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QPages 365-366: Q30: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q30 Solution: In a photoelectric experiment, the collector plate is at 2.0 V…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the sign of the collector potential, assuming a positive voltage always implies a retarding field that reduces electron energy. You must recognize that a positive collector potential relative to the emitter accelerates photoelectrons, adding $eV$ to their kinetic energy rather than subtracting it.

In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper ϕ=4.5 eV\phi = 4.5 \text{ eV}). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.

Step-by-Step Verified Solution

**Answer:** Minimum KE = 0, Maximum KE = 3.7 eV. **Step-by-Step Explanation:** Photon energy: E=1240200=6.2E = \frac{1240}{200} = 6.2 eV. ϕ=4.5\phi = 4.5 eV. KEmax=6.2−4.5=1.7KE_{max} = 6.2 - 4.5 = 1.7 eV. The collector is at 2.0 V relative to emitter. For electrons just emitted, they are accelerated by 2.0 V, gaining 2.0 eV. Minimum KE at collector = 0 (those that lose energy in collisions). Maximum KE at collector = 1.7 + 2.0 = 3.7 eV.

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