A light wave described by E = 60[sin(3 × 10¹⁵t) + sin(12 × 10¹⁵t)] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) _________ eV. (h = 6.6 × 10⁻³⁴ J·s. and e = 1.6 × 10⁻¹⁹C)
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Question

A light wave described by E = 60[sin(3 × 10¹⁵t) + sin(12 × 10¹⁵t)] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) _________ eV. (h = 6.6 × 10⁻³⁴ J·s. and e = 1.6 × 10⁻¹⁹C)

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G1 Physics · Accepted Answer

The wave contains two angular-frequency components: ω₁ = 3 × 10¹⁵ rad·s⁻¹, ω₂ = 12 × 10¹⁵ rad·s⁻¹. The maximum-energy photoelectrons are produced by the larger photon frequency corresponding to ω₂. Photon frequency is ν = ω ──── 2π . Hence, ν₂ = 12 × 10¹⁵ ────────── 2π ≈ 1.91 × 10¹⁵ Hz. The photon energy is E₍γ₎ = hν₂ = 6.6 × 10⁻³⁴ × 1.91 × 10¹⁵ ≈ 1.26 × 10⁻¹⁸ J. In electron-volts, E₍γ₎ = 1.26 × 10⁻¹⁸ ──────────── 1.6 × 10⁻¹⁹ ≈ 7.9 eV. B₍y₎ Einstein's photoelectric equation, Kₘₐₓ = E₍γ₎-φ. Ther…