The momentum of a photon of light of frequency $f$ is . . . . . . .

  • A
    $\frac{h c}{f}$
  • B
    $\frac{h f}{c}$
  • C
    $\frac{h}{c f}$
  • D
    $h c f$

Explore More

Similar Questions

$A$ scientist claims to have a perfect technique in which he can spontaneously convert an electron completely into energy in the laboratory without any other material required. What is the conclusion about this claim from our current understanding of physics?

Energy of a photon whose frequency is $10^{12} \text{ MHz}$ is
[ Planck's constant,$h = 6.63 \times 10^{-34} \text{ Js}, e = 1.6 \times 10^{-19} \text{ C}$ ]

The eye can detect $5 \times 10^4$ photons per square metre per second of green light $(\lambda = 5000 \ \mathring{A})$ while the ear can detect $10^{-13} \ W/m^2$. The factor by which the eye is more sensitive as a power detector than the ear is close to:

Difficult
View Solution

Two sources of light emit $X$-rays of wavelength $1 \ nm$ and visible light of wavelength $500 \ nm$,respectively. Both sources emit light of the same power $200 \ W$. The ratio of the number of photons emitted per second by the $X$-ray source to the number of photons emitted per second by the visible light source is:

$A$ $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu m$. Assuming it to be $25\%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo