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 1 f 7/2
8.31 For the nuclei
12
6 C,
13
6 C,
14
6 C and
15
6 C the ground state configuration of protons
is
1s 1/2 , 1 p 3/2
For neutrons it is
(1s 1/2 , 1 p 3/2 ), (1s 1/2 1 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 , 1D 5/2 ) respectively
The spin and parity assignment for
13
6 C is (1/2)
− and that for
15
6 C, it is
(1/2)
+ .
8.32 Twenty neutrons and twenty protons fill up the third shell. The extra neutron
goes into the 1 f 7/2 state. The spin and parity are determined by this extra
neutron. Therefore the spin is 7/2. The parity is determined by the l - value
which is 3 for the f-state. Hence the parity is (−1)
l
= (−1)
3
= −1.
The model predicts J
π
= (0)
+ for
30
14 Si and J
π
= (1)
+ for
14
7 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
Z min =
A
2 + (a c /2a s )A 2/3
Z min
A
=
1
2 + 0.0156A 2/3
For light nuclei, say A = 10, the second term in the denominators is small,
and z min /A = 0.48. For heavy nuclei, say A = 200, Z min /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 × 64 2/3 = 28.57 or 29
8.35 Δm = (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) plus Δm, that is 6.05. The difference in the masses of mirror nuclei is assumed to be due to difference in proton number. Then equating
the Coulomb energy difference to the mass difference
ΔB = a c
[Z (Z + 1) − Z (Z − 1)]
A 1/3
=
2a c Z
A 1/3
a c =
A
1/3
ΔB
2Z
=
3 × 6.05
2 × 13
= 0.7
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 1 f 7/2
8.31 For the nuclei
12
6 C,
13
6 C,
14
6 C and
15
6 C the ground state configuration of protons
is
1s 1/2 , 1 p 3/2
For neutrons it is
(1s 1/2 , 1 p 3/2 ), (1s 1/2 1 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 , 1D 5/2 ) respectively
The spin and parity assignment for
13
6 C is (1/2)
− and that for
15
6 C, it is
(1/2)
+ .
8.32 Twenty neutrons and twenty protons fill up the third shell. The extra neutron
goes into the 1 f 7/2 state. The spin and parity are determined by this extra
neutron. Therefore the spin is 7/2. The parity is determined by the l - value
which is 3 for the f-state. Hence the parity is (−1)
l
= (−1)
3
= −1.
The model predicts J
π
= (0)
+ for
30
14 Si and J
π
= (1)
+ for
14
7 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
Z min =
A
2 + (a c /2a s )A 2/3
Z min
A
=
1
2 + 0.0156A 2/3
For light nuclei, say A = 10, the second term in the denominators is small,
and z min /A = 0.48. For heavy nuclei, say A = 200, Z min /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 × 64 2/3 = 28.57 or 29
8.35 Δm = (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) plus Δm, that is 6.05. The difference in the masses of mirror nuclei is assumed to be due to difference in proton number. Then equating
the Coulomb energy difference to the mass difference
ΔB = a c
[Z (Z + 1) − Z (Z − 1)]
A 1/3
=
2a c Z
A 1/3
a c =
A
1/3
ΔB
2Z
=
3 × 6.05
2 × 13
= 0.7