1000 Solved Problems in Modern Physics

482 8 Nuclear Physics – II

Σa(graphite)=

1

λa(C)

=

1

2 , 700

= 3. 7 × 10 −^4 cm−^1

Σa=Σa(C)+Σa(B)=Σa(C)+σa(B)N

= 3. 7 × 10 −^4 + 755 × 10 −^24 N(1)

λtr=

λs

1 − 32 A

=

2. 7

1 − 3 ×^212

= 2 .859 cm (2)

The attenuation dependence ofe−^0.^03 ximplies that the diffusion length

L=

1

0. 03

= 33 .33 cm (3)

ButL^2 =

λtrλa

3

=

λtr

3 Σa

(4)

Combining (1), (2), (3), and (4) and solving forN, we find

N= 5. 83 × 1017 boron atoms/cm^3

8.92 Refer to solution of Problem 8.84 with the change of notation.

v 12 =v 1 ∗^2 +vc^2 + 2 v 1 ∗vccosθ∗

=

v 02

(A+1)^2

(A^2 + 1 + 2 Acosθ∗)

where we have used the relations

v 1 ∗=AV 0 /(A+1) andvc=v 0 /(A+1)

We can then write

E 1

E 0

=

A^2 + 2 Acosθ∗+ 1

(A+1)^2

<E 1 /E 0 >=

1

(A+1)^2

∫+ 1

− 1

(A^2 + 2 Acosθ∗+1)

1

2

d(cos∗)

=

(A^2 +1)

(A+1)^2

Therefore,<Ef>=Ei(A

(^2) +1)
(A+1)^2
Let the neutron energy beEnafterncollisions
En
E 0

=

E 1

E 0

.

E 2

E 1

.

E 3

E 2

...

En

En− 1

=

(

E 1

E 0

)n

Therefore 20 ×.025 eV 106 eV=

[

A^2 + 1

(A+1)^2

]n

=

( 145

169

)n

where we have putA=12 for graphite. Taking logarithm on both sides and

solving for n, we obtainn=119. Compare this value withn=115 obtained

from the average logarithmic decrement (Problem 8.85).