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

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$\sqrt{\frac{m_p}{m_e}}$

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(C) The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by:$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$Since both the proton and electron are accelerated through the same potential $V$ and have the same magnitude of charge $q = e$,we have:$\lambda \propto \frac{1}{\sqrt{m}}$Therefore,the ratio of the wavelengths is:$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$

$A$ proton and an electron are accelerated through the same potential difference. The ratio $\frac{\lambda_e}{\lambda_p}$ will be:

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