Question

Calculate the momentum of a particle which has a de Broglie wavelength of $2.25\times {10}^{-10}m$.
$(h=6.6\times {10}^{-34}kg{m}^2{s}^{-1})$

• A. $1.24\times {10}^{-24}kgm{s}^{-1}$.
• B. $3.24\times {10}^{-24}kgm{s}^{-1}$.
• C. $5.64\times {10}^{-24}kgm{s}^{-1}$.
• D. $2.93\times {10}^{-24}kgm{s}^{-1}$.

Answer 1

The correct option is D $2.93\times {10}^{-24}kgm{s}^{-1}$.

Given:
$\text{de Broglie wavelength}\left(\lambda \right)=2.25\times {10}^{-10}m$
$\text{Planck's constant}\left(h\right)=6.6\times {10}^{-34}Js$
Let 'p' be the momentum of the particle.
We know that de Broglie wavelength is given by
$\lambda =\frac{h}{p}$
$\therefore p=\frac{h}{\lambda }=\frac{6.6\times {10}^{-34}}{2.25\times {10}^{-10}}=2.93\times {10}^{-24}kgm{s}^{-1}$