Question

If \( \lambda \) is the incident wavelength and \( \lambda_0 \) is the threshold wavelength for a metal surface, photoelectric effect takes place only if:

• A. \( \lambda \le \lambda_0 \)
• B. \( \lambda \ge \lambda_0 \)
• C. \( \lambda \ge 2\lambda_0 \)
• D. None of these

Hint:

Shorter wavelength (higher frequency) is needed for photoelectric emission.

Answer 1

The Correct Option is A

Concept: Photoelectric effect condition: \[ \nu \ge \nu_0 \] Since \( \nu = \frac{c}{\lambda} \), this implies: \[ \lambda \le \lambda_0 \]

Step 1: Relation between wavelength and frequency.
Higher frequency \(\Rightarrow\) lower wavelength.

Step 2: Conclusion.
\[ \therefore \lambda \le \lambda_0 \]