JEE Advance - Physics (2022 - Paper 1 Online - No. 2)
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 ________ .
Answer
2.325
2.3
- OR
2.33
- OR
Explanation
$$
\begin{aligned}
\mathrm{Q} &=\left(\mathrm{m}_{\mathrm{N}}+\mathrm{m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{H}}-\mathrm{m}_{\mathrm{O}}\right) \times \mathrm{c}^{2} \\
&=(16.006+4.003-1.008-19.003) \times 930 \mathrm{MeV} \\
&=-1.86 \mathrm{MeV} \\
&=1.86 \mathrm{MeV} \text { energy absorbed. }
\end{aligned}
$$
And, $\frac{1}{2} \times \frac{m \times 4 m}{5 m} \times v^{2}=$ max loss in kinetic energy
$$ \begin{aligned} \Rightarrow \frac{1}{2} m v^{2} &=\frac{5}{4} \times Q \\ &=\frac{5}{4} \times(1.86) \mathrm{MeV} \\ &=2.325 \mathrm{MeV} \end{aligned} $$
And, $\frac{1}{2} \times \frac{m \times 4 m}{5 m} \times v^{2}=$ max loss in kinetic energy
$$ \begin{aligned} \Rightarrow \frac{1}{2} m v^{2} &=\frac{5}{4} \times Q \\ &=\frac{5}{4} \times(1.86) \mathrm{MeV} \\ &=2.325 \mathrm{MeV} \end{aligned} $$
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