A photon of wavelength 4400 , mathring{A} is | vedclass

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Question

(A) The momentum $p$ of a photon is given by the formula $p = \frac{h}{\lambda}$.Given $h = 6.6 	imes 10^{-34} \, J \cdot s$ and $\lambda = 4400 	im

Vedclass · Accepted Answer

$5 	imes 10^{-36} \, kg, \, 1.5 	imes 10^{-27} \, kg \cdot m/s$

Vedclass · Answer

(A) The momentum $p$ of a photon is given by the formula $p = \frac{h}{\lambda}$.Given $h = 6.6 	imes 10^{-34} \, J \cdot s$ and $\lambda = 4400 	imes 10^{-10} \, m$.$p = \frac{6.6 	imes 10^{-34}}{4400 	imes 10^{-10}} = 1.5 	imes 10^{-27} \, kg \cdot m/s$.The effective mass $m$ of a photon is given by $m = \frac{p}{c}$,where $c = 3 	imes 10^8 \, m/s$.$m = \frac{1.5 	imes 10^{-27}}{3 	imes 10^8} = 5 	imes 10^{-36} \, kg$.Thus,the effective mass is $5 	imes 10^{-36} \, kg$ and the momentum is $1.5 	imes 10^{-27} \, kg \cdot m/s$.

$A$ photon of wavelength $4400 \, \mathring{A}$ is passing through vacuum. The effective mass and momentum of the photon are respectively

Similar Questions

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Assertion: The energy $(E)$ and momentum $(p)$ of a photon are related by $p = E/c$.
Reason: The photon behaves like a particle.

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