For the same photosensitive metal, light of frequency 6.0×10¹⁴ Hz produces a stopping potential of 0.50 V, while light of frequency 1.0×10¹⁵ Hz produces a stopping potential of 2.16 V. Take e = 1.6×10⁻¹⁹ C. Which pair is closest to Planck's constant h and the metal work function φ?
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Question

For the same photosensitive metal, light of frequency 6.0×10¹⁴ Hz produces a stopping potential of 0.50 V, while light of frequency 1.0×10¹⁵ Hz produces a stopping potential of 2.16 V. Take e = 1.6×10⁻¹⁹ C. Which pair is closest to Planck's constant h and the metal work function φ?

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G1 Physics · Accepted Answer

Einstein's equation written in terms of stopping potential is eV₀ = hν−φ. For two measurements on the same metal: e(V₂−V₁) = h(ν₂−ν₁). Therefore, h ÷ e = (V₂−V₁) ÷ (ν₂−ν₁). Substitute: h ÷ e = (2.16−0.50) ÷ [(1.0×10¹⁵)−(6.0×10¹⁴)] = 1.66 ÷ 4.0×10¹⁴ = 4.15×10⁻¹⁵ V·s. Hence, h = 1.6×10⁻¹⁹×4.15×10⁻¹⁵ = 6.64×10⁻³⁴ J·s. Now use the first measurement. In eV units, φ = (hν₁ ÷ e)−V₁. hν₁ ÷ e = 4.15×10⁻¹⁵×6.0×10¹⁴ = 2.49 eV. Thus, φ = 2.49−0.50 = 1.99 eV. Correct option: A.