Question

The momentum of a particle with de-Broglie wavelength $1\stackrel{\circ }{A}$ is: $(h=6.62\times {10}^{-34}Js)$

• A. $6.62\times {10}^{-24}kgm{s}^{-1}$
• B. $6.62\times {10}^{-24}kgcm{s}^{-1}$
• C. $9.62\times {10}^{-24}kgm{s}^{-1}$
• D. $6.62\times {10}^{-22}kgm{s}^{-1}$

Answer 1

The correct option is A $6.62\times {10}^{-24}kgm{s}^{-1}$

Given:
$\text{de Broglie wavelength}\left(\lambda \right)=1Å={10}^{-10}m$
$\text{Plank's constant}\left(h\right)=6.62\times {10}^{-34}Js$
Let 'p' be the momentum of the particle.
We know that de Broglie wavelength is given by
$\lambda =\frac{h}{p}$
$\therefore p=\frac{h}{\lambda }=\frac{6.62\times {10}^{-34}}{{10}^{-10}}=6.62\times {10}^{-24}kgm{s}^{-1}$.