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
Frequency determines where the graph starts on the negative V-axis; Intensity determines the height of the plateau.
Step 1: Understanding the Question:
The graph relates photocurrent (determined by intensity) and stopping potential (determined by frequency) for four different light sources. Step 2: Key Formula or Approach:
1. Saturation photocurrent \( \propto \) Intensity \( I \).
2. Stopping potential \( V_0 \propto \) Frequency \( f \). (Magnitude wise, higher frequency light requires more negative potential to stop). Step 3: Detailed Explanation:
- Analyzing Intensities: Curves 'c' and 'd' reach the same higher saturation current level, so \( I_c = I_d \). Curves 'a' and 'b' reach a lower saturation level, so \( I_a = I_b \).
- Analyzing Frequencies: Curves 'b' and 'd' start from the same stopping potential (intercept on the negative x-axis), so \( f_b = f_d \). Curves 'a' and 'c' start from another common stopping potential closer to zero.
- Since the stopping potential for 'b' and 'd' is more negative (higher magnitude) than for 'a' and 'c', it follows that \( f_b = f_d > f_a = f_c \).
- Comparing specific options, \( f_b > f_a \) and \( f_b = f_d \) and \( I_c = I_d \) correctly describes the observations. Step 4: Final Answer:
The correct relationship is \( f_b > f_a, f_b = f_d, I_c = I_d \).
