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Question

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

Vedclass · Accepted Answer

$1.23$

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mQV}}$.Substituting the values $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 9.11 	imes 10^{-31} \ kg$,$Q = 1.60 	imes 10^{-19} \ C$,and $V = 100 \ V$:$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.11 	imes 10^{-31} 	imes 1.60 	imes 10^{-19} 	imes 100}}$$\lambda \approx \frac{12.27}{\sqrt{V}} \ \mathring A$For $V = 100 \ V$,$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \ \mathring A \approx 1.23 \ \mathring A$.

The wavelength associated with an electron accelerated through a potential difference of $100 \ V$ is nearly .............. $ \mathring A $

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