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(B) The de Broglie wavelength $\lambda$ for 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 $\lambda$ for an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \; \mathring{A}$.Given $V = 10,000 \; V = 10^4 \; V$.Substituting the value of $V$ in the formula:$\lambda = \frac{12.27}{\sqrt{10^4}} \; \mathring{A} = \frac{12.27}{100} \; \mathring{A} = 0.1227 \; \mathring{A}$.Since $1 \; \mathring{A} = 10^{-10} \; m$,we have $\lambda = 0.1227 	imes 10^{-10} \; m = 12.27 	imes 10^{-12} \; m$.Rounding to the nearest option,the value is approximately $12.2 	imes 10^{-12} \; m$.

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