An electron accelerated through a potential difference V1 has a de-Broglie wavelength of λ. When the potential is changed to V2, its de-Broglie wavelength increases by 50%. The value of V1V2is equal to :

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

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$\frac{9}{4}$

$\frac{2}{3}$

$\frac{4}{9}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$λ = \frac{h}{p}$ and $λ \propto \frac{1}{\sqrt{v}}$

$\frac{λ}{1.5 λ} = \sqrt{(\frac{V_{2}}{V_{1}})} \Rightarrow \frac{V_{2}}{V_{1}} = \frac{4}{9} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{9}{4}$

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