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Direction : The questions 64, 65 and 66 are based on the following paragraph: The work function of caesium metal is 2.14 eV. When light of frequency 6xx10^(14)Hz is incident, on the metal surface, the photoemission of electrons starts. Then max. speed of emitted electron is

Answer»

`12XX10^(4)m//s`
`6xx10^(5)m//s`
`350xx10^(3)m//s`
`10^(8)m//s`

Solution :`1//2 mv_("max")^(2)=0.558xx10^(-19)J`
then `v_("max")=350xx10^(3)m//"sec"`


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