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(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by the relation $p = \sqrt{2mqV}$,where $

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$\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$

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(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by the relation $p = \sqrt{2mqV}$,where $m$ is the mass and $q$ is the charge of the particle.For an electron,$m = m_{e}$ and $q = e$.For an $\alpha$-particle,$m = m_{\alpha}$ and $q = 2e$.Taking the ratio of the momenta:$\frac{p_{e}}{p_{\alpha}} = \frac{\sqrt{2 m_{e} e V}}{\sqrt{2 m_{\alpha} (2e) V}}$$\frac{p_{e}}{p_{\alpha}} = \sqrt{\frac{2 m_{e} e V}{4 m_{\alpha} e V}}$$\frac{p_{e}}{p_{\alpha}} = \sqrt{\frac{m_{e}}{2 m_{\alpha}}}$

The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100 \ V$ is

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