Question 109: Atoms & Nuclei

Question

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ${ }_{7}^{16} \mathrm{~N}+$ ${ }_{2}^{4} \mathrm{He} \rightarrow{ }_{1}^{1} \mathrm{H}+{ }_{8}^{19} \mathrm{O}$ in a laboratory frame is $n$ (in $M e V$. Assume that ${ }_{7}^{16} \mathrm{~N}$ is at rest in the laboratory frame. The masses of ${ }_{7}^{16} \mathrm{~N},{ }_{2}^{4} \mathrm{He},{ }_{1}^{1} \mathrm{H}$ and ${ }_{8}^{19} \mathrm{O}$ can be taken to be $16.006 u, 4.003 u, 1.008 u$ and $19.003 u$, respectively, where $1 u=930 \,\mathrm{MeVc}^{-2}$. The value of $n$ is ________ .

Accepted Answer

$$
\begin{aligned}
\mathrm{Q} &=\left(\mathrm{m}_{\mathrm{N}}+\mathrm{m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{H}}-\mathrm{m}_{\mathrm{O}}ight) 	imes \mathrm{c}^{2} \
&=(16.006+4.003-1.008-19.003) 	imes 930 \mathrm{MeV} \
&=-1.86 \mathrm{MeV} \
&=1.86 \mathrm{MeV} 	ext { energy absorbed. }
\end{aligned}
$$
And, $\frac{1}{2} 	imes \frac{m 	imes 4 m}{5 m} 	imes v^{2}=$ max loss in kinetic energy
$$
\begin{aligned}
\Rightarrow \frac{1}{2} m v^{2} &=\frac{5}{4} 	imes Q \
&=\frac{5}{4} 	imes(1.86) \mathrm{MeV} \
&=2.325 \mathrm{MeV}
\end{aligned}
$$

Atoms & Nuclei

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