The wavelength (in m) of a particle of mass 1 | vedclass

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Question

(A) According to the de Broglie equation,the wavelength $(\lambda)$ is given by $\lambda = \frac{h}{mv}$.Given:Mass $(m)$ = $1.67 	imes 10^{-27} \ kg

Vedclass · Accepted Answer

$1 	imes 10^{-13}$

Vedclass · Answer

(A) According to the de Broglie equation,the wavelength $(\lambda)$ is given by $\lambda = \frac{h}{mv}$.Given:Mass $(m)$ = $1.67 	imes 10^{-27} \ kg$Velocity $(v)$ = $3.97 	imes 10^6 \ m \ s^{-1}$Planck's constant $(h)$ = $6.626 	imes 10^{-34} \ J \ s$Substituting the values:$\lambda = \frac{6.626 	imes 10^{-34}}{1.67 	imes 10^{-27} 	imes 3.97 	imes 10^6}$$\lambda = \frac{6.626 	imes 10^{-34}}{6.63 	imes 10^{-21}}$$\lambda \approx 1 	imes 10^{-13} \ m$.

The wavelength (in $m$) of a particle of mass $1.67 \times 10^{-27} \ kg$ moving with velocity of $3.97 \times 10^6 \ m \ s^{-1}$ is:

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