The momentum of a photon of wavelength lambda | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

$A$ parallel beam of light of wavelength $900 \, nm$ and intensity $100 \, W/m^2$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \, cm^2$ area perpendicular to the beam in one second is:

The momentum of a photon with a wavelength of $5000\,\mathring{A}$ is:

Find the number of photons emitted per second by a $10 \ kW$ radio transmitter operating at a wavelength of $500 \ m$.

$A$ radio station broadcasts at a wavelength of $300 \ m$. If the radiating power of the transmitter is $10 \ kW$,then the number of photons emitted per second is:

If $E$ and $p$ are the energy and momentum of a photon respectively,what happens if the wavelength of the photon is decreased?

Vedclass Test Series

Exam Paper Generator

Online Exam Module