The Question is from neet-physics dual-nature-matter-radiation photocells, and it was created by the author Mr. ANIL KUMAR.G

Question

A photocell is illuminated by a monochromatic light source with intensity $I$ and frequency $
u$.  If the intensity is doubled while keeping the frequency constant, which of the following statements is TRUE regarding the photocurrent and stopping potential?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Photoelectric effect
 **Related Topics**: Intensity, frequency, photocurrent, stopping potential
 **Prerequisites**: Understanding of photons, energy quantization, and the photoelectric equation.
## SOLUTION PATHWAY
**Given Information**:
  Initial intensity: $I$
  Frequency: $
u$ (constant)
  Intensity is doubled: $2I$
**Method & Steps**:
 1. **Understanding the Photoelectric Effect:** The photoelectric effect describes the emission of electrons when light hits a material.  The number of emitted electrons is proportional to the intensity of the light, while the kinetic energy of the emitted electrons is proportional to the frequency of the light.
  Reasoning:  Increasing intensity means more photons strike the surface per unit time, leading to more electrons being ejected.
  Formula/Rule:  $E = h
u$, where $E$ is the energy of a photon, $h$ is Planck's constant, and $
u$ is the frequency.
 2. **Effect of Intensity on Photocurrent:** Photocurrent is the flow of electrons, directly proportional to the number of electrons emitted. Doubling the intensity doubles the number of photons, thus doubling the number of emitted electrons and hence doubling the photocurrent.
  Working:  $I \propto 	ext{number of photons} \propto 	ext{photocurrent}$ 
  Key Point:  The frequency determines the maximum kinetic energy of the emitted electrons, not their number.
 3. **Effect of Intensity on Stopping Potential:** The stopping potential is the voltage needed to stop the most energetic electrons. Since the frequency remains constant, the maximum kinetic energy of the emitted electrons ($K_{max}$) remains the same. Therefore, the stopping potential ($V_s$) remains unchanged. It is given by $K_{max} = eV_s$, where $e$ is the electron charge. 
  Result: The photocurrent doubles, and the stopping potential remains the same.
  Verification: The relationship between intensity and photocurrent is linear, while the relationship between frequency and stopping potential is linear.  Doubling intensity only affects the number of electrons, not their energy.
 ## OPTION ANALYSIS
A ✓: The photocurrent doubles, and the stopping potential remains the same. (Correct)
B :x: The photocurrent remains the same, and the stopping potential doubles. (Incorrect: Intensity affects photocurrent, not stopping potential)
C :x: Both the photocurrent and stopping potential double. (Incorrect: Stopping potential is independent of intensity)
D :x: Both the photocurrent and stopping potential remain the same. (Incorrect: Intensity affects photocurrent)
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Confusing intensity and frequency:  Remember intensity affects the number of emitted electrons, while frequency determines their maximum energy.
  2.Assuming a relationship between intensity and stopping potential:  The stopping potential depends only on the frequency of the light.
 **Quick Tips**:
  1.Visualize the photoelectric effect: Imagine more photons hitting the surface with increased intensity, leading to more electrons being ejected.
  2.Remember the key equations: $E = h
u$ and $K_{max} = eV_s$

Mr. ANIL KUMAR.G · Answer

The correct Option is "The photocurrent doubles, and the stopping potential remains the same."

Mr. ANIL KUMAR.G · Answer

The photocurrent remains the same, and the stopping potential doubles.

Mr. ANIL KUMAR.G · Answer

Both the photocurrent and stopping potential double.

Mr. ANIL KUMAR.G · Answer

Both the photocurrent and stopping potential remain the same.

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