A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

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Aicharya · Accepted Answer

$\lambda_p=\lambda_\alpha$ $(m q V)_p=(m q V)_\alpha$ Potential difference $V_\alpha=\frac{V}{8} \quad\left[\begin{array}{c}\because m_\alpha=4 m_p \ q_\alpha=2 q_p\end{array}ight]$

A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$…

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