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 of finding the number of photoelectrons produced by each incident photon in a photocell operating in saturation mode, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy of a Photon**:
The energy \( E \) of a single photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength of the radiation.

2. **Define Spectral Sensitivity**:
The spectral sensitivity \( J \) of the photocell is given in units of \( \text{A/W} \) (amperes per watt). This represents the current produced per unit power of incident light.

3. **Relate Current to Photoelectrons**:
The current \( I \) produced by the photocell can be expressed as:
\[ I = N_e \cdot e \]
where:
- \( N_e \) is the number of photoelectrons produced per second,
- \( e \) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)).

4. **Power of Incident Photons**:
The power \( P \) of the incident photons can be expressed as:
\[ P = N_p \cdot E \]
where:
- \( N_p \) is the number of incident photons per second,
- \( E \) is the energy of one photon.

5. **Relate Current to Power**:
From the definition of spectral sensitivity, we have:
\[ J = \frac{I}{P} \]
Substituting the expressions for \( I \) and \( P \):
\[ J = \frac{N_e \cdot e}{N_p \cdot \frac{hc}{\lambda}} \]

6. **Rearranging for \( N_e \)**:
We want to find the number of photoelectrons produced per incident photon, which is \( \frac{N_e}{N_p} \). Rearranging the equation gives:
\[ \frac{N_e}{N_p} = \frac{J \cdot \frac{hc}{\lambda}}{e} \]

7. **Final Expression**:
Thus, the number of photoelectrons produced by each incident photon is given by:
\[ \frac{N_e}{N_p} = \frac{hc \cdot J}{\lambda \cdot e} \]

### Final Answer:
The number of photoelectrons produced by each incident photon is:
\[ \frac{N_e}{N_p} = \frac{hc \cdot J}{\lambda \cdot e} \]