8.3 Solutions
457
8.3.7 Shell Model
8.26 A state with quantum number j can accommodate a maximum number of
N j = 2(2 j + 1) nucleons. Now j = l ±
1
2
and for N j = 12, j = 5/2 and
l = 3 or 2. However, because the parity is odd and P = (−1)
l , it follows
that l = 3.
8.27 In the shell model the nuclear spin is predicted as due to excess or deficit of
a particle (proton or neutron) when the shell is filled. Its parity is determined
by the l value of the angular momentum, and is given by (−1)
l . For s- state,
l = 0, p – state, l = 1, d-state l = 2, f-state l = 3 etc.
For the ground state of the nuclei:
7
3 Li: Spin is due to the third proton in P 3/2 state.
Therefore J
π
= (3/2)
−
(∵ l = 1)
16
8 O: This is a doubly magic nucleus, and J
π
= 0
+
17
8 O: Spin is due to the 9th neutron in d 5/2 state.
Therefore J
π
= (5/2)
+
(∵ l = 2)
39
19 K: Spin is due to the proton hole in the d 3/2 state.
Therefore J
π
= (3/2)
+
(∵ l = 2)
45
21 Sc: spin is due to the 21st proton in the f 7/2 state.
Therefore J
π
= (7/2)
−
(∵ l = 3)
8.28 The
15 O nucleus in the 1 p 1/2 shell is an
16 O nucleus deficit in one neutron,
its energy being B(15)–B(16), while
17 O in the 1 f 5/2 shell is an
16 O nucleus
with a surplus neutron, its energy being B(16)–B(17). Thus the gap between
the shells is
E(1 f 5/2 ) − E(1 p 1/2 ) = B(16) − B(17) − [B(15) − B(16)]
= 2 B(16) − B(17) − B(15)
= 2 × 127.6193 − 131.7627 − 111.9556
= 11.52 MeV
8.29 Q = −
2 j − 1
2 j + 2
< r
2
>
For
209 Bi, j = 9/2, < r
2
>=
3
5
R
2
=
3
5
r
2
0 A
2/3
=
3
5
(1.2)
2 (209)
2/3
= 30.42 fm
2
= 0.3 b
Q = −0.22 b
8.30 The spin and parity are determined as in Problem 8.27. The ground state spin
and parity for the following nuclides are: