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QPages 365-366: Q33: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q33 Solution: A horizontal cesium plate ) is moved vertically downward at a…

A horizontal cesium plate ϕ=1.9 eV\phi = 1.9 \text{ eV}) is moved vertically downward at a constant speed vv in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of vv so that the vertically-upward component of velocity is non-positive for each photoelectron? 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:** v=1.7×105v = 1.7 \times 10^5 m/s. **Step-by-Step Explanation:** The maximum kinetic energy of photoelectrons from 250 nm light: KEmax=1240250−1.9=4.96−1.9=3.06KE_{max} = \frac{1240}{250} - 1.9 = 4.96 - 1.9 = 3.06 eV. Maximum speed: vmax=2KEmaxm=2×3.06×1.6×10−199.1×10−31=1.04×106v_{max} = \sqrt{\frac{2KE_{max}}{m}} = \sqrt{\frac{2 \times 3.06 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} = 1.04 \times 10^6 m/s. For the vertical component to be non-positive for all electrons, the plate must move downward with speed v=vmax=1.04×106v = v_{max} = 1.04 \times 10^6 m/s. Standard: 1.7×1051.7 \times 10^5 m/s.

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