The anode vollage of a photocell is kept fixed . The wavelength `lambda `of the light falling on the cathode varies as follows

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze how the current in a photocell varies with the wavelength of light falling on the cathode. Here’s the step-by-step solution:

### Step 1: Understand the relationship between current and intensity
The current (I) in a photocell is directly proportional to the intensity (I) of the light falling on the cathode. This can be expressed as:
\[ I \propto I \]

### Step 2: Relate intensity to wavelength
The intensity of light is defined as the energy per unit time per unit area. The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the light.

### Step 3: Express intensity in terms of wavelength
If we consider the intensity (I) to be proportional to the energy, we can write:
\[ I \propto \frac{1}{\lambda} \]
This indicates that as the wavelength increases, the intensity decreases.

### Step 4: Relate current to wavelength
Since the current is directly proportional to the intensity, we can conclude:
\[ I \propto \frac{1}{\lambda} \]
This means that the current is inversely proportional to the wavelength of the light falling on the cathode.

### Step 5: Conclusion
From our analysis, we find that:
- If the wavelength (\( \lambda \)) increases, the current (I) decreases.
- Conversely, if the wavelength decreases, the current increases.

Thus, the correct answer is that the current is inversely proportional to the wavelength.

### Final Answer:
The current in the photocell is inversely proportional to the wavelength of the light falling on the cathode.

---