A stream of photons of energy 10.6 eV and intensity `2.0 W m^(-2)` is incident on a platinum surface. Area of the surface is `1.0 xx 10^(-4) m^(2)` and its work function is 5.6 eV. 0.53 % of incident photons emit photoelectrons. Find the number of photoelectrons emitted per second and maximum and minimum energies of the emitted photoelectrons in eV. `(1eV = 1.6 xx 10^(-19)J)`

Explanation

If intensity of incident light is I, energy incident on a surface area A is IA. Hence, number of photons incident per second, `n = (IA)/(hf)`
If x% of photons help to emit photoelectrons, number of photoelectron emitted per second,
`N = n xx (x)/(100) = (IAx)/(hf.100)`
GIven, `I = 2.0 W.m^(-2), A = 1.0 xx 10^(-4) m^(2)`,
`hf = 10.6eV = 10.6 xx 1.6 xx 10^(-19) J and x = 0.53`
`:. N = (2.0 xx 1.0 xx 10^(-4) xx 0.53)/(10.6 xx 1.6 xx 10^(-19) xx 100) = 6.25 xx 10^(11)`
Minimum kinetic energy of emitted photoelectron = 0
Maximum kinetic energy, `E_("max") = hf - W_(0) = 10.6 - 5.6 = 5eV`