Question

A neutron in a nuclear reactor makes an elastic head-on collision with the nucleus of a plutonium atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the plutonium nucleus

Answer 1

Answer:

Fraction = 59049/60025

Explanation:

Let m be the mass of the neutron and M be the mass of the plutonium nucleus (at rest)

Now, formula for kinetic energy before collision is;

K_i = ½mu²

Formula for kinetic energy after collision is;

K_f = ½mv²

Where;

u is the velocity of the neutron before collision

v is the velocity of the neutron after collision.

From collision principle where momentum before collision equals momentum after collision, we can say that;

(m - M)u = (m + M)v

Thus,

v = [(m - M)u]/(m + M)

Putting [(m - M)u]/(m + M) for v in the final kinetic energy equation gives;

K_f = ½m([(m - M)u]/(m + M))²

K_f = ½mu²((m - M)²/(m + M)²)

To get the fraction of the neutron's kinetic energy is transferred to the plutonium nucleus, it is simply;

K_f/K_i = [½mu²((m - M)²/(m + M)²)]/½mu²

This gives;

K_f/K_i = ((m - M)²/(m + M)²)

But mass of plutonium = 244m

Thus;

K_f/K_i = ((m - 244m)²/(m + 244m)²)

K_f/K_i = 59049m²/60025m²

K_f/K_i = 59049/60025