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. K.E. of emitted electron is

Direction : The questions 64, 65 and 66 are based on the following par

1.	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. K.E. of emitted electron is
Answer» `0.92xx10^(-19)J` `0.558xx10^(-19)J` `1.2xx10^(-19)J` `1.7xx10^(-19)J` SOLUTION :`phi_(0)`(WORK function) `=2.14eV=3.42xx10^(-19)J` `K.E.=hv-phi_(0)` `=(6.63xx10^(-34)xx6xx10^(14)-3.42xx10^(-19))J` `=0.558xx10^(-19)J`

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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. K.E. of emitted electron is

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