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

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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 $p = \sqrt{2mqV}$,where $m$ is the mass and $q$ is the charge of the particle.Since $V$ is the same for both particles,$p \propto \sqrt{mq}$.For an electron,$m = m_e$ and $q = e$.For an $\alpha$-particle,$m = m_\alpha$ and $q = 2e$.Therefore,the ratio of the momenta is $\frac{p_e}{p_\alpha} = \sqrt{\frac{m_e \cdot e}{m_\alpha \cdot 2e}} = \sqrt{\frac{m_e}{2m_\alpha}}$.

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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