Question

The momentum of a particle having a de-Broglie wavelength of $0.1\mathring{\text{A}}$ is (approximately),

• A. $6.63\times {10}^{-23}{\text{kg m sec}}^{-1}$
• B. $6.63\times {10}^{-22}{\text{kg m sec}}^{-1}$
• C. $6.63\times {10}^{-20}{\text{kg m sec}}^{-1}$
• D. $6.63\times {10}^{-24}{\text{kg m sec}}^{-1}$

Answer 1

The correct option is A $6.63\times {10}^{-23}{\text{kg m sec}}^{-1}$

The de-Broglie wavelength is given by,
$\lambda =\frac{h}{p}$
$\Rightarrow p=\frac{h}{\lambda }$
$\therefore p=\frac{6.63\times {10}^{-34}}{{10}^{-11}}=6.63\times {10}^{-23}{\text{kg m sec}}^{-1}$

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

Key concept: Particle travelling with certain velocity behaves as both wave and particle and de-Broglie wavelength formula can be used to calculate the same. $\begin{array}{c}\text{Why this question?}\\ \text{Simple application of de-Broglie equation.}\end{array}$