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When a proton is accelerated through a potential difference of 4000 volt, its de Broglie wavelength is $λ$ . What will be the de Broglie wavelength of a deuteron when accelerated through a potential difference of 8000 volt?

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$\frac{λ}{\sqrt{2}}$

$\sqrt{2} λ$

$\frac{λ}{2 \sqrt{2}}$

$\frac{λ}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{p^{2}}{2 m} = Q . V \Rightarrow p = \sqrt{2 m Q V} \\ ∴ λ = \frac{h}{p} = \frac{h}{\sqrt{2 m Q V}} \end{array}$

$∴ \frac{λ_{d}}{λ_{p}} = \sqrt{\frac{m_{p}}{m_{d}} \frac{Q_{p}}{Q_{d}} . \frac{V_{p}}{V_{d}}} = \sqrt{\frac{1}{2} .1. \frac{4000}{8000}} = \frac{1}{2}$

$\Rightarrow λ_{d} = \frac{λ}{2}$

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