1000 Solved Problems in Modern Physics

8.3 Solutions 477

Number of^235 Uatoms/cm^3 ,

N=

N 0 ρ

A

×

0. 7

100

=

6 × 1023 × 19

238

×

0. 7

100

= 3. 353 × 1020

Σa=σaN= 550 × 10 −^24 × 3. 353 × 1020 = 0 .1844 cm−^1

Ifφis the neutron flux and 200 MeV is released per fission, then energy

released in 1 cm^3 /s will be 200φΣaMeV or 200× 0. 1844 × 1. 6 × 10 −^13 φ=

5. 9 × 10 −^12 φ= 7 .648 Thereforeφ= 1. 3 × 1012 /cm^2 −s

8.84 Let the Lab kinetic energy of neutron beE 1 andE 2 before and after the scat-
tering. The neutron and proton mass is approximately identical.
The neutron velocity in the CMS, isv∗ 1 =v 1 /2 as the masses of projectile
and target are nearly identical. As the scattering is elastic, the velocity of the
neutronv 2 ∗=v∗ 1 in magnitude. Let the neutron be scattered at angleθ∗in
the CMS. The velocityv 2 ∗is combined withvcto yieldv 2 in the Lab. From
Fig. 8.13,

Fig. 8.13Kinematics of
scattering

v 22 =v∗ 22 +v^2 c+ 2 v 2 ∗vccosθ∗ (1)

v 22 =v^21 (1+cosθ∗)/ 2

(∵vc=v∗ 2 =v 1 /2)

or

E 2 =

E 1

2

(1+cosθ∗)(2)

Let the neutrons scattered between the anglesθ∗andθ∗+dθ∗in the CMS

appear with energy betweenE 2 andE 2 +dEin the LS.

Differentiating (2), holdingE 1 as constant.

dE 2 =

E 1

2

d cosθ∗ (3)

The mean value

〈

ln

E 1

E 2

〉

=

∫

ln

(

E 1

E 2

)

2 πsinθ∗dθ∗

∫

2 πsinθ∗dθ∗

(4)