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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 ; $c= 3 \times 10^8 m/s)$

$\begin{array}{1 1} 12 \times 10^{18} \\ 10 \times 10^{18} \\ 12 \times 10^{17} \\ 12 \times 10^{16} \end{array} $

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