Question

A $5$ watt source emits monochromatic light of wavelength $5000\,?$. When placed $0.5\, m$ away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of $1.0 \,m$, the number of photo electrons liberated will:

• A. be reduced by a factor of 8
• B. be reduced by a factor of 16
• C. be reduced by a factor of 2
• D. be reduced by a factor of 4

Answer 1

The Correct Option is D

To solve this problem, we need to understand how the intensity of light and the resulting photoelectron emission are affected by the distance from the light source. The intensity of light is the power per unit area and follows the inverse square law, which states that the intensity of light from a point source is inversely proportional to the square of the distance from the source.

1. **Initial Setup**:

• The power of the light source is given as 5 \text{ watt}.
• The initial distance from the source is 0.5 \text{ m}.

2. **Intensity at Initial Distance**:

The intensity I_1 at a distance r_1 = 0.5 \text{ m} is calculated as:

I_1 = \frac{P}{4\pi r_1^2}

Substituting the values:

I_1 = \frac{5}{4\pi (0.5)^2}

3. **Intensity at New Distance**:

When the light source is moved to r_2 = 1.0 \text{ m}, the new intensity I_2 is:

I_2 = \frac{P}{4\pi r_2^2}

Substituting the values:

I_2 = \frac{5}{4\pi (1.0)^2}

4. **Ratio of Intensities**:

The ratio of the initial intensity to the new intensity gives the factor by which the intensity is reduced:

\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}

Substituting values:

\frac{I_2}{I_1} = \frac{(0.5)^2}{(1.0)^2} = \frac{0.25}{1} = \frac{1}{4}

5. **Conclusion**:

Since the number of photoelectrons liberated is directly proportional to the intensity of the light, moving the light source from 0.5 \text{ m} to 1.0 \text{ m} reduces the number of photoelectrons emitted by a factor of 4.