When a photosensitive surface is irradiated by lights of wavelengths ' λ₁ ' and ' λ₂ ', kinetic energies of the emitted photoelectrons is ' E₁ ' and ' E₂ ' respectively. The work function of the photosensitive surface is (a) ((E₂λ₂ - E₁λ₁))/((λ₂ - λ₁)) (b) ((E₁λ₁ + E₂λ₂))/((λ₂ - λ₁)) (c) ((E₁λ₁ - E₂λ₂))/((λ₂ - λ₁)) (d) ((E₂λ₂ + E₁λ₁))/((λ₁ - λ₂))

Answer: (c). E₁ = (hc)/(λ₁) - φ and E₂ = (hc)/(λ₂) - φ, so hc = λ₁(E₁ + φ) = λ₂(E₂ + φ) φ(λ₁ - λ₂) = E₂λ₂ - E₁λ₁ φ = (E₁λ₁ - E₂λ₂)/(λ₂ - λ₁) So the answer is (c).

When a photosensitive surface is irradiated by lights of wavelengths… | MHT CET 2024 (10 May, Shift 2)

Q63MHT CET 2024 · 10 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation

When a photosensitive surface is irradiated by lights of wavelengths ' $\lambda_1$ ' and ' $\lambda_2$ ', kinetic energies of the emitted photoelectrons is ' $E_1$ ' and ' $E_2$ ' respectively. The work function of the photosensitive surface is

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Answer: (c)

$E_1 = \frac{hc}{\lambda_1} - \phi$ and $E_2 = \frac{hc}{\lambda_2} - \phi$ , so $hc = \lambda_1(E_1 + \phi) = \lambda_2(E_2 + \phi)$
$\phi(\lambda_1 - \lambda_2) = E_2\lambda_2 - E_1\lambda_1$
$\phi = \frac{E_1\lambda_1 - E_2\lambda_2}{\lambda_2 - \lambda_1}$
So the answer is (c).

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