The work function of cesium is 2.14 eV. Find (a) the threshold frequency for cesium, and (b) the wavelength of the incident light if the photo current is brought to zero by a stopping potential 0.60 V. Given `h=6.63xx10^(-34)Js`.

Explanation

(a) For the cut-off or threshold frequency, the energy h `v_ 0` of the incident radiation must be equal to work function `phi_0`, so that
`v_0=phi_0/h=(2.14eV )/(6.63xx10^(-34) Js)`
`=(2.14xx1.6xx10^(-19)J)/(6.63xx10^(-34)J s)=5.16xx10^(14) Hz`
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e `V_0` by theretarding potential `V_0`. Einstein’s Photoelectric equation is
`eV_0=hv-phi_0=(hc)/lambda-phi_0`
or , `lambda=hc//(eV_0+phi_0)`
`=((6.63xx10^(-34) Js)xx(3xx10^(8) m//s))/((0.60 eV+2.14 eV))`
`=(19.89xx10^(-26)Jm)/((2.74 eV))`
`lambda=(19.89xx10^(-26) J m )/(2.74xx1.6xx10^(-19)J)=454 nm`