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A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity is non-positive for each photoelectron?

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Aditi Chauhan
Sol. When λ = 250 nm Energy of photon = hc/λ = 1240/250 = 4.96 ev ∴ K.E. – hc/λ = 1240/250 = 4.96 ev Velocity to be non positive for each photo electron The minimum value of velocity of plate should be = velocity of photo electron ∴ Velocity of photo electron = √2Ke/m = √3.06/9.1 * 10^-31 = √3.06 * 1.6 * 10^-19/9.1 * 10^-31 = 1.04 * 10^6 m/sec.
Last Activity: 11 Years ago
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