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 power of the source P is 1.5 W and distance of the foil from the source is 3.5 m, the energy received by an electron per second is

Options

A. a.`7.65xx10^(-23)` W
B. b.`7.65xx10^(-28)` W
C. c.`7.65xx10^(-15)` W
D. d.`7.65xx10^(-10)` W

Answer

To find the energy received by an electron per second from an isotropic point source emitting power \( P \), we can follow these steps:

### Step 1: Understand the setup
We have a point source emitting power \( P = 1.5 \, \text{W} \) at a distance \( r = 3.5 \, \text{m} \) from the source. The energy is absorbed by an electron located within a circular area with a radius of \( 5 \times 10^{-11} \, \text{m} \).

### Step 2: Calculate the intensity of the source
The intensity \( I \) of the source at a distance \( r \) is given by the formula:
\[ I = \frac{P}{A} \]
where \( A \) is the surface area of a sphere with radius \( r \):
\[ A = 4 \pi r^2 \]
Substituting \( r = 3.5 \, \text{m} \):
\[ A = 4 \pi (3.5)^2 = 4 \pi \times 12.25 = 49 \pi \, \text{m}^2 \]
Now, substituting \( A \) into the intensity formula:
\[ I = \frac{1.5}{49 \pi} \]

### Step 3: Calculate the energy received by the electron
The energy received by the electron per second can be calculated using the intensity and the area \( A_e \) of the circular path where the electron absorbs energy:
\[ A_e = \pi (5 \times 10^{-11})^2 \]
Now, substituting this into the energy received formula:
\[ E = I \cdot A_e = \left(\frac{1.5}{49 \pi}\right) \cdot \left(\pi (5 \times 10^{-11})^2\right) \]
The \( \pi \) cancels out:
\[ E = \frac{1.5 \cdot (5 \times 10^{-11})^2}{49} \]

### Step 4: Calculate the numerical value
Calculating \( (5 \times 10^{-11})^2 = 25 \times 10^{-22} = 2.5 \times 10^{-21} \):
\[ E = \frac{1.5 \cdot 2.5 \times 10^{-21}}{49} = \frac{3.75 \times 10^{-21}}{49} \]
Now, dividing:
\[ E \approx 7.65 \times 10^{-23} \, \text{J} \]

### Final Answer
The energy received by an electron per second is approximately:
\[ E \approx 7.65 \times 10^{-23} \, \text{J} \]