Q. If the linear momentum of a particle is $2.2 \times 10^{4}\ kg$-$ms^{- 1}$, then what will be its de-Broglie wavelength? (Take $h = 6.6 \times 10^{- 34}\ Js)$
A) $3 \times 10^{- 29}$m
B) $3 \times 10^{- 29}$nm
C) $6 \times 10^{- 29}$m
D) $6 \times 10^{- 29}$nm
Explanation
(b) Given, the linear momentum of particle $(p)$
$= 2.2 \times 10^{4}kg - {ms}^{- 1}$
$\ h = 6.6 \times 10^{- 34}$ JS
The de-Broglie wavelength of particle
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{h}{p}$
$\lambda = \frac{6.6 \times 10^{- 34}}{2.2 \times 10^{4}}$
Or $\lambda = 3 \times 10^{- 38}$ m
Or $\lambda = 3 \times 10^{- 29}$ mm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.