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:

For a free electron, doubling the kinetic energy results in the de-Broglie wavelength being reduced by a factor of \( \sqrt{2} \).

Answer 1

The Correct Option is C

Step 1: de-Broglie wavelength formula.
The de-Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck’s constant and \( p \) is the momentum of the electron. Step 2: Relationship between kinetic energy and momentum.
The kinetic energy \( K \) of the electron is related to its momentum \( p \) by: \[ K = \frac{p^2}{2m} \] where \( m \) is the mass of the electron. If the kinetic energy doubles, the momentum increases by a factor of \( \sqrt{2} \). Therefore, the wavelength decreases by a factor of \( \sqrt{2} \).
Final Answer: \[ \boxed{\sqrt{2}} \]