Question

A beam of light falls on a metal surface such that photo-electrons are generated. If the power of the light source starts to decrease linearly with time, then the variation of the photocurrent \(I\) and magnitude of the stopping potential \(|V|\) with time is best represented by :

• A. \(I=\text{constant},\; |V|=\text{constant}\)
• B. \(I\) decreases linearly with time, \(|V|\) remains constant
• C. \(I\) decreases linearly with time, \(|V|\) also decreases linearly with time
• D. \(I=\text{constant},\; |V|\) decreases linearly with time

Hint:

Photocurrent depends on intensity of incident light. Stopping potential depends on frequency, not intensity. A decrease in intensity reduces the number of emitted electrons. The maximum kinetic energy remains unchanged if frequency remains constant.

Answer 1

The Correct Option is B

Step 1: Identify the two quantities.
We track the photocurrent $I$ and the magnitude of the stopping potential $|V|$ as the source power falls linearly with time.
Step 2: Power sets the intensity.
The light intensity reaching the metal is proportional to the source power. So if power falls linearly, intensity falls linearly too.
Step 3: Photocurrent depends on intensity.
More photons per second eject more electrons per second, so $I$ is proportional to intensity. Hence $I$ decreases linearly with time.
Step 4: Stopping potential depends on energy, not intensity.
The stopping potential is fixed by the maximum kinetic energy of the electrons, $eV_s = K_{max}$.
Step 5: Use Einstein's equation.
Here $K_{max} = h\nu - \phi$, which depends only on the frequency $\nu$ and the work function $\phi$. Lowering the power does not change the frequency.
\[ K_{max} = h\nu - \phi = \text{constant} \]
Step 6: Conclusion.
Since $\nu$ is unchanged, $|V|$ stays constant while $I$ falls linearly. This is option (B).
\[ \boxed{I \text{ decreases linearly with time, } |V| \text{ remains constant}} \]