What will be the de-Broglie wavelength of an | vedclass

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Question

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} \ J \ s$$m = 9

Vedclass · Accepted Answer

$6.068 	imes 10^{-9} \ m$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} \ J \ s$$m = 9.1 	imes 10^{-31} \ kg$$v = 1.2 	imes 10^5 \ m \ s^{-1}$Substituting the values:$\lambda = \frac{6.626 	imes 10^{-34}}{9.1 	imes 10^{-31} 	imes 1.2 	imes 10^5}$$\lambda = \frac{6.626 	imes 10^{-34}}{10.92 	imes 10^{-26}}$$\lambda \approx 0.6068 	imes 10^{-8} \ m = 6.068 	imes 10^{-9} \ m$.

What will be the de-Broglie wavelength of an electron moving with a velocity of $1.2 \times 10^5 \ m \ s^{-1}$?

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