A metal surface has a work function $phi=2.4$ eV. Light of wavelength 350 nm is shone on it with a power of 6 mW uniformly over an area of 2 cm${}^2$.

Step-by-step solution

1. Compute the photon energy $E_{\text{ph}}=hu$, where $u=c/\lambda$: $E_{\text{ph}}=\frac{6.626\times 10^{-34},\text{J⋅s}\times 3\times 10^8,\text{m/s}}{350\times 10^{-9},\text{m}}=5.677\times 10^{-19},\text{J}=3.548,\text{eV}$.
2. Compute the maximum kinetic energy: $K_{max}=E_{\text{ph}}-\varphi =3.548-2.4=1.148,\text{eV}=1.84\times 10^{-19},\text{J}$.
3. Compute photon flux: ${\Phi }_{\text{ph}}=\frac{P}{E_{\text{ph}}}=\frac{6\times 10^{-3},\text{J/s}}{5.677\times 10^{-19},\text{J}}=1.058\times 10^{16},\text{photons/s}$.
4. Assuming one electron emitted per photon (100 % efficiency), the number of electrons per second is the same: $1.06\times 10^{16},\text{electrons/s}$.

Concept tested: photoelectric effect, work function, photon flux

Published 20 September 2026 · How this solution was prepared and checked