Question

If the de Broglie wavelength associated with an electron is $0.1227$ nm, then its accelerating potential is

• A. $64$ V
• B. $200$ V
• C. $100$ V
• D. $160$ V
• E. $36$ V

Hint:

For electrons: - $\lambda(\text{nm}) = \frac{1.227}{\sqrt{V}}$ - Easy shortcut for quick calculations

Answer 1

The Correct Option is C

Concept: For an electron: \[ \lambda = \frac{h}{\sqrt{2meV}} \] Simplified relation: \[ \lambda(\text{nm}) = \frac{1.227}{\sqrt{V}} \]

Step 1: Substitute given wavelength.
\[ 0.1227 = \frac{1.227}{\sqrt{V}} \]

Step 2: Solve for $V$.
\[ \sqrt{V} = \frac{1.227}{0.1227} = 10 \] \[ V = 100\ \text{V} \]