A 40 W ultraviolet light source of wavelength `2480 A`. Illuminates a magnesium (mg) surface placed 2 m away. Determine the number of photons emitted from the surface per second and the number incident on unit area of Mg surface per second . The photoelectric work function for Mg is `3.68eV`. Calculate the kinetic energy of the fastest electrons ejected from the surface. Determine the maximum wavelength for which the photoelectric effects can be observed with Mg surface.

Explanation

Energy of each photon, `epsilon = (hc)/(lambda)`
`= (6.6 xx 10^(-34) xx 3 xx 10^(8))/(2480 xx 10^(-10)) J = 8.0 xx 10^(-19) J`
Number of photons emitted per second
`N = (P)/(epsilon) = (40)/(8.0 xx 10^(-19)) = 5.0 xx 1019 s^(-1)`
From Einstein's photoelectric equation ,
`N_(i) = (N)/(4 pi r^(2)) = (5.0 xx 10^(19))/(4 xx 3.14 xx (2)^(2))`
`(As r = 2 m) = 9.95 xx 10^(13) s^(-1)`
From Einstein's photoelectron equation.
`E_(k) = hv - W`
`epsilon = hv = (hc)/(lambda) = 8.0 xx 10^(-19) J`
`= (8.0 xx 10^(-19))/(1.6 xx 10^(-19)) e V = 5.0 e V`
`:. E_(k) = 5.0 e V - 3.68 e V = 1.32 e V`
Threshold (or maximum) wavelength for photoelectrons emission,
`lambda_(0) = (hc)/(W) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(3.68 xx 1.6 xx 10^(-19)) m`
`= 3.373 xx 10^(-7) = 3373 Å`