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

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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 $\;(\large\frac{1}{2}mu^2)_{min}=2 \times 10.2 eV=20.4 eV$

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