Question

If \( \lambda_0 \) and \( \lambda \) are respectively the threshold wavelength and wavelength of incident light, the velocity of photoelectrons ejected from the metal surface is:

• A. \( \sqrt{\frac{2h}{m} (\lambda_0 - \lambda)} \)
• B. \( \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right)} \)
• C. \( \sqrt{\frac{2hc}{m} (\lambda_0 - \lambda)} \)
• D. \( \sqrt{\frac{2h}{m} \left( \frac{1}{\lambda_0} - \frac{1}{\lambda} \right)} \)

Hint:

The velocity of emitted photoelectrons is given by \( v = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right)} \).

Answer 1

The Correct Option is B

Step 1: Apply Einstein’s Photoelectric Equation \[ \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \] Step 2: Compute Electron Velocity \[ v = \sqrt{\frac{2}{m} \left( \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \right)} \] \[ = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0 \lambda} \right)} \] Thus, the correct answer is \( \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right)} \).