A photo - cell is being operated in saturation mode . Electromagnetic radiations of wavelength `lamda` m fall upon the photo - cell . The corresponding spectral sensitivity of photocell is `JA W^(-1)` . The number of photo - electrons produced by each incident photon, is

Options

A. `(hc)/(2elamdaJ)`
B. `(hc)/(elamda)`
C. `(hc)/(elamda)`
D. `(2hc)/(elamdaJ)`

Answer

To solve the problem, we need to determine the number of photo-electrons produced by each incident photon when electromagnetic radiation of wavelength \( \lambda \) falls upon a photo-cell operating in saturation mode.

### Step-by-Step Solution:

1. **Understand the Energy of a Photon**:
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Identify the Spectral Sensitivity**:
The spectral sensitivity of the photocell is given as \( J \) A/W, which means that for every watt of power incident on the photocell, a current of \( J \) amperes is produced.

3. **Calculate the Number of Photons Incident per Second**:
If we have a power \( P \) (in watts), the number of photons \( N \) incident per second can be calculated as:
\[ N = \frac{P}{E} = \frac{P \lambda}{hc} \]
This means for a power of 1 watt, the number of photons incident per second is:
\[ N = \frac{\lambda}{hc} \]

4. **Relate Current to Number of Electrons**:
The current \( J \) produced by the photocell is related to the number of electrons \( n \) emitted per second by the equation:
\[ J = n \cdot e \]
where \( e \) is the charge of an electron. Therefore, the number of electrons emitted per second is:
\[ n = \frac{J}{e} \]

5. **Determine the Number of Electrons per Photon**:
To find the number of photo-electrons produced by each incident photon, we can use the relationship established:
\[ \text{Number of electrons per photon} = \frac{n}{N} \]
Substituting the values we have:
\[ \text{Number of electrons per photon} = \frac{J/e}{\lambda/hc} = \frac{J \cdot hc}{e \cdot \lambda} \]

6. **Final Expression**:
Thus, the number of photo-electrons produced by each incident photon is given by:
\[ n = \frac{hcJ}{e\lambda} \]

### Conclusion:
The number of photo-electrons produced by each incident photon is \( \frac{hcJ}{e\lambda} \).