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

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

3.2x10^7

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia