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Question

(A) 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

$10^{4}$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A}$.Given $\lambda = 1.227 	imes 10^{-2} \, nm$.Since $1 \, nm = 10 \, \mathring{A}$,we have $\lambda = 1.227 	imes 10^{-2} 	imes 10 \, \mathring{A} = 0.1227 \, \mathring{A}$.Substituting this into the formula: $0.1227 = \frac{12.27}{\sqrt{V}}$.Rearranging for $\sqrt{V}$: $\sqrt{V} = \frac{12.27}{0.1227} = 100 = 10^{2}$.Squaring both sides: $V = (10^{2})^{2} = 10^{4} \, V$.

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 $......V$.

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