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An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{- 2} nm$ , the potential difference is

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10²V

10³ V

10⁴ V

10 V

answer is C.

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Detailed Solution

$\begin{array}{l} λ = \frac{h}{mv} \\ = \frac{h}{\sqrt{2 vqm}} \\ \Rightarrow λ = \frac{12 .27}{\sqrt{v}} A^{0} \\ 1 .227 \times 10^{- 11} = \frac{12 .27}{\sqrt{v}} \times 10^{- 10} \\ \sqrt{v} = 10^{2} \Rightarrow V = 10^{4} volt \end{array}$

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