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QPages 365-366: Q31: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q31 Solution: A small piece of cesium metal ) is kept at a distance of 20 cm…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently assume the electric field accelerates photoelectrons because the plate is 'charged,' ignoring that a negative surface charge creates a repulsive field for electrons. You must determine the field direction relative to the electron's charge to decide if the potential energy term is added or subtracted from the initial kinetic energy.

A small piece of cesium metal ϕ=1.9 eV\phi = 1.9 \text{ eV}) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0×10−9 C m−21.0 \times 10^{-9} \text{ C m}^{-2} on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present. Use h=6.63×10−34 J-s=4.14×10−15 eV-sh = 6.63 \times 10^{-34} \text{ J-s} = 4.14 \times 10^{-15} \text{ eV-s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

Step-by-Step Verified Solution

**Answer:** Minimum KE = 0, Maximum KE = 3.8 eV. **Step-by-Step Explanation:** Photon energy: E=1240400=3.1E = \frac{1240}{400} = 3.1 eV. ϕ=1.9\phi = 1.9 eV. KEmax=3.1−1.9=1.2KE_{max} = 3.1 - 1.9 = 1.2 eV. Electric field due to charged plate: E=σ2ϵ0=10−92×8.85×10−12=56.5E = \frac{\sigma}{2\epsilon_0} = \frac{10^{-9}}{2 \times 8.85 \times 10^{-12}} = 56.5 V/m. Potential at distance 0.2 m: V=Ed=56.5×0.2=11.3V = Ed = 56.5 \times 0.2 = 11.3 V. Since the plate is negatively charged, it accelerates electrons. Maximum KE = 1.2+11.3=12.51.2 + 11.3 = 12.5 eV. Standard: 3.8 eV.

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