Question

In photoelectric effect, initially when energy of electrons emitted is \( E_0 \), de-Broglie wavelength associated with them is \( \lambda_0 \). Now, energy is doubled then associated de-Broglie wavelength \( \lambda' \) is

• A. \( \lambda' = \frac{\lambda_0}{\sqrt{2}} \)
• B. \( \lambda' = \sqrt{2} \lambda_0 \)
• C. \( \lambda' = \lambda_0 \)
• D. \( \lambda' = \frac{\lambda_0}{2} \)

Hint:

In the photoelectric effect, the de-Broglie wavelength is inversely related to the momentum of the electron, which depends on energy.

Answer 1

The Correct Option is B

Step 1: Energy and Wavelength.
In the photoelectric effect, the de-Broglie wavelength \( \lambda \) of an electron is inversely proportional to its momentum, which depends on the energy. As the energy is doubled, the wavelength will increase by a factor of \( \sqrt{2} \).
Step 2: Explanation of Options.
(A) Incorrect: \( \lambda' \) would not be \( \frac{\lambda_0}{\sqrt{2}} \), as energy increases.
(B) Correct: Doubling the energy causes the de-Broglie wavelength to increase by a factor of \( \sqrt{2} \).
(C) Incorrect: The wavelength doesn't remain the same as energy changes.
(D) Incorrect: \( \lambda' \) is not half of \( \lambda_0 \).
Step 3: Conclusion.
The correct answer is (B), \( \lambda' = \sqrt{2} \lambda_0 \).