An electron accelerated through a potential difference has a de-Broglie wavelength of . When the potential is changed to , its de-Broglie wavelength increases by 50%. The value of is equal to [2023]

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Solution

Q.
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 [2023]

1 $3$

2 $\frac{9}{4}$

3 $\frac{3}{2}$

4 $4$

Ans.
(2)

$K.E. = \frac{P^{2}}{2 m}, P = \frac{h}{λ}$

$e V_{1} = \frac{{(\frac{h}{λ})}^{2}}{2 m}$

$e V_{2} = \frac{{(\frac{h}{1.5 λ})}^{2}}{2 m}$

$\frac{V_{1}}{V_{2}} = {(1.5)}^{2} = \frac{9}{4}$

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