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

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

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(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by $p = \sqrt{2mK}$,where $K = qV$ is the kinetic energy.Thus,$p = \sqrt{2mqV}$.Since $V$ is the same for both particles,the ratio of momenta is $\frac{p_e}{p_\alpha} = \sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}}$.For an electron,$q_e = e$. For an $\alpha$-particle,$q_\alpha = 2e$.Substituting these values,we get $\frac{p_e}{p_\alpha} = \sqrt{\frac{m_e \cdot e}{m_\alpha \cdot 2e}} = \sqrt{\frac{m_e}{2m_\alpha}}$.

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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