Question

A photo-cell is illuminated by a source of light, which is placed at a distance d from the cell. If the distance becomes d/2. Then number of electrons emitted per second will be
1. Remain same
2. Four times
3. Two times
4. One-fourth

Answer 1

Correct Answer - Option 2 : Four times

CONCEPT:

• Intensity is the amount of energy received per unit area.
• The kinetic energy of photoelectrons is directly proportional to the intensity of incident radiation, as the intensity of incident radiation increases the kinetic energy of photoelectrons increases.
• The intensity I can be written as

\(\Rightarrow I = \frac{E}{At}\)

Where A = area, t= time, E = Energy incident

EXPLANATION :

Given - r₁ = d, r₂ =d/2

• In the above equation, A can be substituted as A = 4π r2

\(⇒ I = \frac{E}{4\pi r^{2}t}\)

• The above equation can be rewritten for two intensities from the same source  as 

\(⇒ \frac{I_{1}}{I_{2}} = \frac{\frac{E}{4\pi r_{1}^{2}t}}{\frac{E}{4\pi r_{2}^{2}t}} = \frac{r_{2}^{2}}{r_{1}^{2}}\)

Where r = is the distance from the source

• The ratio between intensities can be written as

\(\Rightarrow \frac{I_{1}}{I_{2}} = \frac{(\frac{d}{2})^{2}}{d^{2}} = \frac{1}{4}\)

\(⇒ \frac{I_{1}}{I_{2}}= \frac{1}{4}\)

⇒ I₂ = 4I₁

• Hence, option 2 is the answer