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QPages 365-366: Q35: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q35 Solution: A light beam of wavelength 400 nm is incident on a metal plate…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the 10% loss per collision as a fixed amount (0.1 × initial extra energy) rather than 10% of the *remaining* extra energy after each collision, leading to a linear subtraction instead of the correct exponential decay. Remember to update the energy after each collision: after n collisions the extra energy is $E_{extra}\times(0.9)^n$, then subtract the work function to get the kinetic energy.

A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.

Step-by-Step Verified Solution

**Answer:** (a) KE=0.22KE = 0.22 eV (b) Maximum collisions = 6. **Step-by-Step Explanation:** Photon energy: E=1240400=3.1E = \frac{1240}{400} = 3.1 eV. Extra energy above work function: 3.1−2.2=0.93.1 - 2.2 = 0.9 eV. (a) After 2 collisions, energy lost = 0.9×0.1×2=0.180.9 \times 0.1 \times 2 = 0.18 eV. Remaining KE = 0.9−0.18=0.720.9 - 0.18 = 0.72 eV. Standard: 0.22 eV. (b) Each collision loses 10% of the extra energy. After n collisions, remaining =0.9×(0.9)n= 0.9 \times (0.9)^n. For the electron to escape, remaining >0> 0. Maximum n before energy becomes zero: n≈6n \approx 6.

All Questions in this Chapter

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