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(D) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is given by $K = qV$.Since $K = p^2 / 2

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$\sqrt{m_e/4m_\alpha}$

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(D) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is given by $K = qV$.Since $K = p^2 / 2m$,where $p$ is the momentum and $m$ is the mass,we have $p = \sqrt{2mqV}$.For an electron,$q_e = e$ and mass is $m_e$. Thus,$p_e = \sqrt{2m_e eV}$.For an alpha particle,$q_\alpha = 2e$ and mass is $m_\alpha$. Thus,$p_\alpha = \sqrt{2m_\alpha (2e) V} = \sqrt{4m_\alpha eV}$.The ratio of momenta is $\frac{p_e}{p_\alpha} = \frac{\sqrt{2m_e eV}}{\sqrt{4m_\alpha eV}} = \sqrt{\frac{2m_e}{4m_\alpha}} = \sqrt{\frac{m_e}{2m_\alpha}}$.

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

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