Photocurrent and collector potential
"Photocurrent and collector potential" is a free, interactive Class 12 Physics simulation for NEET and JEE, built on the lesson "Saturation current and stopping potential" in Dual Nature of Radiation and Matter. Change the intensity and the frequency. The faint curve is where you started. Watch the saturation level and the point where the current reaches zero, −V₀. Intensity raises the saturation current; frequency moves the stopping potential. Kmax = eV₀, and below the threshold frequency there is no current at all. JEE Main asked 40 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Saturation current and stopping potential in Dual Nature of Radiation and Matter. Change the values and watch what happens.
In JEE Main
42 shifts analysed
40 questions from Dual Nature of Radiation and Matter across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §11.4.2
- Keep the frequency ν and intensity fixed and make A more and more positive. The photocurrent grows, then levels off: at a large enough accelerating potential every electron emitted by C reaches A. This largest current is the saturation current, and raising the potential further does not increase it.
- Now make A negative with respect to C. The electrons are pushed back, and only the more energetic ones reach A, so the current falls quickly.
- At one sharply defined negative potential of A the current becomes zero. This smallest retarding potential that stops the photocurrent, for a given frequency, is the cut-off or stopping potential V₀.
- The photoelectrons do not all leave with the same energy. The stopping potential is just enough to turn back the fastest of them, so the maximum kinetic energy is Kmax = eV₀ (Eq. 11.1).
- Repeat with the same frequency at higher intensities I₂ and I₃ (I₃ > I₂ > I₁). The saturation current rises in proportion to the intensity, but the stopping potential stays exactly the same (Fig. 11.3).
- So, for light of a given frequency, the stopping potential, and with it Kmax, is independent of the intensity. Kmax depends on the light source and the emitter material, not on how bright the light is.
- Exercise 11.3: a cut-off voltage of 1.5 V means Kmax = 1.5 eV = 1.5 × 1.6 × 10⁻¹⁹ J = 2.4 × 10⁻¹⁹ J.
Questions
3 answers
- What does the "Photocurrent and collector potential" simulation show?
- Change the intensity and the frequency. The faint curve is where you started. Watch the saturation level and the point where the current reaches zero, −V₀.
- Does JEE Main ask questions from Dual Nature of Radiation and Matter?
- Yes. JEE Main asked 40 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Saturation current and stopping potential" in Dual Nature of Radiation and Matter (Class 12 Physics).
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