The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $\mathrm{P}$ is $\mathrm{d}$. The distance of closest approach of alpha particle to nucleus, if the linear momentum of the alpha particle is $1.5 \mathrm{P}$

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Aicharya · Accepted Answer

At sufficient distance, electric potential energy is zero. So, $\mathrm{U}_1=0$

at (1) $\mathrm{K}_1=\frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{P}^2}{2 \mathrm{~m}}$ at (2) $\mathrm{K}_2=0$ $\mathrm{U}_2=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_\alpha \mathrm{q}_{\mathrm{n}}}{\mathrm{d}_{\mathrm{c}}}$ By energy conservation $\begin{aligned} & \mathrm{U}_1+\mathrm{K}_1=\mathrm{U}_2+\mathrm{K}_2 \\ & 0+\frac{\mathrm{p}^2}{2 \mathrm{~m}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_\alpha \mathrm{q}_{\mathrm{n}}}{\mathrm{d}_{\mathrm{c}}}+0\end{aligned}$ Distance of closest approach, $d_c \propto \frac{1}{p^2}$ $\begin{aligned} & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{p}_1^2}{\mathrm{p}_2^2} \\ & =\mathrm{d} \frac{\mathrm{p}^2}{(1.5 \mathrm{p})^2}=\frac{4 \mathrm{~d}}{9}\end{aligned}$

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the…

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