When a certain metallic surface is illuminated with monochromatic light wavelength λ, the stopping potential for photoelectric current is 4V₀. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V₀. The threshold wavelength for this surface for photoelectric effect is (a) 9λ (b) (λ)/(9) (c) 3λ (d) (λ)/(3)

Answer: (a). (hc)/(λ) - φ = 4eV₀ and (hc)/(3λ) - φ = eV₀ Multiply the second by 4 and equate: (4hc)/(3λ) - 4φ = (hc)/(λ) - φ 3φ = (hc)/(3λ) ⇒ φ = (hc)/(9λ) λ₀ = (hc)/(φ) = 9λ. So the answer is (a) 9λ.

When a certain metallic surface is illuminated with monochromatic… | MHT CET 2024 (15 May, Shift 2)

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Photoelectric Effect: Einstein's Equation

When a certain metallic surface is illuminated with monochromatic light wavelength $\lambda$ , the stopping potential for photoelectric current is $4V_0$ . When the same surface is illuminated with light of wavelength $3\lambda$ , the stopping potential is $V_0$ . The threshold wavelength for this surface for photoelectric effect is

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Answer: (a)

$\frac{hc}{\lambda} - \phi = 4eV_0$ and $\frac{hc}{3\lambda} - \phi = eV_0$
Multiply the second by 4 and equate: $\frac{4hc}{3\lambda} - 4\phi = \frac{hc}{\lambda} - \phi$
$3\phi = \frac{hc}{3\lambda} \Rightarrow \phi = \frac{hc}{9\lambda}$
$\lambda_0 = \frac{hc}{\phi} = 9\lambda$ . So the answer is (a) $9\lambda$ .

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