When a beam of 10.6 eV photons of intensity 2.0 ~W / m ^2 falls on a platinum surface of area 1.0 10⁻⁴ ~m ^2 and work function 5.6 eV . 0.53 % of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is

Step-by-step solution

Energy of incident photons, aligned E₁ & =10.6 eV =10.6 1.6 10⁻¹⁹ ~J & =16.96 10⁻¹⁹ ~J aligned Energy incident per unit area per unit time (intensity) =2 ~J Number of photons incident on unit area in unit time aligned & = 2 16.96 10⁻¹⁹ & =1.18 10¹⁸ aligned Therefore, number of photons incident per unit time on given area. aligned (1.0 10⁻⁴ ~m ^2 ) & & = (1.18 10¹⁸ ) (1.0 10⁻⁴ ) & =1.18 10¹⁴ aligned But only 0.53 % of incident, photons emit photoelectrons. Number of photoelectrons emitted per second, aligned & n= (