Question

A neutron is moving with velocity $u$. It collides head on and elastically with an atom of mass number $A$. If the initial kinetic energy of the neutrons is $E$, then how much kinetic energy will be retained by the neutron after collision?
(Assume fission doesn’t takes place)
A. ${\left[ {\dfrac{A}{{\left( {A + 1} \right)}}} \right]^2}E$
B. $\left[ {\dfrac{A}{{{{\left( {A + 1} \right)}^2}}}} \right]E$
C. ${\left[ {\dfrac{{\left( {A - 1} \right)}}{{\left( {A + 1} \right)}}} \right]^2}E$
D. $\left[ {\dfrac{{\left( {A - 1} \right)}}{{{{\left( {A + 1} \right)}^2}}}} \right]E$

Answer 1

Hint: Assume the neutron and the atom to be system. You can apply conservation of momentum if the particles collide elastically. In an atom, there are some protons and neutrons and the mass number of an atom is the sum of the number of protons and neutrons.
The kinetic energy of a particle of mass \[m\] moving with velocity $v$ is given by $\dfrac{1}{2}m{v^2}$ .

Complete step by step answer:
Let us first consider the neutron and the atom to be systems.
In a head on elastic collision, the initial and final velocities of the colliding particles lie along the same line and the linear momentum is conserved for the system.
Let the mass of the neutron is $m$. In an atom, there are some protons and neutrons. Let the no. of neutrons and protons in the atom be $x$ and $y$ respectively. The mass number of an atom is the sum of number of protons and neutrons i.e. $A = x + y$ and mass of the atom will be $mx + my = mA$ as the mass of proton and neutron is almost the same.
Let the neutron acquire a velocity ${v_1}$ towards left and the atom acquire a velocity ${v_2}$ towards right after the collision. Now we apply conservation of momentum. Initially, the momentum of the atom will be zero as it is at rest.
$mu = m\left( { - {v_1}} \right) + mA{v_2}$
On simplifying we have
${v_2} = \dfrac{{u + {v_1}}}{A}$ ……(i)
Now as we know that the coefficient of restitution $e$ is the ratio of velocity of separation after collision to the velocity of approach of the particles before collision and $e = 1$ for elastic collision. In this case, velocity of separation after collision will be equal to ${v_1} + {v_2}$ and velocity of approach before collision will be equal to $u$ .
Therefore, $\dfrac{{{v_1} + {v_2}}}{u} = 1$ which implies $u = {v_1} + {v_2}$ .
Now, substituting the value of ${v_2}$ from equation (i) in the above equation we have
$u = {v_1} + \dfrac{{u + {v_1}}}{A}$
On simplifying we have
$Au = A{v_1} + u + {v_1}$
On further solving we get the value of ${v_1}$ as
${v_1} = \dfrac{{\left( {A - 1} \right)}}{{\left( {A + 1} \right)}}u$
We know that the kinetic energy of a particle of mass \[m\] moving with velocity $v$ is given by $\dfrac{1}{2}m{v^2}$ .
So, the kinetic energy of the neutron after the collision will be given by ${E_f} = \dfrac{1}{2}m{v_1}^2 = \dfrac{1}{2}m{u^2}{\left[ {\dfrac{{A - 1}}{{A + 1}}} \right]^2}$ ……(ii)
As given in the question that the initial kinetic energy of the neutrons is $E$. So,
$E = \dfrac{1}{2}m{u^2}$
Substituting this value in equation (ii) we have
${E_f} = {\left[ {\dfrac{{\left( {A - 1} \right)}}{{\left( {A + 1} \right)}}} \right]^2}E$
Hence, option C is correct.

Note: All the forces involved in an elastic collision are conservative in nature. So, both the energy and momentum is conserved in case of elastic collision. On the other hand, some of the forces involved in an inelastic collision are non-conservative. So, only momentum is conserved in case of an inelastic collision.