When a beam of 10.6 eV photons of intensity 2.0 W/m² falls on a platinum surface of area 1.0 * 10 ^-4 m² and work functions 5.6 eV. 0.53% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV). Take 1 eV = 1.6 * 10 ^-19 J.

Energy of incident photon,

                E_i = 10.6 eV

                   =10.6 × 1.6 × 10^–19 J

                   = 16.96 × 10^–19 J

        Energy incident per unit per unit time (intensity) = 2J

        \ Number of photons incident on unit area in unit time

                        = 2/16.96 * 10 ^-19

                        = 1.18 × 10¹⁴

Therefore, number of photons incident per unit time on given area
(1.0 × 10^–4 m²)

                = (1.18 × 10¹⁸) (1.0 × 10^–4)

                = 1.18 × 10¹⁴

But only 0.53% of incident photons emit photoelectrons

\ Number of photoelectrons emitted per second (n)

                n = (0.53/100) (1.18 × 10¹⁴)

                n = 6.25 × 10¹¹

                K_min = 0

and   K_max = E_i – work function

                = (10.6 – 5.6) eV = 5.0 eV

\      K_max = 5.0 eV