An electron of mass m_{e} and a proton of mas | vedclass

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(A) The kinetic energy $K$ acquired by a charged particle accelerated through a potential difference $\Delta V$ is given by $K = q \Delta V$.Since bot

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$\sqrt{\frac{m_{p}}{m_{e}}}$

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(A) The kinetic energy $K$ acquired by a charged particle accelerated through a potential difference $\Delta V$ is given by $K = q \Delta V$.Since both the electron and the proton have the same magnitude of charge $e$,their kinetic energies are equal: $K_{e} = K_{p} = e \Delta V$.The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.For the electron,$\lambda_{e} = \frac{h}{\sqrt{2m_{e}(e \Delta V)}}$.For the proton,$\lambda_{p} = \frac{h}{\sqrt{2m_{p}(e \Delta V)}}$.Taking the ratio of the two wavelengths:$\frac{\lambda_{e}}{\lambda_{p}} = \frac{\frac{h}{\sqrt{2m_{e}(e \Delta V)}}}{\frac{h}{\sqrt{2m_{p}(e \Delta V)}}} = \sqrt{\frac{m_{p}}{m_{e}}}$.

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

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