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(B) Both the electron and the proton are accelerated through the same potential difference $V$,so they acquire the same kinetic energy $K = eV$.The mo

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$\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}$

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(B) Both the electron and the proton are accelerated through the same potential difference $V$,so they acquire the same kinetic energy $K = eV$.The momentum $P$ of a particle is related to its kinetic energy $K$ by the formula $P = \sqrt{2mK}$.For the electron,the momentum is $P_{e} = \sqrt{2m_{e}K}$.For the proton,the momentum is $P_{p} = \sqrt{2m_{p}K}$.The de-Broglie wavelength is given by $\lambda = \frac{h}{P}$.Therefore,the ratio of the wavelengths is $\frac{\lambda_{p}}{\lambda_{e}} = \frac{h/P_{p}}{h/P_{e}} = \frac{P_{e}}{P_{p}}$.Substituting the expressions for momentum: $\frac{\lambda_{p}}{\lambda_{e}} = \frac{\sqrt{2m_{e}K}}{\sqrt{2m_{p}K}} = \sqrt{\frac{m_{e}}{m_{p}}} = \left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}$.

An electron and a proton are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{p}$ to $\lambda_{e}$ is $[m_{e} = \text{mass of electron}, m_{p} = \text{mass of proton}]$

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