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

1. 1:2
2. 2:1
3. 4:1
4. 1:4

Correct answer: 2:1

Solution

The work function (ϕ) is inversely proportional to the wavelength (λ) of light required for photoemission, as ϕ = hc/λ. For the first emitter, λ₁ = 300 nm, and for the second emitter, λ₂ = 600 nm. The ratio of work functions is ϕ₁/ϕ₂ = λ₂/λ₁ = 600/300 = 2:1.

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