In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function φ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 × 10 - 34 Js, c = 3 × 108 ms ⁻¹) (a) 1.3 V (b) 1.1 V (c) 1.9 V (d) 0.6 V

Answer: (a). eV₀ = (hc)/(λ) - φ, so Δ V₀ = hc/e((1)/(λ₁) - (1)/(λ₂)). hc/e = 6.63 × 10⁻³⁴ × 3 × 10⁸1.6 × 10⁻¹⁹ = 1243 V nm Δ V₀ = 1243(1/280 - 1/400) = 1243 × 1.07 × 10⁻³ ≈ 1.3 V So the answer is (a) 1.3 V.

In a photoelectric experiment ultraviolet light of wavelength 280 nm… | JEE Main 2021 (26 Aug, Shift 1)

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Photoelectric Effect: Einstein's Equation

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function $\phi$ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 $\times$ 10 $-$ 34 Js, c = 3 $\times$ 108 ms ${}^{-1}$ )

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Answer: (a)

$eV_0 = \frac{hc}{\lambda} - \phi$ , so $\Delta V_0 = \frac{hc}{e}\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$ .
$\frac{hc}{e} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19}} = 1243$ V nm
$\Delta V_0 = 1243\left(\frac{1}{280} - \frac{1}{400}\right) = 1243 \times 1.07 \times 10^{-3} \approx 1.3$ V
So the answer is (a) 1.3 V.

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