The cathode of a photoelectric cell is changed such that the work function changes from `(W_(1) to W_(2) (W_(2) gt W_(1))`. If the current before and after change are `I_(1)` and `I_(2)`, all other conditions remaining unchanged , then (assuming `hv gt W_(2)`)

Options

A. `I_(1)` = `I_(2)`
B. `I_(1) lt I_(2)`
C. `I_(1) gt I_(2)`
D. `I_(1) lt I_(2) lt 2I_(1)`

Answer

To solve the problem, we need to analyze how the change in work function affects the current in a photoelectric cell while keeping other conditions constant.

### Step-by-Step Solution:

1. **Understanding Work Function**: The work function (W) is the minimum energy required to remove an electron from the surface of a material. In this case, we have two work functions: W1 (before the change) and W2 (after the change), with W2 > W1.

2. **Photoelectric Effect Condition**: The condition given is that the energy of the incident photons (hv) is greater than W2. This means that the photons have enough energy to emit electrons from the cathode even after the work function has increased.

3. **Current in the Photoelectric Cell**: The current (I) in a photoelectric cell is directly proportional to the number of electrons emitted from the cathode. The number of emitted electrons depends on the intensity of the incident light, not on the work function or frequency, as long as the energy of the photons is sufficient to overcome the work function.

4. **Intensity and Current Relationship**: Since the intensity of the incident light is unchanged in this scenario, the number of emitted electrons remains constant. Therefore, the current before the change (I1) and after the change (I2) will be the same.

5. **Conclusion**: Since all other conditions remain unchanged and the intensity is constant, we conclude that:
\[ I_1 = I_2 \]

### Final Answer:
Thus, the current before and after the change in work function will be equal:
\[ I_1 = I_2 \]