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# The nuclear reaction $\;n + _{5}^{10}B \to _{3}^{7}{Li}+_{2}^{4}He\;$ is observed to occur when very slow - moving neutrons $\;(M_{n}=1.0087 u)\;$ strike a boron atom at rest . For a particular reaction in which $\;K_{n} \approx 0\;,$ the helium $\;(M_{He}=4.0026 u)\;$ is observed to have a speed of $\;9.30\times 10^{6} m/s\;.$ Then the kinetic energy of $\;(lithium\;M_{li}=7.016 u)\;$

$(a)\;2.04 MeV\qquad(b)\;1.02 MeV\qquad(c)\;0.04 MeV\qquad(d)\;3.2 MeV$

Answer : (b) $\;1.02 MeV$
Explanation :
Since neutron and boron are both initially at rest , total momentum before reaction is zero , and after ward is also zero
$H_{Li} \nu_{Li}=M_{He} V_{Li}$
We solve this for $\;V_{Li}\;$ and substitute it in to equation for kinetic energy .We can use classical kinetic energy .
Thus $\;K_{Li}=\large\frac{1}{2}\;M_{Li}V_{Li}^{2}=\large\frac{1}{2} M_{Li}\;(\large\frac{M_{He} V_{He}}{M_{Li}})^2$
$=\large\frac{M_{He}^2 V_{He}^2}{2 M_{Li}}$
$K_{Li}=\large\frac{(4.0026 u)^2\;(1.66\times 10^{-27}\;Kg/w)\;(9.30 \times10^{6} m/s)^2}{2(7.016 u)\;(1.66\times10^{-27} Kg/u)}$
$=1.64 \times 10^{-13} J=1.02 MeV\;.$