The momentum of a photon of wavelength lambda | vedclass

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Question

(A) The energy of a photon is given by $E = h
u$,where $h$ is Planck's constant and $
u$ is the frequency.According to Einstein's mass-energy equiva

Vedclass · Accepted Answer

$\frac{h}{\lambda}$

Vedclass · Answer

(A) The energy of a photon is given by $E = h
u$,where $h$ is Planck's constant and $
u$ is the frequency.According to Einstein's mass-energy equivalence,$E = mc^2$,where $m$ is the relativistic mass and $c$ is the speed of light.Equating the two expressions for energy: $mc^2 = h
u$.Since the momentum $p$ of a photon is defined as $p = mc$,we can substitute $m = \frac{p}{c}$ into the equation: $p = \frac{h
u}{c}$.Using the relation $
u = \frac{c}{\lambda}$,we substitute for $
u$: $p = \frac{h(c/\lambda)}{c} = \frac{h}{\lambda}$.Therefore,the momentum of a photon is $\frac{h}{\lambda}$.

The momentum of a photon of wavelength $\lambda$ is

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