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A proton when accelerated through a potential difference of V volt has a wavelength $λ$ associated with it. An $α$ -particle in order to have the same $λ$ must be accelerated through a potential difference of

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(V/8) volt

V volt

2V volt

4V volt

answer is D.

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Detailed Solution

$\begin{array}{l} λ_{P} = λ_{α} \\ or \frac{h}{\sqrt{2 m_{p} Q_{p} V_{p}}} = \frac{h}{\sqrt{2 m_{α} Q_{α} V_{α}}} \end{array}$

$\begin{array}{l} ∴ m_{p} Q_{p} V_{p} = m_{α} Q_{α} V_{α} \\ ∴ V_{α} = (\frac{m_{p}}{m_{α}}) (\frac{Q_{p}}{Q_{α}}) V = (\frac{1}{4}) (\frac{1}{2}) V = \frac{V}{8} \end{array}$

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