in an experimental set up to study the photoelectric effect a point soure fo light of power `3.2xx10^(-3)` W was taken. The source can emit monoenergetic photons of energy 5eV and is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work-function 3.0 eV. The radius of the sphere is `r = 8..10^(-3)` m. The efficiency of photoelectric emission is one for every `10^6` incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swepts away after the emission).
Time after which photoelectric emission stops is

Options

A. 100s
B. 121s
C. 111s
D. 141s

Explanation

To solve the problem step by step, we will follow the outlined procedure to find the time after which photoelectric emission stops.

### Step 1: Calculate the intensity of light on the sphere
The intensity \( I \) of light at a distance \( d \) from a point source is given by the formula:

\[ I = \frac{P}{4 \pi d^2} \]

Where:
- \( P = 3.2 \times 10^{-3} \) W (power of the light source)
- \( d = 0.8 \) m (distance from the light source to the sphere)

Substituting the values:

\[ I = \frac{3.2 \times 10^{-3}}{4 \pi (0.8)^2} \]

Calculating \( I \):

\[ I = \frac{3.2 \times 10^{-3}}{4 \pi (0.64)} \approx \frac{3.2 \times 10^{-3}}{8.042} \approx 3.97 \times 10^{-4} \text{ W/m}^2 \]

### Step 2: Calculate the energy intercepted by the sphere
The energy \( E_1 \) intercepted by the sphere can be calculated using the formula:

\[ E_1 = I \cdot A \]

Where \( A \) is the surface area of the sphere:

\[ A = 4 \pi r^2 \]

Given \( r = 8 \times 10^{-3} \) m, we find:

\[ A = 4 \pi (8 \times 10^{-3})^2 = 4 \pi (64 \times 10^{-6}) \approx 8.04 \times 10^{-4} \text{ m}^2 \]

Now substituting for \( E_1 \):

\[ E_1 = (3.97 \times 10^{-4}) \cdot (8.04 \times 10^{-4}) \approx 3.2 \times 10^{-7} \text{ J} \]

### Step 3: Calculate the energy of a single photon
The energy of a single photon \( E \) is given as 5 eV. Converting this to Joules:

\[ E = 5 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 8.0 \times 10^{-19} \text{ J} \]

### Step 4: Calculate the number of photoelectrons emitted
The number of photoelectrons \( n_1 \) emitted can be calculated using:

\[ n_1 = \frac{E_1}{E} \]

Substituting the values:

\[ n_1 = \frac{3.2 \times 10^{-7}}{8.0 \times 10^{-19}} \approx 4.0 \times 10^{11} \]

### Step 5: Adjust for the efficiency of photoelectric emission
Given that the efficiency of photoelectric emission is 1 for every \( 10^6 \) incident photons, the number of emitted photoelectrons \( n \) is:

\[ n = \frac{n_1}{10^6} = \frac{4.0 \times 10^{11}}{10^6} = 4.0 \times 10^5 \]

### Step 6: Calculate the stopping potential
The stopping potential \( V_0 \) is given by:

\[ V_0 = E - \text{Work function} = 5 \text{ eV} - 3 \text{ eV} = 2 \text{ V} \]

### Step 7: Calculate the time after which photoelectric emission stops
Using the formula for the stopping potential:

\[ V_0 = \frac{Q}{4 \pi \epsilon_0 R} \]

Where \( Q = n \cdot e \) (total charge from emitted electrons):

\[ Q = n \cdot e = (4.0 \times 10^5) \cdot (1.6 \times 10^{-19}) \approx 6.4 \times 10^{-14} \text{ C} \]

Substituting into the equation for \( V_0 \):

\[ 2 = \frac{6.4 \times 10^{-14}}{4 \pi (8.85 \times 10^{-12}) (8 \times 10^{-3})} \]

Solving for \( t \):

\[ t = \frac{V_0 \cdot 4 \pi \epsilon_0 R}{n \cdot e} \]

Substituting the known values:

\[ t = \frac{2 \cdot 4 \cdot 3.14 \cdot 8.85 \times 10^{-12} \cdot 0.008}{4.0 \times 10^5 \cdot 1.6 \times 10^{-19}} \]

Calculating \( t \):

\[ t \approx 111 \text{ seconds} \]

### Final Answer
The time after which photoelectric emission stops is approximately **111 seconds**.