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: {Using de-Broglie’s Equation}
The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{p} \] Since momentum \( p \) is related to kinetic energy: \[ p = \sqrt{2mK} \] Step 2: {Effect of Doubling Kinetic Energy}
If \( K \) is doubled: \[ p' = \sqrt{2m (2K)} = \sqrt{2} p \] Since \( \lambda \propto \frac{1}{p} \), we get: \[ \lambda' = \frac{\lambda}{\sqrt{2}} \] Thus, the correct answer is \( \frac{1}{\sqrt{2}} \).