Question

Calculate the momentum of a particle which has a de Broglie wavelength of $4.5\times {10}^{-10}m$.
$(\text{Planck's constant},h=6.6\times {10}^{-34}kg{m}^2{s}^{-1})$

• A. $1.24\times {10}^{-24}kgm{s}^{-1}$
• B. $3.24\times {10}^{-24}kgm{s}^{-1}$
• C. $1.46\times {10}^{-24}kgm{s}^{-1}$
• D. $2.93\times {10}^{-24}kgm{s}^{-1}$

Answer 1

The correct option is C $1.46\times {10}^{-24}kgm{s}^{-1}$

Given:
$\text{de Broglie wavelength}\left(\lambda \right)=4.5\times {10}^{-10}m$
$\text{Planck's constant}\left(h\right)=6.6\times {10}^{-34}Js$
Let 'p' be the momentum of the particle.
We know that de Broglie wavelength is given by,
$\lambda =\frac{h}{p}$
$\therefore p=\frac{h}{\lambda }=\frac{6.6\times {10}^{-34}}{4.5\times {10}^{-10}}=1.46\times {10}^{-24}kgm{s}^{-1}$