A metal surface has a work function of 2.8 eV. Light of wavelength 350 nm is shone on it with a power of 4 mW over an area of 1 cm². Calculate (i) the maximum kinetic energy of the emitted electrons, and (ii) the number of electrons emitted per second.

Step-by-step solution

1. (i) Compute photon energy: $E_{\gamma }=hc/\lambda =1240,\text{eV⋅nm}/350,\text{nm}=3.543$ eV.
2. Subtract work function: $K_{max}=E_{gamma}-phi=3.543-2.8=0.743$ eV.
3. (ii) Convert power to joules per second: $P=4$ mW=0.004 J s⁻¹.
4. Convert photon energy to joules: $E_{\gamma }=3.543\times 1.602\times 10^{-19}=5.673\times 10^{-19}$ J.
5. Calculate photons per second: $N_{\text{ph}}=P/E_{\gamma }=0.004/5.673\times 10^{-19}=7.05\times 10^{15}$.
6. Assuming one electron per photon (quantum efficiency 1), electrons per second = $7.05\times 10^{15}\approx 7.0\times 10^{15}$.

Concept tested: photoelectric effect calculations

Published 17 September 2026 · How this solution was prepared and checked