A neutron collides elastically with the stationary nucleus of an atom of mass number A . If the collision is perfectly elastic, then after the collision the fraction of the initial kinetic energy retained by the neutron, is

Question

A neutron collides elastically with the stationary nucleus of an atom of mass number A . If the collision is perfectly elastic, then after the collision the fraction of the initial kinetic energy retained by the neutron, is (A) ((1 - A/A + 1))2 (B) ((A + 1/A - 1))2 (C) ((A - 1/A))2 (D) ((A + 1/A))2

Tardigrade Admin · Accepted Answer

Let us assume the mass of the neutron is

m_{n} = m

and mass of the nucleus is

m_{A} = A m

Also, let us assume that the initial and the final velocities of the neutron are

v_{i}

and

v_{f}

respectively.
Using the formula of final velocity for a perfectly elastic collision we get,

v_{f} = \frac{( m - A m )}{( m + A m )} v_{i} = (\frac{1 - A}{1 + A}) v_{i}

\Rightarrow \frac{v _{f}}{v _{i}} = (\frac{1 - A}{1 + A})

\Rightarrow \frac{K _{f ina l}}{K _{ini t ia l}} = (\frac{1 - A}{1 + A})^{2}

Q. A neutron collides elastically with the stationary nucleus of an atom of mass number $A$ . If the collision is perfectly elastic, then after the collision the fraction of the initial kinetic energy retained by the neutron, is

$(\frac{1 - A}{A + 1})^{2}$

$(\frac{A + 1}{A - 1})^{2}$

$(\frac{A - 1}{A})^{2}$

$(\frac{A + 1}{A})^{2}$

Q. A neutron collides elastically with the stationary nucleus of an atom of mass number A . If the collision is perfectly elastic, then after the collision the fraction of the initial kinetic energy retained by the neutron, is

(A+11−A​)2

(A−1A+1​)2

(AA−1​)2

(AA+1​)2

Q. A neutron collides elastically with the stationary nucleus of an atom of mass number $A$ . If the collision is perfectly elastic, then after the collision the fraction of the initial kinetic energy retained by the neutron, is

$(\frac{1 - A}{A + 1})^{2}$

$(\frac{A + 1}{A - 1})^{2}$

$(\frac{A - 1}{A})^{2}$

$(\frac{A + 1}{A})^{2}$