Find the de Broglie wavelength of an electron | vedclass

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Question

(A) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{3mk_BT}}$Given values are:Mass of electron,$m = 9.11 	imes

Vedclass · Accepted Answer

$6.2 	imes 10^{-9} \ m$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{3mk_BT}}$Given values are:Mass of electron,$m = 9.11 	imes 10^{-31} \ kg$Boltzmann constant,$k_B = 1.38 	imes 10^{-23} \ J/K$Temperature,$T = 27 + 273 = 300 \ K$Planck's constant,$h = 6.63 	imes 10^{-34} \ J \cdot s$Substituting these values into the formula:$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{3 	imes (9.11 	imes 10^{-31}) 	imes (1.38 	imes 10^{-23}) 	imes 300}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{11293.62 	imes 10^{-54}}}$$\lambda = \frac{6.63 	imes 10^{-34}}{1.0627 	imes 10^{-25}}$$\lambda \approx 6.24 	imes 10^{-9} \ m$Thus,the correct option is $A$.

Find the de Broglie wavelength of an electron in a metal at $27^{\circ}C$.

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