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(A) The momentum of a photon is given by $p = \frac{h}{\lambda}$.For an electron,the momentum is $p = mv$.Equating the two,we get $mv = \frac{h}{\lamb

Vedclass · Accepted Answer

$1098$

Vedclass · Answer

(A) The momentum of a photon is given by $p = \frac{h}{\lambda}$.For an electron,the momentum is $p = mv$.Equating the two,we get $mv = \frac{h}{\lambda}$,which implies $v = \frac{h}{m \lambda}$.Given:$h = 6.63 	imes 10^{-34} \ J \cdot s$$m = 9.1 	imes 10^{-31} \ kg$$\lambda = 663 \ nm = 663 	imes 10^{-9} \ m = 6.63 	imes 10^{-7} \ m$Substituting the values:$v = \frac{6.63 	imes 10^{-34}}{(9.1 	imes 10^{-31}) 	imes (6.63 	imes 10^{-7})}$$v = \frac{10^{-34}}{9.1 	imes 10^{-31} 	imes 10^{-7}} = \frac{10^{-34}}{9.1 	imes 10^{-38}}$$v = \frac{10^4}{9.1} \approx 1098.9 \ m/s$.Thus,the velocity is approximately $1098 \ m/s$.

With what velocity must an electron travel,so that its momentum is equal to that of a photon of wavelength $663 \ nm$ (in $m/s$)?

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