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The wavelength of electron accelerated from rest through a potential difference of $x k V$ is $6.14 P M$ . The Value of x is very close to ___

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answer is C.

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Detailed Solution

$λ = \frac{h}{\sqrt{2 m (K E)}} = \frac{h}{\sqrt{2 m q v}} 6.14 \times 10^{- 12} = \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} k g \times 1.6 \times 10^{- 19} c \times V}} V \approx 40000 v o l t s \approx 40 K . v o l t s$

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