Light of wavelength 310 nm falls on a metal of work function 2.5 eV. Taking hc = 1240 eV nm, find the maximum kinetic energy of the photoelectrons, the stopping potential and the threshold wavelength.
- Photon energy = hc/λ = 1240 ÷ 310 = 4.0 eV.
- KEmax = 4.0 − 2.5 = 1.5 eV, so the stopping potential is 1.5 V.
- Threshold wavelength = hc/φ = 1240 ÷ 2.5 = 496 nm; longer wavelengths free no electrons.
Answer: 1.5 eV; 1.5 V; 496 nm.