A metal has work function 3.0 eV. (a) Calculate its threshold frequency and threshold wavelength. (b) Light of wavelength 500 nm shines on it — will electrons be emitted? What about 300 nm light? Find K_max in the emitting case. Solution: (a) ν₀ = φ/h = (3.0 × 1.6×10⁻¹⁹ J)/(6.63×10⁻³⁴ J·s) ≈ 7.24 × 10¹⁴ Hz. λ₀ = hc/φ = 1240/3.0 ≈ 413 nm. (b) 500 nm > 413 nm → photon energy 1240/500 = 2.48 eV < 3.0 eV → NO emission. 300 nm < 413 nm → photon energy 1240/300 = 4.13 eV > 3.0 eV → YES. K_max = 4.13 − 3.0 = 1.13 eV.