When a photosensitive surface is irradiated by light of wavelengths ' λ₁ ' and ' λ₂ ', kinetic energies of emitted photoelectrons are 'E ₁ ' and 'E ₂ ' respectively. The work function of photosensitive surface is (a) ((λ₁E₁ - λ₂E₂))/((λ₂ - λ₁)) (b) ((λ₁E₁ + λ₂E₂))/((λ₂ - λ₁)) (c) ((λ₁E₂ - λ₂E₁))/((λ₂ - λ₁)) (d) ((λ₁E₂ + λ₂E₁))/((λ₂ - λ₁))

Answer: (a). (hc)/(λ₁) = φ + E₁ and (hc)/(λ₂) = φ + E₂ Equate hc: λ₁(φ + E₁) = λ₂(φ + E₂) φ(λ₂ - λ₁) = λ₁E₁ - λ₂E₂ So the answer is (a) φ = (λ₁E₁ - λ₂E₂)/(λ₂ - λ₁).

When a photosensitive surface is irradiated by light of wavelengths '… | MHT CET 2021 (20 Sep, Shift 1)

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Photoelectric Effect: Einstein's Equation

When a photosensitive surface is irradiated by light of wavelengths ' $\lambda_1$ ' and ' $\lambda_2$ ', kinetic energies of emitted photoelectrons are 'E $_1$ ' and 'E $_2$ ' respectively. The work function of photosensitive surface is

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Answer: (a)

$\frac{hc}{\lambda_1} = \phi + E_1$ and $\frac{hc}{\lambda_2} = \phi + E_2$
Equate $hc$ : $\lambda_1(\phi + E_1) = \lambda_2(\phi + E_2)$
$\phi(\lambda_2 - \lambda_1) = \lambda_1E_1 - \lambda_2E_2$
So the answer is (a) $\phi = \frac{\lambda_1E_1 - \lambda_2E_2}{\lambda_2 - \lambda_1}$ .

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