Question

In photoelectric effect, if wavelength changes from \(\lambda\) to \(3\lambda/4\), how does max speed change?

• A. \((3/4)^{1/2}\)
• B. \((4/3)^{1/2}\)
• C. less than \((4/3)^{1/2}\)
• D. greater than \((4/3)^{1/2}\)

Hint:

In photoelectric effect, subtracting work function makes increase non-linear.

Answer 1

The Correct Option is D

Understanding the Concept: \[ K_{\max} = \frac{hc}{\lambda} - \phi,\quad v \propto \sqrt{K} \]

Step 1: New wavelength smaller $\Rightarrow$ energy increases \[ \frac{1}{\lambda'} = \frac{4}{3\lambda} \Rightarrow \text{energy increases more than } \frac{4}{3} \]

Step 2: Since \(\phi\) constant, increase in KE is more than linear \[ v>\sqrt{\frac{4}{3}} \]