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
