Given, intensity, I=2W∕m2 Area of metallic surface (A)=1×10−4m2 Hence, power incident over metallic surface area =I×A =2×1×10−4W =2×10−4W ∴ Energy incident over metallic surface per second =2×10−4J Energy required to produce photoelectrons per second =0.53% of 2×10−4 =
0.53×2×10−4
100
=1.06×10−6J Hence, number of photoelectrons emitted per second