The equation lambda = frac{1.227}{x} text{ nm | vedclass

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.For an electron a

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$\sqrt{V}$

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.For an electron accelerated through a potential $V$,the kinetic energy $K = eV$,where $e$ is the charge of the electron.Thus,$\lambda = \frac{h}{\sqrt{2meV}}$.Substituting the values of Planck's constant $h = 6.626 	imes 10^{-34} 	ext{ J s}$,mass of electron $m = 9.11 	imes 10^{-31} 	ext{ kg}$,and charge $e = 1.602 	imes 10^{-19} 	ext{ C}$:$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 9.11 	imes 10^{-31} 	imes 1.602 	imes 10^{-19} 	imes V}} 	ext{ m}$.Calculating this gives $\lambda \approx \frac{1.227 	imes 10^{-9}}{\sqrt{V}} 	ext{ m} = \frac{1.227}{\sqrt{V}} 	ext{ nm}$.Comparing this with $\lambda = \frac{1.227}{x} 	ext{ nm}$,we find $x = \sqrt{V}$.

The equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:

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