De Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is

λ = \frac{12.3}{\sqrt{h}} A

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λ = \frac{12.26}{\sqrt{V}} A

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λ = \frac{12.3}{\sqrt{E}} A

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λ = \frac{12.3}{\sqrt{m}} A

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Solution

The correct option is B $λ = \frac{12.26}{\sqrt{V}} A$
According to De broglie wavelength of an electron after being accelerated by a potential difference of $V$ volt from rest is
$\begin{matrix} λ = \frac{h}{p} = \frac{h}{\sqrt{2 m e v}} H e n c e, λ = \frac{12.26}{\sqrt{v}} \end{matrix}$
Hence, the option $B$ is the correct answer.

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De Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is

The correct option is B λ=12.26√VAAccording to De broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is λ=hp=h√2mevHence,λ=12.26√vHence, the option B is the correct answer.