When a light of wavelength 300 nm falls on a photoelectric emitter, photo electrons are emitted. For another emitter light of wavelength 600 nm is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is (a) 1: 2 (b) 2: 1 (c) 4: 1 (d) 1: 4

Answer: (b). Treat each wavelength as the threshold: φ = (hc)/(λ₀), so φ ∝ (1)/(λ₀) (φ₁)/(φ₂) = (600)/(300) = 2 So the answer is (b) 2:1.

When a light of wavelength 300 nm falls on a photoelectric emitter… | MHT CET 2023 (13 May, Shift 1)

Q70MHT CET 2023 · 13 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation

When a light of wavelength $300\text{ nm}$ falls on a photoelectric emitter, photo electrons are emitted. For another emitter light of wavelength $600\text{ nm}$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

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Answer: (b)

Treat each wavelength as the threshold: $\phi = \dfrac{hc}{\lambda_0}$ , so $\phi \propto \dfrac{1}{\lambda_0}$
$\dfrac{\phi_1}{\phi_2} = \dfrac{600}{300} = 2$
So the answer is (b) $2:1$ .

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