1000 Solved Problems in Modern Physics

458 8 Nuclear Physics – II

3

2 He : J

π=(1/2)+ The state due to neutron hole is 1s 1 / 2

20

10 Ne : J

π=(0)+ The protons and neutrons complete the sub-shell.

27

13 Al : J

π=(5/2)+ The state of an extra neutron is 1d 5 / 2

41

21 Sc : J

π=(7/2)− The state of an extra proton is 1f

7 / 2

8.31 For the nuclei^126 C,^136 C,^146 C and^156 C the ground state configuration of protons

is

(

1 s 1 / 2 , 1 p 3 / 2

)

For neutrons it is

(1s 1 / 2 , 1 p 3 / 2 ),(1s 1 / 21 p 3 / 2 , 1 p 1 / 2 ),(1s 1 / 2 , 1 p 3 / 2 , 1 p 1 / 2 ) and

(1s 1 / 2 , 1 p 3 / 2 , 1 p 1 / 2 , 1 D 5 / 2 ) respectively

The spin and parity assignment for^136 Cis(1/2)−and that for^156 C, it is

(1/2)+.

8.32 Twenty neutrons and twenty protons fill up the third shell. The extra neutron

goes into the 1f 7 / 2 state. The spin and parity are determined by this extra

neutron. Therefore the spin is 7/2. The parity is determined by thel- value

which is 3 for the f-state. Hence the parity is (−1)l=(−1)^3 =−1.

The model predicts Jπ=(0)+for^3014 Si and Jπ=(1)+for^147 N nuclides.

8.3.8 Liquid Drop Model ................................

8.33 Using the mass formula one can deduce the atomic number of the most stable

isobar. It is given by

Zmin=

A

2 +(ac/ 2 as)A^2 /^3

Zmin

A

=

1

2 + 0. 0156 A^2 /^3

For light nuclei, sayA=10, the second term in the denominators is small,

andzmin/A= 0 .48. For heavy nuclei, sayA=200,Zmin/A= 0. 4

8.34 The most stable isobar is given by

Z 0 =

A

2 + 0 .015 A^2 /^3

=

64

2 + 0. 015 × 642 /^3

= 28 .57 or 29

8.35Δm=(26. 986704 − 26 .981539)× 931. 5 = 4 .76 MeV.

The difference in binding energy is due to mass difference of neutron and

proton. (1.29 MeV) plusΔm, that is 6.05. The difference in the masses of mir-

ror nuclei is assumed to be due to difference in proton number. Then equating

the Coulomb energy difference to the mass difference

ΔB=ac

[Z(Z+1)−Z(Z−1)]

A^1 /^3

=

2 acZ

A^1 /^3

ac=

A^1 /^3 ΔB

2 Z

=

3 × 6. 05

2 × 13

= 0. 7