A neutron travelling with a velocity v and kinetic energy E collides elastically head on with the nucleus of an - Physics - System Of Particles And Rotational Motion

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

Kranav Sharma · Accepted Answer

Hello!

Let say that a neutron with mass 'm₁' and initial speed 'u₁' strikes the nucleus (which is at rest, so u₂ = 0) and undergoes a perfectly elastic collision. The final velocity (v₁) of the neutron would be given as (from combining the Laws of Conservation of Kinetic energy and Momentum)

$v_{1} = \frac{(m_{1} - m_{2}) u_{1} + 2 m_{2} u_{2}}{(m_{1} + m_{2})}$

In this case, u₂ = 0, thus $v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{(m_{1} + m_{2})}$

or $\frac{v_{1}}{u_{1}} = \frac{(m_{1} - m_{2})}{(m_{1} + m_{2})}$

taking m₁ common and canceling it , we get

$\frac{v_{1}}{u_{1}} = \frac{(1 - \frac{m_{2}}{m_{1}})}{(1 + \frac{m_{2}}{m_{1}})}$

or $\frac{v_{1}}{u_{1}} = \frac{(1 - A)}{(1 + A)}$

here A is the mass number of the nucleus.

Now, the fraction of kinetic energy retained would be

$K . E . = \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m u_{1}^{2}} = \frac{v_{1}^{2}}{u_{1}^{2}}$

or

$K . E . = \frac{{(1 - A)}^{2}}{{(1 + A)}^{2}} = {(\frac{1 - A}{1 + A})}^{2} = {(\frac{A - 1}{A + 1})}^{2}$

So, the answer would be option (1)

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

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