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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 neutron is $\mathrm{E}$, then how much kinetic energy will be retained by the neutron after reflection?
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The correct answer is:
$\left(\frac{A-1}{A+1}\right)^{2} E$
Fraction retained by nucleus
$\left(\frac{\Delta \mathrm{k}}{\mathrm{k}}\right)_{\text {retained }}=\left(\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right)^{2}=\left(\frac{\mathrm{A}-1}{\mathrm{~A}+1}\right)^{2}$
After collision kinetic energy retained by neutron $\left(\frac{\mathrm{A}-1}{\mathrm{~A}+1}\right)^{2} \mathrm{E}$
$\left(\frac{\Delta \mathrm{k}}{\mathrm{k}}\right)_{\text {retained }}=\left(\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right)^{2}=\left(\frac{\mathrm{A}-1}{\mathrm{~A}+1}\right)^{2}$
After collision kinetic energy retained by neutron $\left(\frac{\mathrm{A}-1}{\mathrm{~A}+1}\right)^{2} \mathrm{E}$
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