An electron accelerated through a potential difference has a de-Broglie wavelength of $λ$ . When the potential is changed to V₂. Its de-Broglie wavelength increases by 50%. The value of $(\frac{V_{1}}{V_{2}})$ is equal to

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3/2

9/4

answer is D.

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Detailed Solution

de-Broglie wavelength $= \frac{h}{mv} = \frac{h}{\sqrt{2 m (ev)}}$
$\Rightarrow\propto \frac{1}{\sqrt{V}} \frac{λ_{2}}{λ_{1}} = \sqrt{\frac{V_{1}}{V_{2}}}$

$\Rightarrow \frac{V_{1}}{V_{2}} = {(\frac{λ_{2}}{λ_{1}})}^{2} = \frac{9}{4}$

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