Question

If de Broglie wavelength associated with an electron is $3\times {10}^{-10}\text{m}$, then the momentum of the electron will be -

• A. $2.2\times {10}^{-22}{\text{kg ms}}^{-1}$
• B. $2.2\times {10}^{-23}{\text{kg ms}}^{-1}$
• C. $2.2\times {10}^{-24}{\text{kg ms}}^{-1}$
• D. $2.2\times {10}^{-25}{\text{kg ms}}^{-1}$

Answer 1

The correct option is C $2.2\times {10}^{-24}{\text{kg ms}}^{-1}$

Given:

$\lambda =3\times {10}^{-10}\text{m}$

We know that, momentum,

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

$\Rightarrow p=\frac{6.6\times {10}^{-34}}{3\times {10}^{-10}}$

$\Rightarrow p=2.2\times {10}^{-24}{\text{kg ms}}^{-1}$

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