forbidden
conserves B, L e , L µ , S, Q, T & T 3 ; proceeds via the strong interaction
violates conservation of energy so forbidden
27. [K 18.7]. Analyse the following decays according to their quark content:
( a)
( b)
Ω
K
−
+
→ π
0
→ Λ + K
+
+ π
−
0
( c)
( d)
Ξ
−
0
→ Λ + π
+
−
Λc → p + K
0
( a)
s → u + W
( b)
( c)
Ω
K
Ξ
−
−
−
= sss;
Λ
−
& W
+
0
= us;
π = ud ; π
= dss;
Λ
0
0
= uds;
K
−
−
→ u + d
−
= uds;
π
= us sss → uds + us ∴ s → d & a uu pair created
= uu + dd ; us → ud + ( uu + dd );
= ud dss → uds + ud
s → u + W
∴ s → u + W
→ u + d
+
0
+
+
( d)
Λ c = udc;
p = uud;
K = ds udc → uud + ds ∴c
→ s + W & W → u + d
−
+
&
&
W
−
W
+
→ u + d

28. [K 18.8]. Analyse the following reactions according to their quark content:
( a)
( b)
K
−
+ p → Ω
−
+ K
+
+
+ K
0
0
0
( c)
+ p → Ξ
+ K
p + p → p + π + Λ + K
( d)
π + n → ∆ + π
K
−
−
−
−
+
0
( a)
us + uud → sss + us + ds ∴ uu → ss + ss
( b)
uud + uud → uud + ud + uds + ds ∴ create dd
−
− +
( c)
K + p → Ξ + K
us + uud → dss + us ∴ uu → ss
( d)
K
−
+ p → Ω
+ K
+
p + p → p + π + Λ
−
−
π + n → ∆
−
0
+ π
+ K
+ K
ud + udd → ddd + ( uu + dd ) ∴ create dd
+
0
0
0
+ ss
29. [K 17.4]. Find which of the following reactions are forbidden by one or more
conservation laws. Give all violated laws in each case.
+
( a)
K + n → Σ + π
( b)
−
+ 0
π + n → K + Λ
( c)
−
0
K + p → n + Λ
+
0
( d)
( e)
π
−
π
+ p → Σ
+ p → Ξ
( f ) d + d →α
+ π
−
+
+ K
−
0
−
+ K
+
+ K
0
(a) S; (b) Q & T 3 ; (c) T & T 3 ; (d) S & T 3 ; (e) S & T 3 and (f) T