Question:
In photoelectric effect, if wavelength changes from \(\lambda\) to \(3\lambda/4\), how does max speed change?
In photoelectric effect, if wavelength changes from \(\lambda\) to \(3\lambda/4\), how does max speed change?
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In photoelectric effect, subtracting work function makes increase non-linear.
Updated On: Apr 23, 2026
- \((3/4)^{1/2}\)
- \((4/3)^{1/2}\)
- less than \((4/3)^{1/2}\)
- greater than \((4/3)^{1/2}\)
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The Correct Option is D
Solution and Explanation
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 \[ v>\sqrt{\frac{4}{3}} \]
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 \[ v>\sqrt{\frac{4}{3}} \]
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