The curves `(a) , (b) , ( c )` and `(d)` show the variation between the applied potential difference `(V)` and the photoelectric current `(i)` , at two different intensities of light `(I_(1) gt I_(2))`. In which figure is the correct variation shown ?

Options

A.
B.
C.
D.

Answer

To determine the correct variation between the applied potential difference (V) and the photoelectric current (i) for two different intensities of light (I1 > I2), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**: The photoelectric effect states that when light of sufficient frequency (or energy) shines on a metal surface, it can eject electrons from that surface. The energy of the incident photons is given by \( E = h \nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the light.

2. **Energy Equation**: The energy of the incident photon can be expressed as:
\[ E = \text{Work Function} + \text{Kinetic Energy of ejected electrons} \]
This can be rewritten as:
\[ h \nu = \phi + \frac{1}{2} mv^2 \]
where \( \phi \) is the work function of the metal.

3. **Stopping Potential**: The stopping potential \( V \) can be derived from the energy conservation equation:
\[ eV = \frac{1}{2} mv^2 \]
This implies that the stopping potential is related to the maximum kinetic energy of the emitted electrons.

4. **Cut-off Voltage**: The cut-off voltage \( V_0 \) can be expressed as:
\[ V_0 = \frac{h}{e} \nu - \frac{h}{e} \nu_0 \]
where \( \nu_0 \) is the threshold frequency. This shows that \( V_0 \) depends on the frequency of the incident light, not on its intensity.

5. **Effect of Intensity**: The intensity of light affects the number of photons incident on the surface. A higher intensity (I1 > I2) means more photons are hitting the surface, which results in more emitted electrons and thus a higher photoelectric current \( i \).

6. **Current vs. Potential Difference**: Since the stopping potential \( V \) is independent of intensity, both intensities will have the same cut-off voltage. However, the photoelectric current will be higher for the higher intensity (I1) than for the lower intensity (I2).

7. **Analyzing the Graphs**: Given the above understanding, we can analyze the provided curves:
- The correct graph should show that both intensities (I1 and I2) have the same cut-off voltage (V) but different currents (i), with the current for I1 being greater than that for I2.

8. **Conclusion**: Based on the analysis, the correct variation is shown in figure (b), where both curves intersect at the same potential difference but have different photoelectric currents.