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(C) The de Broglie wavelength $\lambda$ of a particle of charge $q$ and mass $m$ accelerated through a potential difference $V$ is given by $\lambda =

Vedclass · Accepted Answer

$2 \sqrt{2}: 1$

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(C) The de Broglie wavelength $\lambda$ of a particle of charge $q$ and mass $m$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Since $h$ and $V$ are constant for both particles, $\lambda \propto \frac{1}{\sqrt{mq}}$.For a proton $(p)$: mass $m_p = m$, charge $q_p = e$.For an alpha particle $(\alpha)$: mass $m_{\alpha} = 4m$, charge $q_{\alpha} = 2e$.The ratio of wavelengths is $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{4m \cdot 2e}{m \cdot e}} = \sqrt{8} = 2\sqrt{2}$.Thus, the ratio $\lambda_p : \lambda_{\alpha} = 2\sqrt{2} : 1$.

If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is

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