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Question

(B) 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

$12.2 	imes 10^{-12} \ m$

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(B) 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 $V = 10,000 \ V$,we have $\sqrt{V} = \sqrt{10,000} = 100$.Substituting the values: $\lambda = \frac{12.27}{100} \ \mathring{A} = 0.1227 \ \mathring{A}$.Since $1 \ \mathring{A} = 10^{-10} \ m$,we have $\lambda = 0.1227 	imes 10^{-10} \ m = 1.227 	imes 10^{-11} \ m$.Rounding to the nearest option,we get $12.2 	imes 10^{-12} \ m$.

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

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