472 8 Nuclear Physics – II

where the first term on the right hand side denotes the absorption rate of neu-

trons in^23 Na, and if it is assumed that each neutron thus absorbed produces a

(^24) Na atom, then this also represents the production rate of (^24) Na. The second
term represents the decay rate, so that dQ/dtdenotes the rate of change of
atoms of^24 Na.
The saturation activity is obtained by setting dQ/dt= 0. Then
λQs=φΣa=φσa
Noρ
A

1011 × 536 × 10 −^27 × 6. 02 × 1023 × 0. 97

23

= 1. 36 × 109 s−^1

8.76 At equilibrium number of^198 Au atoms is

Qs=

φσaNoWT 1 / 2

0. 0693 A

=

1012 × 98 × 10 −^24 × 6. 02 × 1023 × 0. 1 × 2. 7 × 3 , 600

0. 693 × 197

= 4. 2 × 1014

Activity=Qsλ=

Qs× 0. 693

T 1 / 2

=

4. 2 × 1014 × 0. 693

2. 7 × 3 , 600

= 3 × 1012 s−^1

8.77 Consider a binary fission, that is a heavy nucleus of mass numberAand atomic
numberZbreaking into two equal fragments each characterized byA 2 andZ 2.
In this problem the only terms in the mass formula which are relevant are the
Coulomb term and the surface tension term.

M(A,Z)=

acZ^2

A^1 /^3

+asA^2 /^3

Energy released is equal to the difference in energy of the parent nucleus

and that of the two fragments

Q=M(Z,A)− 2 M

(

Z

2

,

A

2

)

=

(

asA^2 /^3 +ac

Z^2

A^1 /^3

)

− 2

[

as

(

A

2

) 2 / 3

+ac

(Z/2)^2

(A/2)^1 /^3

]

=asA^2 /^3 (1− 21 /^3 )+ac

Z^2

A^1 /^3

(

1 −

1

22 /^3

)

InsertingA= 238 ,Z= 92 ,ac= 0 .59 andas=14 we findQ∼=160 MeV.

8.78 The diffusion equation for the steady state in the absence of sources at the
point of interest is
∇^2 φ−K^2 φ=0(1)
whereK^2 = 3 Σa/λtr