The work function of caesium is 2.14 eV . Find (a) the threshold frequency for caesium , and ( b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.

Explanation

For the cut - off or threshold frequency, the energy `hv_(0)` of the incident radiation must be equal to work function ` phi_(0)` , so that ` v_(0)= (phi_(0))/h = (2.14 eV)/(6.63 xx 10^(-34)JKs)`
` = ((2.14 xx 1.6 xx 10^-19)J)/(6.63 xx10^(-34)Js)= 5.16 xx 10^(14) Hz`
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinectic energy of the emitted photoelectrons equals the potential energy e `V_(0)` by the retarding potential `V_(0)` Einstein's Photoelectric equation is
` eV_(0) = hv - phi_(0) = (hc)/lambda - phi_(0)`
or ` lambda = hc //(eV_(0)+phi_(0))`
` = ((6.63 xx 10^(-34)Js)xx(3xx10^(8)m//s))/((0.60eV+2.14eV))`
` = (19.89 xx 10^(-26)Jm)/(2.74eV)`
` lambda =(19.89 xx 10^(-26)Jm)/(2.74 xx1.6 xx10^(-19)J)=454 " nm"`