1000 Solved Problems in Modern Physics

456 8 Nuclear Physics – II

8.22 The mass-energy equation for the positron decay gives

MNa=MNe+ 2 me+

Tmax+Tγ

931. 5

= 21. 991385 + 2 × 0. 511 +

0. 542 + 1. 277

931. 5

= 23 .015338 amu

8.23 For electron capture^7 Be+e−→^7 Li+νe,

Q=(mBe−mLi)× 931. 5 =(7. 016929 − 7 .016004)× 931. 5 = 0 .8616 MeV

which is positive

For positron emission the Q-value must be atleast 1.02 meV which is not

available. Therefore, positron emission is not possible.

8.3.6 Fermi Gas Model ................................

8.24U=

3

5

AEF (1)

p=−

∂U

∂V

=

3

5

A

∂EF

∂V

(2)

From Fermi gas model

A=KVE^3 F/^2 (3)

Differentiating (3) with respect toV

3

2

V

√

EF

∂EF

∂V

+EF^3 /^2 = 0

whence

∂EF

∂V

=−

2

3

EF

V

(4)

Using (4) in (2)

p=

2

5

A

V

EF=

2

5

ρNEF

whereρN=A/V, is the nucleon density.

8.25 The Fermi momentum forN=Z=A/2, is

pF(n)=pF(p)=(/r 0 )(9π/8)^1 /^3

cpF=(c/r 0 )(9π/8)^1 /^3 =(197.3MeV.fm/ 1 .3fm)(9π/8)^1 /^3 =231 MeV

pF=231 MeV/c

EF=pF^2 /2M=(231)^2 /(2×940)=28 MeV

IfBis the binding energy of a nucleon

V=EF+B= 28 + 8 =36 MeV