Q. The wavelength of de-Broglie wave is 2$\mu$m, then its momentum is $(h = 6.63 \times \ \ 10^{- 34}\ Js)$
A) $3.315 \times 10^{- 28}\ \ kg - {ms}^{- 1}$
B) $1.66 \times 10^{- 28}\ kg - {ms}^{- 1}$
C) $4.97 \times 10^{- 28}\ kg - {ms}^{- 1}$
D) $9.9 \times 10^{- 28}\ kg - {ms}^{- 1}$
Explanation
(a) According to de-Broglie hypothesis
$\lambda = \frac{h}{p}$
Where, $h$ is Planck’s constant.
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p = \frac{h}{\lambda}$
Given, $h = 6.63 \times 10^{- 34}$ J-s
$\lambda = 2\mu m = 2 \times 10^{- 6}\ m$
$\therefore\ \ \ \ \ \ \ \ \ \ \ p = \frac{6.63 \times 10^{- 34}}{2 \times 10^{- 6}}$
=3.315$\times 10^{- 28}$ kg-${ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.