A neutron travelling with a velocity v and ki | vedclass

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(A) For a perfectly elastic head-on collision between a particle of mass $m_1$ moving with velocity $v_1$ and a particle of mass $m_2$ at rest,the fin

Vedclass · Accepted Answer

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

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(A) For a perfectly elastic head-on collision between a particle of mass $m_1$ moving with velocity $v_1$ and a particle of mass $m_2$ at rest,the final velocity $v_1'$ of the first particle is given by:$v_1' = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) v_1$Here,the mass of the neutron $m_1 = 1$ and the mass of the nucleus $m_2 = A$.Substituting these values,the final velocity of the neutron is:$v_1' = \left( \frac{1 - A}{1 + A} \right) v$The initial kinetic energy of the neutron is $E = \frac{1}{2} m_1 v^2$.The final kinetic energy of the neutron is $E' = \frac{1}{2} m_1 (v_1')^2$.The fraction of energy retained is $\frac{E'}{E} = \frac{\frac{1}{2} m_1 (v_1')^2}{\frac{1}{2} m_1 v^2} = \left( \frac{v_1'}{v} \right)^2$.Substituting $v_1'$,we get:$\frac{E'}{E} = \left( \frac{1 - A}{1 + A} \right)^2 = \left( \frac{A - 1}{A + 1} \right)^2$.

$A$ neutron travelling with a velocity $v$ and kinetic energy $E$ collides perfectly 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

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