Light rays of wave length $6000 A^{\circ}$ and photon intensity $39.6 \;W/m^2$ is incident on metal surface. If only $1 \%$ of photon incident on surface emit photo electrons, than the number of electrons omitted per second per unit area from the surface will be : $(h= 6.64 \times 10^{-34}$ Is ; $c= 3 \times 10^8 m/s)$