The nuclear reaction,

$n + {}_5{}^{10}B \to {}_3{}^{7}Li + {}_2{}^{4}He$

is observed to occur even when very slow-moving neutrons $\left(M_n=1.0087 amu\right)$ strike a boron atom at rest. For a particular reaction in which $K_n=0$, the helium $\left(M_{He}=4.0026 amu\right)$ is observed to have a speed of $9.30\times {10}^6 m/s$. Determine the kinetic energy of the lithium $\left(M_{Li}=7.0160 amu\right)$ in MeV .

Since the neutron and boron are both initially at rest, the total momentum before the reaction is zero and afterward is also zero. Therefore, $M_{Li}{v}_{Li}=M_{He}{v}_{He}$.

We solve this for $v_{Li}$ and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistic formulas, because $v_{He}\left(=9.30\times {10}^6 m/s\right)$ is not close to the speed of light $c$ and $v_{Li}$ will be even less since $M_{Li}>M_{He}$. Thus we can write :

$K_{Li}=\frac{1}{2}{M}_{Li}{v}_{Li}^2=\frac{1}{2}{M}_{Li}{\left(\frac{M_{He}{v}_{He}}{M_{Li}}\right)}^2=\frac{M_{He}^2{v}_{He}^2}{2M_{Li}}$

We put in numbers, changing the mass in $u$ to kg and recalling that $1.60\times {10}^{13} J = 1 MeV$ :

$\begin{array}{rcl}{K}_{Li}& =& \frac{{(4.0026)}^2(1.66\times {10}^{-27}){(9.30\times {10}^6)}^2}{2(7.0160)(1.66\times {10}^{-27})}\\ & =& 1.64\times {10}^{-13} J = 1.02 MeV\end{array}$