The work fuction of caseium 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 0.60V. Given `h=6.63xx10^(-34)Js.`

Explanation

(a) `v_0=(phi_0)/h=(2.14eV)/(6.63xx10^(-34)Js)`
`=(2.14xx1.6xx10^(-19)J)/(6.63xx10^(-34)Js)=5.16xx10^(14)Hz`
(b) `eV_0=(hc)/lambda-phi_0 or lambda=(hc)/((eV_0+phi_0))`
`:. lambda=((6.63xx10^(-34))xx(3xx10^8m//s))/((exx0.6V+2.14eV))`
`= (19.89xx10^(-26)Jm)/(2.74xx1.6xx10^(-19)J)~~454nm`