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(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Given

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$4 : 1$

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(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Given the ratio of wavelengths $\frac{\lambda_{p}}{\lambda_{d}} = \frac{1}{\sqrt{2}}$.Using the formula,we have $\frac{\lambda_{p}}{\lambda_{d}} = \sqrt{\frac{m_{d} q_{d} V_{d}}{m_{p} q_{p} V_{p}}}$.For a proton,$m_{p} = m$ and $q_{p} = e$. For a deuteron,$m_{d} = 2m$ and $q_{d} = e$.Substituting these values: $\frac{1}{\sqrt{2}} = \sqrt{\frac{(2m)(e)V_{d}}{(m)(e)V_{p}}} = \sqrt{\frac{2V_{d}}{V_{p}}}$.Squaring both sides: $\frac{1}{2} = \frac{2V_{d}}{V_{p}}$.Therefore,$\frac{V_{p}}{V_{d}} = 4$,which is $4 : 1$.

The ratio of wavelengths of a proton and a deuteron accelerated by potentials $V_{p}$ and $V_{d}$ is $1 : \sqrt{2}$. Then,the ratio of $V_{p}$ to $V_{d}$ will be:

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