Question

The momentum of a particle with de Broglie wavelength $2\stackrel{\circ }{A}$ is:
$(\text{Planck's constant},h=6.62\times {10}^{-34}Js)$

• A. $3.31\times {10}^{-24}kgm{s}^{-1}$
• B. $3.31\times {10}^{-24}kgcm{s}^{-1}$
• 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 $3.31\times {10}^{-24}kgm{s}^{-1}$

Given:
$\text{de Broglie wavelength}\left(\lambda \right)=2Å=2\times {10}^{-10}m$
$\text{Planck's constant}\left(h\right)=6.62\times {10}^{-34}Js$

Let 'p' be the momentum of the particle.
We know that de Broglie wavelength is given by,
$\lambda =\frac{h}{p}$
$\therefore p=\frac{h}{\lambda }=\frac{6.62\times {10}^{-34}}{2\times {10}^{-10}}=3.31\times {10}^{-24}kgm{s}^{-1}$