Question

Graphite (carbon with atomic mass of 12 atomic mass units (amu)) has some desirable properties as a neutron moderator in a nuclear reactor. The neutron (approximately 1 amu) needs to be slowed in order for the reaction to proceed. Assuming each collision is elastic, how many times must the neutron collide with a carbon atom (at rest) until its final speed is less than 25% of its initial speed?

Answer 1

Answer on Question #75141-Physics-Other

Graphite (carbon with atomic mass of 12 atomic mass units (amu)) has some desirable properties as a neutron moderator in a nuclear reactor. The neutron (approximately 1 amu) needs to be slowed in order for the reaction to proceed. Assuming each collision is elastic, how many times must the neutron collide with a carbon atom (at rest) until its final speed is less than 25% of its initial speed?

Solution

1) From the conservation of momentum:

$$mV = -mV' + 12mv \rightarrow v = \frac{V + V'}{12}$$

The collision is elastic, so

$$\frac{mV^2}{2} = \frac{mV'^2}{2} + 12m\frac{v^2}{2}$$ $$\frac{V^2}{2} = \frac{V'^2}{2} + 12\frac{1}{2}\left(\frac{V + V'}{12}\right)^2$$ $$V^2 = V'^2 + \frac{1}{12}\left(V^2 + 2VV' + V'^2\right)$$ $$\frac{13}{12}V'^2 + \frac{1}{6}VV' - \frac{11}{12}V^2 = 0$$

Two solutions:

$$V' = -V$$ $$V' = \frac{11}{13}V$$

First solution is impossible for magnitude of velocity.

2)

$$V^{(n)} = \left(\frac{11}{13}\right)^n V$$ $$KE \sim V^2.$$

Thus,

$$KE^{(n)} = \left(\frac{11}{13}\right)^{2n} KE < 0.25 KE$$ $$\left(\frac{11}{13}\right)^{2n} < 0.25$$ $$n > \frac {1 \ln 0 . 2 5}{2 \ln \left(\frac {1 1}{1 3}\right)} = 4. 1 5.$$

Answer: 5 times.

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