The wavelength of an electron of kinetic ener | vedclass

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Question

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.Substituting the given values:$\lambda = \frac{6

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.Substituting the given values:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes (9 	imes 10^{-31} \ kg) 	imes (4.50 	imes 10^{-29} \ J)}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{81 	imes 10^{-60}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{9 	imes 10^{-30}}$$\lambda = 0.733 	imes 10^{-4} \ m = 7.33 	imes 10^{-5} \ m$.The nearest integer is $7$.

The wavelength of an electron of kinetic energy $4.50 \times 10^{-29} \ J$ is $... \times 10^{-5} \ m$. (Nearest integer) Given: mass of electron is $9 \times 10^{-31} \ kg$,$h = 6.6 \times 10^{-34} \ J \ s$.

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