A photo-cell is illuminated by a source of light which is placed at a distance d from the cell . If the distance become d/2 . Then number of electrons emitted per seconed will be

Options

A. Remain same
B. Four times
C. Two times
D. One-fourth

Answer

To solve the problem, we need to understand the relationship between the intensity of light and the number of electrons emitted from a photocell. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between intensity and distance
The intensity \( I \) of light from a point source is inversely proportional to the square of the distance \( d \) from the source. This can be expressed mathematically as:
\[ I \propto \frac{1}{d^2} \]

### Step 2: Define the initial and new conditions
Let’s denote:
- The initial distance from the light source to the photocell as \( d \).
- The initial intensity of light at this distance as \( I_1 \).
- The new distance as \( \frac{d}{2} \).
- The new intensity at this distance as \( I_2 \).

### Step 3: Calculate the new intensity
Using the inverse square law, we can express the intensities:
\[ I_1 \propto \frac{1}{d^2} \]
\[ I_2 \propto \frac{1}{\left(\frac{d}{2}\right)^2} = \frac{1}{\frac{d^2}{4}} = \frac{4}{d^2} \]

### Step 4: Find the ratio of the intensities
Now, we can find the ratio of the new intensity to the initial intensity:
\[ \frac{I_2}{I_1} = \frac{\frac{4}{d^2}}{\frac{1}{d^2}} = 4 \]

### Step 5: Relate intensity to the number of emitted electrons
The number of electrons emitted per second from the photocell is directly proportional to the intensity of the light:
\[ N_1 \propto I_1 \quad \text{and} \quad N_2 \propto I_2 \]
Where \( N_1 \) is the number of electrons emitted at distance \( d \) and \( N_2 \) is the number emitted at distance \( \frac{d}{2} \).

### Step 6: Find the relationship between the number of emitted electrons
From the intensity ratio, we can relate the number of emitted electrons:
\[ \frac{N_2}{N_1} = \frac{I_2}{I_1} = 4 \]
Thus, we can express \( N_2 \) in terms of \( N_1 \):
\[ N_2 = 4 N_1 \]

### Conclusion
Therefore, when the distance is reduced to \( \frac{d}{2} \), the number of electrons emitted per second will be **4 times** the number emitted when the distance was \( d \).

### Final Answer
**The number of electrons emitted per second will be 4 times the original number.**

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