Question

If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with the electrons change?

• A. Wavelength is decreased to $\dfrac{1}{3}$ times.
• B. Wavelength is increased to $\dfrac{1}{2}$ times.
• C. Wavelength is increased to $\dfrac{1}{\sqrt{2}}$ times.
• D. Wavelength is decreased to $\dfrac{1}{\sqrt{2}}$ times.

Hint:

Higher accelerating voltage increases momentum and hence decreases de-Broglie wavelength.

Answer 1

The Correct Option is D

Step 1: de-Broglie wavelength formula.
For an electron accelerated through potential difference $V$:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]
Step 2: Relation between wavelength and potential difference.
\[ \lambda \propto \frac{1}{\sqrt{V}} \]
Step 3: Effect of doubling the potential difference.
If $V$ is doubled:
\[ \lambda' = \frac{\lambda}{\sqrt{2}} \]
Step 4: Conclusion.
The de-Broglie wavelength decreases to $\dfrac{1}{\sqrt{2}}$ times its original value.