AMU Engineering 2018 solved paper Solved Paper

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
=

1.06×10−6

10.6×1.6×10−19

=6.25×1011