Q. A photon of wavelength $4400\ \mathring{\mathrm{A}}$ is passing through vacuum. The effective mass and momentum of the photon are respectively
A) $5 \times 10^{- 36}kg,1.5 \times 10^{- 27}kg$-$m/s$
B) $5 \times 10^{- 35}kg,1.5 \times 10^{- 26}kg$-$m/s$
C) Zero,$\ 1.5 \times 10^{- 26}kg$-$m/s$
D) $5 \times 10^{- 36}kg,1.67 \times 10^{- 43}kg$-$m/s$
Explanation
(a) $p = \frac{h}{\lambda} = \frac{6.6 \times 10^{- 34}}{4400 \times 10^{- 10}} = 1.5 \times 10^{- 27}kg.m/s$
and mass $m = \frac{p}{c} = \frac{1.5 \times 10^{- 27}}{3 \times 10^{8}} = 5 \times 10^{- 36}kg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.