Question

The wavelength of de-Broglie wave is $2\mu \text{m}$, then its momentum is, $h=6.63\times {10}^{-34}\text{J-s}$

• A. $3.315\times {10}^{-28}\text{kg-m/s}$
• B. $1.66\times {10}^{-28}\text{kg-m/s}$
• C. $4.97\times {10}^{-28}\text{kg-m/s}$
• D. $9.9\times {10}^{-28}\text{kg-m/s}$

Answer 1

The correct option is A $3.315\times {10}^{-28}\text{kg-m/s}$

The de Broglie wavelength is given by,

$\lambda =\frac{h}{p}$

$\Rightarrow p=\frac{h}{\lambda }=\frac{6.63\times {10}^{-34}}{2\times {10}^{-6}}$

$=3.315\times {10}^{-28}\text{kg-m/s}$

Hence, $\left(A\right)$ is the correct answer.