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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 are accelerated through the same potential difference $V$ and both have the same magnitude of charge $q = e$,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto \frac{1}{\sqrt{m}}$.For the electron: $\lambda_e = \lambda = \frac{k}{\sqrt{m}}$,where $k$ is a constant.For the proton: $\lambda_p = \frac{k}{\sqrt{M}}$.Dividing the two equations: $\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m}}{\sqrt{M}} = \sqrt{\frac{m}{M}}$.Therefore,$\lambda_p = \lambda \sqrt{\frac{m}{M}}$.

An electron of mass $m$ when accelerated through a potential difference $V$ has a 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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