# A neutron moving with speed V makes a head on collision with a hydrogen atom in ground state kept at a rest. Then find the minimum kinetic energy of neutron for which inelastic (completely of partially ) collision may take place . (mass of neutron =mass of hydrogen =$\;1.67\times10^{-27} Kg$) .

$(a)\;17.2 eV \qquad(b)\;20.4 eV\qquad(c)\;19.6 eV\qquad(d)\;12 eV$

Answer : $\;20.4 eV$
Explanation :
Collision between neutron and hydrogen atom will be inelastic if a part of kinetic energy is used to excite atom .
If $\;u_{1}\;$ and $\;u_{2}\;$ are speed of neutron and hydrogen atom after collision , then
$mu=mu_{1}+mu_{2}---(1)$
$\large\frac{1}{2}\;mu^2=\large\frac{1}{2}\;mu_{1}^2+\large\frac{1}{2}\;mu_{2}^2+\bigtriangleup \varepsilon$---(2)
$u^2=u_{1}^2+(u-u_{1})^2+\large\frac{2 \bigtriangleup \varepsilon}{m}$
$u_{1}^2-uu_{1}+\large\frac{2 \bigtriangleup \varepsilon}{m}=0$
For $\;u_{1}\;$ to be real
$u^2-\large\frac{4 \bigtriangleup \varepsilon}{m} \geq 0$
$\large\frac{mu^2}{2} \geq 2 \times\bigtriangleup \varepsilon \qquad [\bigtriangleup \varepsilon=10.2eV]$
Thus $\;(\large\frac{1}{2}mu^2)_{min}=2 \times 10.2 eV=20.4 eV$
edited Mar 25, 2014 by yamini.v