Light rays of wavelengths 6000 Å and of photon intensity 39.6 watt/m2 is incident on a metal surface. If only one percent of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be : (h = 6.64 × 10–34 J-s, velocity of light = 3 × 108 m/s)
Answer & explanation
Correct answer: option 3
$I=\frac{nhc}{Atλ}$
$39.6=\frac{n×6.6×10^{-34}×3×10^8}{1×1×6000×10^{-10}}$