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The ratio of momentum of an electron and $\alpha$-particle which are accelerated from rest by a potential difference of $100 \mathrm{~V}$ is
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Verified Answer
The correct answer is:
$\sqrt{\left(\mathrm{m}_{\mathrm{e}} / 2 \mathrm{~m}_{\alpha}\right)}$
As we know, $\frac{1}{2} m v^{2}=e V$
\begin{aligned}
\Rightarrow \mathrm{v} &=\sqrt{\frac{2 e V}{m}} \\
\mathrm{p} &=\mathrm{mv}=\sqrt{2 m e V}
\end{aligned}
Now, $\mathrm{p}_{\mathrm{e}}=\sqrt{2 m_{e} \times e \times 100}$
and $\mathrm{p}_{\alpha}=\sqrt{4 m_{\alpha} \times e \times 100}$
$\therefore \quad \frac{p_{e}}{p_{\alpha}}=\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$
\begin{aligned}
\Rightarrow \mathrm{v} &=\sqrt{\frac{2 e V}{m}} \\
\mathrm{p} &=\mathrm{mv}=\sqrt{2 m e V}
\end{aligned}
Now, $\mathrm{p}_{\mathrm{e}}=\sqrt{2 m_{e} \times e \times 100}$
and $\mathrm{p}_{\alpha}=\sqrt{4 m_{\alpha} \times e \times 100}$
$\therefore \quad \frac{p_{e}}{p_{\alpha}}=\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$
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