Ultraviolet light of wavelength 300 nm and intensity 1.0 W m - 2 falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 c m 2 of the surface is nearly

Step-by-step solution

Energy incident over 1   c m 2 = 1.0 × 10 - 4   J ; Energy of photon E = h c λ Total no of photons falling = 1.0 × 10 − 4 h c λ = 10 − 4 × 300 × 10 − 9 6.6 × 10 − 34 × 3 × 10 8 No of photoelectrons emitted per second = 1 100 × 10 − 4 × 300 × 10 − 9 6.6 × 10 − 34 × 3 × 10 8 = 1.51 × 10 12