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(B) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula: $\lambda = \

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$\lambda \sqrt{\frac{m}{M}}$

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(B) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.Since both the electron and the proton have the same magnitude of charge $q = e$ and are accelerated through the same potential difference $V$,we have $\lambda \propto \frac{1}{\sqrt{m}}$.For an electron: $\lambda_e = \lambda = \frac{h}{\sqrt{2meV}}$.For a proton: $\lambda_p = \frac{h}{\sqrt{2MeV}}$.Dividing the two equations: $\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m}{M}}$.Therefore,$\lambda_p = \lambda \sqrt{\frac{m}{M}}$.

An electron of mass $m$ when accelerated through a potential difference $V$ has de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be:

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