1000 Solved Problems in Modern Physics

8.3 Solutions 457

8.3.7 Shell Model ....................................

8.26 A state with quantum numberjcan accommodate a maximum number of
Nj=2(2j+1) nucleons. Nowj=l±^12 and forNj= 12 ,j= 5 /2 and
l=3 or 2. However, because the parity is odd andP =(−1)l, it follows
thatl=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 thelvalue of the angular momentum, and is given by (−1)l. For s- state,
l=0, p – state,l=1, d-statel=2, f-statel=3etc.
For the ground state of the nuclei:
7
3 Li: Spin is due to the third proton in P^3 /^2 state.
ThereforeJπ=(3/2)− (∵l=1)

16

8 O: This is a doubly magic nucleus, andJ

π= 0 +

17

8 O: Spin is due to the 9th neutron in d^5 /^2 state.

ThereforeJπ=(5/2)+ (∵l=2)

39

19 K: Spin is due to the proton hole in thed^3 /^2 state.

ThereforeJπ=(3/2)+ (∵l=2)

45

21 Sc: spin is due to the 21st proton in thef^7 /^2 state.

ThereforeJπ=(7/2)− (∵l=3)

8.28 The^15 O nucleus in the 1p 1 / 2 shell is an^16 O nucleus deficit in one neutron,
its energy being B(15)–B(16), while^17 Ointhe1f 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 >=^35 R^2 =^35 r 02 A^2 /^3 =^35 (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: