When light of wavelength ' λ ' is incident on a photosensitive surface, the stopping potential is 'V'. When a light of wavelength 1.5λ is incident on the same surface, the stopping potential is ' V/4 '. Threshold wavelength for the surface is (a) 6/5λ (b) 7.5/4λ (c) 7.5/9λ (d) 9/5λ

Answer: (d). (hc)/(λ) - φ = eV and (hc)/(1.5λ) - φ = eV/4 Multiply the second by 4 and equate: (8hc)/(3λ) - 4φ = (hc)/(λ) - φ (5hc)/(3λ) = 3φ ⇒ φ = (5hc)/(9λ) λ₀ = (hc)/(φ) = 9/5λ. So the answer is (d).

When light of wavelength ' λ ' is incident on a photosensitive… | MHT CET 2026 (19 Apr, Shift 2)

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Photoelectric Effect: Einstein's Equation

When light of wavelength ' $\lambda$ ' is incident on a photosensitive surface, the stopping potential is 'V'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is ' $\frac{V}{4}$ '. Threshold wavelength for the surface is

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Answer: (d)

$\frac{hc}{\lambda} - \phi = eV$ and $\frac{hc}{1.5\lambda} - \phi = \frac{eV}{4}$
Multiply the second by 4 and equate: $\frac{8hc}{3\lambda} - 4\phi = \frac{hc}{\lambda} - \phi$
$\frac{5hc}{3\lambda} = 3\phi \Rightarrow \phi = \frac{5hc}{9\lambda}$
$\lambda_0 = \frac{hc}{\phi} = \frac{9}{5}\lambda$ . So the answer is (d).

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