Question

Calculate the momentum of particle whose de Broglie wavelength is $2^{o}A$.

• A. $3.313\times {10}^{-24}gm{s}^{-1}$
• B. $3.313\times {10}^{-24}kgm{s}^{-1}$
• C. $33.13\times {10}^{-20}kgm{s}^{-1}$
• D. none of these

Answer 1

The correct option is B $3.313\times {10}^{-24}kgm{s}^{-1}$

De-broglie wavelength $\left(\lambda \right)$ = $\frac{h}{mv}=\frac{h}{p}$

Given , $\left(\lambda \right)=2\times {10}^{-10}m$

$p=\frac{h}{\lambda }$ where h $=6.626$ x ${10}^{-34}$ Js.

$p=\frac{6.626x{10}^{-34}}{2x{10}^{-10}}kgm/s$

$p=3.313x{10}^{-24}kgm/s$

Hence, the momentum of a particle whose de-broglie wavelength is $3.313\times {10}^{-24}kgm/s$