Photoelectric Effect

$\lambda = 4\times10^{-7} m,\ \nu = \frac{c}{\lambda} = 0.75\times10^{15} s^{-1}$

$E_{photon} = h \nu = 5\times 10^{-19} J = 3.1\ eV > 2.25\ eV$ hence photon can kick out electron,

$\frac{N_{photons}}{m^2 s} = \frac{10^{-9} J}{Ephoton} = \frac{10^{-9}}{5e-19} = \frac{2\times10^9}{m^2 s},$

3% of photons will kick out electrons, hence $\frac{N_{electrons}}{m^2 s} = 0.03 * \frac{2\times 10^9}{m^2 s} = 6\times 10^7 \frac{electrons}{m^2 s}$