The work function of caesium metal is 2.14 eV . When light of frequency `6xx 10^(14)` Hz is incident on the metal surface , photomission of electrons occurs . What is the
stopping potential and

Explanation

Given, `phi_(0)=2.14 eV, v =6xx10^(14)Hz`
`KE_("max")=hv-phi_(0)=(6.63xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))-2.14 therefore KE_("max")=0.35 eV`
(b) `KE_("max")=eV_(0)rArr 0.35 eV=eV_(0) therefore V_(0)=0.35 V`