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A proton and an $α$ - particle are accelerated through the same potential difference. The ratio of their Broglie wavelengths is

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$\frac{1}{\sqrt{2}}$

$2 \sqrt{2}$

$2$

$\sqrt{2}$

answer is C.

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Detailed Solution

$λ = \frac{h}{m v}$

$\Rightarrow λ = \frac{h}{m \sqrt{\frac{2 q V}{m}}} [∵ q V = \frac{1}{2} m v^{2}]$

$\Rightarrow λ = \frac{h}{\sqrt{2 m q V}}$

$\Rightarrow \frac{λ_{p}}{λ_{α}} = \sqrt{\frac{m_{α} q_{α}}{m_{p} q_{p}}} = \sqrt{\frac{4}{1} \times \frac{2}{1}} = 2 \sqrt{2}$

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