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(D) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{12.27}{

Vedclass · Accepted Answer

$0.04$

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(D) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{12.27}{\sqrt{V}} \ \mathring{A} = \frac{12.27 	imes 10^{-10}}{\sqrt{V}} \ 	ext{m}$.Given $V = 900 \ V$.Substituting the value of $V$ in the formula:$\lambda = \frac{12.27 	imes 10^{-10}}{\sqrt{900}} \ 	ext{m}$.$\lambda = \frac{12.27 	imes 10^{-10}}{30} \ 	ext{m}$.$\lambda = 0.409 	imes 10^{-10} \ 	ext{m}$.$\lambda \approx 0.04 	imes 10^{-9} \ 	ext{m} = 0.04 \ 	ext{nm}$.

The de Broglie wavelength of an electron accelerated between two plates having a potential difference of $900 \ V$ is nearly. (in $nm$)

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