Question

The momentum of particle whose de Broglie wavelength is $2^{o}A$ is:

• A. $6.6266\times {10}^{-34}kgm{s}^{-1}$
• B. $3.313\times {10}^{-34}kgm{s}^{-1}$
• C. $3.313\times {10}^{-24}kgm{s}^{-1}$
• D. $1.656\times {10}^{-24}kgm{s}^{-1}$

Answer 1

Answer: (C)

$3.313\times {10}^{-24}kgm{s}^{-1}$

The formula for de-Broglie wavelength = $\lambda =\frac{h}{p}$

where $h=6.6\times {10}^{-34}kg{m}^2{s}^{-1}$ and p = momentum.

Given that $\lambda =2A^{\circ }=2\times {10}^{-10}m$

Therefore p = $\frac{6.6\times {10}^{-34}kg{m}^2{s}^{-1}}{2\times {10}^{-10}m}=3.313\times {10}^{-24}kgm{s}^{-1}$