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 \( \frac{1}{\sqrt{2}} \).

Answer 2

Step 1: de-Broglie Wavelength Formula
The de-Broglie wavelength \( \lambda \) of a particle is given by: \[ \lambda = \frac{h}{p} \] Where:

• \( h \) is Planck’s constant
• \( p \) is the momentum of the particle

For a particle like an electron, momentum is related to kinetic energy \( K \) by: \[ p = \sqrt{2mK} \] So, \[ \lambda = \frac{h}{\sqrt{2mK}} \]

 

Step 2: Effect of Doubling Kinetic Energy
Let the initial kinetic energy be \( K \), and the new kinetic energy be \( 2K \). Then:

• Initial wavelength: \[ \lambda_1 = \frac{h}{\sqrt{2mK}} \]
• New wavelength: \[ \lambda_2 = \frac{h}{\sqrt{2m(2K)}} = \frac{h}{\sqrt{4mK}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mK}} = \frac{\lambda_1}{\sqrt{2}} \]

 

Step 3: Final Answer
If the kinetic energy of the electron doubles, its de-Broglie wavelength becomes: \[ \frac{1}{\sqrt{2}} \text{ times the original} \]

Correct Option: Option 4 — \( \frac{1}{\sqrt{2}} \)