# A neutron in a reactor makes an elastic head on collision with the nucleus of an atom initially at rest.?

A neutron in a reactor makes an elastic head on collision with the nucleus of an atom initially at rest.

Assume: The mass of the atomic nucleus is

about 14.4 the mass of the neutron.

a. What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?

b. If the initial kinetic energy of the neutron is

1.34 × 10−13 J, find its final kinetic energy.

Answer in units of J.

### 1 Answer

- nyphdinmdLv 78 months ago
Let Kni = initial kinetic energy of the neutron and let Pni = neutron's initial momentum

We can write Kni = Pni^2/(2m) where m = mass of neutron

After collision, momentum is conserved so P = Pnf + Pa where Pa = momentum of atom

Also kineitc energy is conserved (elastic collision) so

K = Kni = Knf + Ka --> Pni^2/(2m) = Pnf^2/(2m) + Pa^2/(28.8m) (used ma = 14.4m)

Pnf = Pni - Pa from conservation of momentum so (multiplying out 2m)

Pni^2 = (Pni - Pa)^2 + 1/14.4 Pa^2

Pni^2 = Pni^2 - 2Pni Pa + Pa^2 + Pa^2/14.4

0 = 15.4/4 Pa^2 - 2Pni Pa = 15.4/4 Pa - 2 Pni --> Pa = 8/15.4 Pni

So Ka/Kni = Pa^2/(2*14.4*m)/(Pni^2/(2m)) = (8/15.4)^2/(14.4) = 1.87%

Kfi = 1.34x10^-13 J

SInce atom takes 1.87% of Kni --> Knf = (1 - 0.0187)Kni = 1.315 x10^-13 J

- Log in to reply to the answers