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When a light of wavelength 300 nm falls on the photoelectric emitter, photoelectrons are just liberated. For another emitter, however light of 600 nm wavelength is sufficient for creating photoemission. What is the ratio of the work function ϕ 0 of the two emitters?

Options

  1. A1 : 4
  2. B4 : 1
  3. C2 : 1
  4. D1 : 2

Correct answer

C. 2 : 1

Step-by-step solution

K.E max = hc λ - ϕ ϕ 1 = 1240 ev-nm 300 nm = 4.13 eV ϕ 2 = 1240 ev-nm 600 nm = 2.06 eV ∴ ϕ 1 ϕ 2 = 4 2 = 2 1

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