Question

If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor: 
 

• A. 2
• B. \( \frac{1}{2} \)
• C. \(\sqrt{2} \)
• D. \( \frac{1}{\sqrt{2}} \)

Hint:

The de-Broglie wavelength decreases as the kinetic energy increases, following the inverse square root relation.

Answer 1

The Correct Option is D

Step 1: {Applying de-Broglie's Equation}
De-Broglie's wavelength is calculated using:\[\lambda = \frac{h}{p}\]Momentum \( p \) is linked to kinetic energy by:\[p = \sqrt{2mK}\]Step 2: {Impact of Doubled Kinetic Energy}
When \( K \) is doubled:\[p' = \sqrt{2m (2K)} = \sqrt{2} p\]Given that \( \lambda \propto \frac{1}{p} \), the new wavelength is:\[\lambda' = \frac{\lambda}{\sqrt{2}}\]Therefore, the correct value is \( \frac{1}{\sqrt{2}} \).