Calculate the de Broglie wavelength of an ele | vedclass

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$.Given:Kinetic energy $K = 200 \ eV = 200 	imes 1.6 \

Vedclass · Accepted Answer

$8.25 	imes 10^{-11} \ m$

Vedclass · Answer

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$.Given:Kinetic energy $K = 200 \ eV = 200 	imes 1.6 	imes 10^{-19} \ J = 3.2 	imes 10^{-17} \ J$.Mass of electron $m = 1 	imes 10^{-30} \ kg$.Planck's constant $h = 6.6 	imes 10^{-34} \ Js$.Substituting the values:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes (1 	imes 10^{-30}) 	imes (3.2 	imes 10^{-17})}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{6.4 	imes 10^{-47}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{64 	imes 10^{-48}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{8 	imes 10^{-24}}$$\lambda = 0.825 	imes 10^{-10} \ m = 8.25 	imes 10^{-11} \ m$.Therefore,the correct option is $B$.

Calculate the de Broglie wavelength of an electron with a kinetic energy of $200 \ eV$. [Mass of electron $= 1 \times 10^{-30} \ kg$,charge on electron $= 1.6 \times 10^{-19} \ C$,Planck's constant $(h) = 6.6 \times 10^{-34} \ Js$].

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