When light of wavelength 2200 `Å` falls on Cu, photo electrons are emitted from it. Find (i) the threshold wavelength and (ii) the stopping potential. Given : the work function for Cu is `phi_(0)` = 4.7 eV.

Explanation

(i) The threshold wavelength is given by
`lambda_(0) = (hc)/(phi_(0)) = (6.626 xx 10^(-43) xx 3 xx 10^(8))/(4.7 xx 1.6 xx 10^(-19)) = 2643 Å`
(ii) Energy of the photon of wavelength 2200 `Å` is
`E = (hc)/(lambda) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(2200 xx 10^(-10))`
`= 9.035 xx 10^(-19)` J = 5.65 eV
We know that kinetic energy of fastest photo electron is
`K_(max) = h upsilon - phi_(0) = 5.65 - 4.7 = 0.95 eV`
From equation, `K_(max) = eV_(0)`
`V_(0) = (K_(max))/(e) = (0.95 xx 1.6 xx 10^(-19))/(1.6 xx 10^(-19))`
Therefore, stopping potential = 0.95 V