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 stopping potential of 0.60 V.

Explanation

Here,work function,
`phi_(0)`=2.14 eV `=3.424xx10^(19)`J
`h=6.63xx10^(-34)Js`
`1eV=1.6xx10^(-19)J`
`C=3.0xx10^(8)m//s`
Stopping potential `V_(0)=0.60`V
`therefore=V_(0)=?,lambda=?`
(a)Work function,
`phi=hv_(0)`
`therefore V_(0)(phi_(0))/(h)=(3.424xx10^(-19))/(6.63xx10^(-34))`
`therefore V_(0)=0.51644xx10^(15)`
`therefore V_(0)~~5.16xx10^(14)Hz`
(b) Photoelectric equation,
`eV_(0)=hv-phi_(0)`
`therefore eV_(0)=hv-phi_(0)`
`therefore eV_(0)=(hc)/(lambda)-phi_(0)`
`therefore eV_(0)+phi_(0)=(hc)/(lambda)`
`therefore lambda=(hc)/(eV_(0)+phi_(0))`
=`(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx0.60+3.424xx10^(-19))`
=`(19.89xx10^(-26))/(0.96xx10^(-19)+3.424xx10^(-19))` `=4.5384xx10^(-7)`
`therefore ~~ 454xx10^(-9)m=454nm`