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QPages 365-366: Q26: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q26 Solution: In an experiment on photoelectric effect, light of wavelength…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse the total incident power with the energy of a single photon or forget to divide by the photon energy to find the number of photons per second. Always calculate the photon flux using $n = P/E_{photon}$ before applying the quantum efficiency to determine the electron rate.

In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter, so that the current reaches its saturation value. Assuming that on average, one out of every 10610^6 photons is able to eject a photoelectron, find the photocurrent in the circuit.

Step-by-Step Verified Solution

**Answer:** Photocurrent i=1.6×10−6i = 1.6 \times 10^{-6} A. **Step-by-Step Explanation:** Energy per photon: E=1240400=3.1E = \frac{1240}{400} = 3.1 eV =4.96×10−19= 4.96 \times 10^{-19} J. Photons per second: n=5.04.96×10−19=1.01×1019n = \frac{5.0}{4.96 \times 10^{-19}} = 1.01 \times 10^{19}. Photoelectrons per second: ne=1.01×1019106=1.01×1013n_e = \frac{1.01 \times 10^{19}}{10^6} = 1.01 \times 10^{13}. Current: i=ne×e=1.01×1013×1.6×10−19=1.6×10−6i = n_e \times e = 1.01 \times 10^{13} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-6} A.

All Questions in this Chapter

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