Two metallic plates `A and B` , each of area `5 xx 10m` are placed parallel to each other at a separation of 1 cm . Plate B carries a positive charge of 33.7 pc . A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0, so that 10 photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 10 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV . Electric field between the plates at the end of 10 seconds is

Options

A. `5xx10^(6)`
B. `7xx10^(7)`
C. `5xx10^(7)`
D. `9xx10^(6)`

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Identify the Given Data
- Area of each plate, \( A = 5 \times 10^{-4} \, \text{m}^2 \)
- Separation between the plates, \( d = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Charge on plate B, \( Q = 33.7 \, \text{pC} = 33.7 \times 10^{-12} \, \text{C} \)
- Energy of each photon, \( E = 5 \, \text{eV} \)
- Work function of plate A, \( \phi = 2 \, \text{eV} \)
- Rate of incident photons, \( n = 10 \, \text{photons/m}^2/\text{s} \)
- Time, \( t = 10 \, \text{s} \)

### Step 2: Calculate the Total Number of Incident Photons
The total number of photons incident on plate A in 10 seconds can be calculated as follows:
\[ \text{Total photons} = \text{Rate of incident photons} \times \text{Area} \times \text{Time} \]
\[ \text{Total photons} = 10 \, \text{photons/m}^2/\text{s} \times 5 \times 10^{-4} \, \text{m}^2 \times 10 \, \text{s} \]
\[ = 10 \times 5 \times 10^{-4} \times 10 = 5 \times 10^{-3} \, \text{photons} \]

### Step 3: Calculate the Number of Photoelectrons Emitted
According to the problem, one photoelectron is emitted for every 10 incident photons. Therefore, the number of photoelectrons emitted can be calculated as:
\[ \text{Number of photoelectrons} = \frac{\text{Total photons}}{10} \]
\[ = \frac{5 \times 10^{-3}}{10} = 5 \times 10^{-4} \, \text{photoelectrons} \]

### Step 4: Calculate the Charge of the Emitted Photoelectrons
The charge of the emitted photoelectrons can be calculated using the charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \):
\[ \text{Total charge} = \text{Number of photoelectrons} \times e \]
\[ = 5 \times 10^{-4} \times 1.6 \times 10^{-19} \, \text{C} \]
\[ = 8 \times 10^{-24} \, \text{C} \]

### Step 5: Calculate the Total Charge on Plate B
The total charge on plate B after collecting the emitted photoelectrons can be calculated as:
\[ \text{Total charge on B} = Q + \text{Charge of emitted photoelectrons} \]
\[ = 33.7 \times 10^{-12} + 8 \times 10^{-24} \approx 33.7 \times 10^{-12} \, \text{C} \quad (\text{since } 8 \times 10^{-24} \text{ is negligible}) \]

### Step 6: Calculate the Electric Field Between the Plates
The electric field \( E \) between two plates is given by:
\[ E = \frac{Q}{\varepsilon_0 A} \]
Where \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \) is the permittivity of free space. Substituting the values:
\[ E = \frac{33.7 \times 10^{-12}}{8.85 \times 10^{-12} \times 5 \times 10^{-4}} \]
Calculating this gives:
\[ E \approx \frac{33.7}{8.85 \times 5} \times 10^{4} \approx 0.76 \times 10^{4} \, \text{N/C} = 7600 \, \text{N/C} \]

### Final Answer
The electric field between the plates at the end of 10 seconds is approximately \( 7600 \, \text{N/C} \).

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