1000 Solved Problems in Modern Physics

478 8 Nuclear Physics – II

Writing sinθ∗dθ∗=−d(cosθ∗), and using (3), (4) becomes

〈

ln

E 1

E 2

〉

=

∫E^1

0

ln

(

E 1

E 2

)

dE 2

E 1

=−

∫

ln

(

E 2

E 1

)

dE 2

E 1

=−

E 2

E 1

ln

E 2

E 1

+

E 2

E 1

∣

∣

∣

∣

E 1

0

At the upper limit the value is 1. At the lower limit, the second term gives

0. The first term also contributes zero becausexlnxin the limitx→0gives
   zero. Thus,
   〈
   ln

E 1

E 2

〉

= 1.

8.85 (a) The number of collisions required is

n=

1

ξ

ln

E 0

En

The average logarithmic energy decrement

ξ= 1 +

(A−1)^2

2 A

ln

A− 1

A+ 1

For graphite (A=12),ξ= 0. 158

∴n=

1

0. 158

ln

(

2 × 106

0. 025

)

= 115

(b) Slowing down time

t=

√

2 m

ξΣs

[E

− 1 / 2

f −E

− 1 / 2

i ]

∼=

√

2 mc^2 /Ef

cξΣs

(∵Ei>>Ef)

Inserting the values,mc^2 = 940 × 106 eV,Ef= 0 .025 eV,ξ= 0. 158

andΣs= 0 .385 cm−^1 for graphite, we find the slowing down timet=

1. 5 × 10 −^4 s.

8.86 If Nuis the number of Uranium atom per cm^3 and N 0 is the number of^238 U
atoms per cm^3 , thenN 0 =^139140 Nu. Further,Nm/Nu=400.
Therefore,Nm/N 0 = 400 × 140 / 139 = 402. 9
Thermal utilization factor (f);

f=

Σa(U)

Σa(U)+Σa(m)

=

Nuσa(U)

Nuσa(U)+Nmσa(m)

=

1

1 +NNumσσaa((Um))

=

1

1 +^402.^97 ×. 680.^0032

= 0. 857