An electron is accelerated through a potential difference of $10, 000 V$ . Its de Broglie wavelength is (nearly) $(m_{e} = 9 \times 10^{- 31} k g)$

12.2 \times 10^{- 13} m

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12.2 \times 10^{- 12} m

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12.2 \times 10^{- 14} m

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12.2 n m

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Solution

The correct option is B $12.2 \times 10^{- 12} m$
$λ = \sqrt{\frac{150}{V}} A^{0} = \sqrt{\frac{150}{10^{4}}} A^{0} = 12.2 \times 10^{- 12} m$

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