The graph shows the variation of photocurrent with anode potential for four different radiations. If $I$ denotes intensity and $f$ denotes frequency, which of the following is correct?
Show Hint
- $f_{b}>f_{a}, f_{b}=f_{d}, I_{c}=I_{d}$
- $f_{b}=f_{a}, f_{b}>f_{c}, I_{c}>I_{d}$
- $f_{b}<f_{a}, f_{b}<f_{c}, I_{c}<I_{d}$
- $f_{b}<f_{a}, f_{b}>f_{c}, I_{c}=I_{d}$
The Correct Option is A
Solution and Explanation
- Stopping potential depends on frequency ($V_s \propto f$). - Saturation current depends on intensity ($I_{photo} \propto I$).
Step 2: Analysis
- Curves 'b' and 'd' meet at the same stopping potential, so $f_b = f_d$. - Curve 'a' has a lower stopping potential than 'b', so $f_b > f_a$. - Curves 'c' and 'd' have the same saturation current, so $I_c = I_d$.
Step 3: Conclusion
This matching logic corresponds to option (A). Final Answer: (A)
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