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