1.

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 kinetic energy of the emitted electrons

Answer» 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`


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