A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius `5xx10^(-11)`m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction.
if work function of the metal of the foil is 2.2 eV, the time taken by electron to come out is nearly

Options

A. a.1.3 minutes
B. b.1.3 hours
C. c.`1.3xx10^(-3)` hour
D. d.`1.3 mus`

Answer

To solve the problem, we need to calculate the time taken by an electron to come out of a metal foil when it absorbs energy from an isotropic point source. The work function of the metal is given as 2.2 eV. We will follow these steps:

### Step-by-Step Solution:

1. **Convert Work Function to Joules**:
The work function (Φ) is given in electron volts (eV). We need to convert it to joules (J) using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).
\[ \Phi = 2.2 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.52 \times 10^{-19} \text{ J} \]

2. **Calculate Energy Received per Second**:
From the video transcript, it is mentioned that the energy received by the electron per second (P) is \(7.6 \times 10^{-23} \text{ W}\).

3. **Calculate the Time Taken for Photoelectric Emission**:
The time taken (Δt) for the electron to absorb enough energy to overcome the work function can be calculated using the formula:
\[ \Delta t = \frac{\Phi}{P} \]
Substituting the values we have:
\[ \Delta t = \frac{3.52 \times 10^{-19} \text{ J}}{7.6 \times 10^{-23} \text{ W}} \]

4. **Perform the Calculation**:
\[ \Delta t = \frac{3.52 \times 10^{-19}}{7.6 \times 10^{-23}} \approx 4.63 \times 10^{3} \text{ seconds} \]

5. **Convert Seconds to Hours**:
To convert seconds to hours, we divide by 3600 (the number of seconds in an hour):
\[ \Delta t \approx \frac{4.63 \times 10^{3}}{3600} \approx 1.28 \text{ hours} \]

6. **Final Result**:
Thus, the time taken by the electron to come out is approximately:
\[ \Delta t \approx 1.28 \text{ hours} \approx 1.3 \text{ hours} \]

### Summary:
The time taken by the electron to come out from the metal foil is nearly **1.3 hours**.