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
maximum speed of the emitted photoelectrons ?

Explanation

Given, `phi_(0)=2.14 eV, v =6xx10^(14)Hz`
`KE_("max")=(1)/(2)m upsilon_("max")^(2) rArr upsilon_("max")^(2)=(2KE_("max"))/(m)=(2xx0.35xx1.6xx10^(-19))/(9.1xx10^(-31))(because e = 1.6 xx 10^(-19) C)`
`upsilon_("max")^(2)=0.123 xx 10^(12) rArr upsilon_("max")=sqrt(1230xx10^(8))=35.071xx10^(4)m//s therefore upsilon_("max")=350.71 km//s`.