Question

An electron is accelerated through a potential difference of $500\text{V}$. What is its de-Broglie wavelength?

• A. $0.55\mathring{\text{A}}$
• B. $0.055\mathring{\text{A}}$
• C. $0.25\mathring{\text{A}}$
• D. $0.025\mathring{\text{A}}$

Answer 1

The correct option is A $0.55\mathring{\text{A}}$

We know that,
$\lambda =\frac{h}{\sqrt{2MeV}}=\frac{12.27}{\sqrt{V}}\left(\text{in}\mathring{\text{A}}\right)$

$\therefore \lambda =\frac{12.27}{\sqrt{500}}=0.55\mathring{\text{A}}$

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

Key concept : Take $\frac{h}{\sqrt{2Me}}=1.227\times {10}^{-9}$ SI units $\therefore$ The formula of wavelength simplifies to, $\lambda =\frac{12.27}{\sqrt{V}}\left(\text{in}\mathring{\text{A}}\right)$