Light of wavelength 331 nm is incident on a metal surface and produces photoelectrons with stopping potential 0.2 V. Taking h = 6.62×10⁻³⁴ J·s, e = 1.6×10⁻¹⁹ C and c = 3×10⁸ m·s⁻¹, the work function of the metal is α×10⁻¹⁹ J. What is α?
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Question

Light of wavelength 331 nm is incident on a metal surface and produces photoelectrons with stopping potential 0.2 V. Taking h = 6.62×10⁻³⁴ J·s, e = 1.6×10⁻¹⁹ C and c = 3×10⁸ m·s⁻¹, the work function of the metal is α×10⁻¹⁹ J. What is α?

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G1 Physics · Accepted Answer

Einstein's photoelectric equation is hc ÷ λ = φ+eV₀. Therefore, φ = hc ÷ λ - eV₀. Substitute: φ = [6.62×10⁻³⁴×3×10⁸ ÷ (331×10⁻⁹)] - [1.6×10⁻¹⁹×0.2]. The photon energy is approximately 6.00×10⁻¹⁹ J, and the stopping-energy term is 0.32×10⁻¹⁹ J. Thus, φ ≈ 5.68×10⁻¹⁹ J. Therefore α = 5.68. Correct option: C.