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The ratio of momenta of an electron and an $\alpha$ -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:
$\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$
Momentum $p=m v$ and $v=\sqrt{\frac{2 q V}{m}}$
$$
\begin{array}{c}
p=\sqrt{2 q m V} \\
p=\sqrt{q m} \\
\frac{p_{e}}{p_{\alpha}}=\sqrt{\frac{e m_{e}}{2 e m_{\alpha}}}
\end{array}
$$
$\frac{p_{e}}{p_{\alpha}}=\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$
$$
\begin{array}{c}
p=\sqrt{2 q m V} \\
p=\sqrt{q m} \\
\frac{p_{e}}{p_{\alpha}}=\sqrt{\frac{e m_{e}}{2 e m_{\alpha}}}
\end{array}
$$
$\frac{p_{e}}{p_{\alpha}}=\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$
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