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(D) The de Broglie wavelength of an electron is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.Since the electron is accelerated from rest,$KE = qV$.Thu

Vedclass · Accepted Answer

$6$

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(D) The de Broglie wavelength of an electron is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.Since the electron is accelerated from rest,$KE = qV$.Thus,$\lambda = \frac{h}{\sqrt{2mqV}}$.Substituting the given values:$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 1.6 	imes 10^{-19} 	imes 40 	imes 10^{3}}}$.$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{116.48 	imes 10^{-47}}} = \frac{6.63 	imes 10^{-34}}{\sqrt{11.648 	imes 10^{-46}}}$.$\lambda = \frac{6.63 	imes 10^{-34}}{3.413 	imes 10^{-23}} \approx 1.94 	imes 10^{-11} \, m$ (Wait,re-calculating).Using the shortcut formula: $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A} = \frac{12.27}{\sqrt{40000}} \, \mathring{A} = \frac{12.27}{200} \, \mathring{A} = 0.06135 \, \mathring{A}$.Converting to meters: $0.06135 	imes 10^{-10} \, m = 6.135 	imes 10^{-12} \, m$.Rounding to the nearest integer,$X = 6$.

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