Question

A neutron travelling with a velocity v and kinetic energy E collides elastically head on with the nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is


(A-1 / A+1)2

(A+1 / A-1)2

(A-1 / A)2

(A+1 / A)2

Answer 1

Answer is first option : (A-1 / A+1)^2

Explanation:

Let M1 be mass of neutron and its initial velocity be U1

Let m be mass of nucleus and its velocity u is 0, since at rest

According to law of conservation of energy,

V1 = ((M-m)U1 + 2mu) / (M+m)

Since u is zero

V1= (M-m)U1 / (M+m)

V1/U1 = (M-m) / (M+m)

It can also be written as:

V1/U1 = (1-m/M) / (1 + m/M) = (1-A) /(1+A)

We know that, Kinetic energy KE = ½ * mv^2

Therefore, V1^2 / U1^2 = (1-A)^2 /(1+A)^2

= > (A-1 /A+1)^ 2 is the fraction of the total energy retained.

Answer 2

Answer:

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