The photo electric threshold for a metal surface is 5800 Å. Calculate the kinetic energy of a photoelectron, when a light of wavelength 4500 Å is incident on the metal surface.

Explanation

Given, `lambda_(0) = 5800" Å" = 5.8 xx10^(-7) m`
` lambda= 4500" Å"=4.5xx10^(-7)m`
`E = hv = (hc)/ lambda `
`= (6.625 xx 10 ^(-34)xx 3 xx10^(8) )/(4.5xx10^(-7))=4.416 xx 10 ^(-19)`J
Work function of the photo metal is given by
`w = (hc)/lambda_(0) = (6.625xx10^(-34)xx3xx10^(8))/(5.8xx10^(-7)) = 3.426 xx 10^(-19) ` J
From Einstein photo electric equation,
`E = w + E_(k)`
i.e., `E_(k) = (4.416 - 3.426)xx 10^(-19)=0.99xx10^(-19)` J.