Question

A 2.00MeV neutron is emitted in a fission reactor. If it loses half its kinetic energy in each collision with a moderatoratom, how many collisions does it undergo as it becomes a thermal neutron, with energy 0.039 eV ?

Answer 1

A 2.00MeV neutron is emitted in a fission reactor. If it loses half its kinetic energy in each collision with a moderator atom, it will undergo 26 collisions as it becomes a thermal neutron, with energy 0.039 eV

How to find the number of collisions moderator atom will undergo?

After $n$ collisions, the kinetic energy of the neutron become :

$E=\frac{Ei}{2^n}$

where, $E_i$ is the initial kinetic energy and $n$ is the number of collisions.

We can rewrite the expression as:

$2^n=\frac{E_i}{E}$

Taking the natural log both sides, we get:

$n=\frac{ln\frac{E_i}{E} }{2}$

Now, from the given data, we have that $E_i=2MeV$ and $E=0.039eV$.

Substituting the values and solving, we get

$n=\frac{ln\frac{2\times 10^6}{0.039} }{ln2} =26$

Hence, the number of collisions are 26.

To learn more about fission reactor, refer to:

brainly.com/question/23276812

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