1.

A metal plate of area 1xx10^(-4)m^(2) is illuminated by a radiation 16 mW//m^(2).The work function of the metal is 5eV.The energy of the incident photons is 10 eV and only 10% of it produces photo-electrons .The number of emitted photoelectrons per second and their maximum energy respectively will be

Answer»

`10^(10)` and 5 eV
`10^(12)` and 4 eV
`10^(11)` and 5 eV
`10^(14)` and 10 eV

Solution :E=Iat `[because I=(E )/(At)]`
`=16xx10^(-3)xx1xx10^(-4)xx1=16xx10^(-7)J`
`impliesE. 10&` of `E=(16xx10^(-7)xx10)/(100)`
`therefore E.=16xx10^(-8)J`
Let number of electron emitted be n,
`impliesn=(E.)/(E )=(16x10^(-8))/(10xx1.6xx10^(-19))`
`therefore n=10^(11)`
Maximum energy =`hv-hv_(0)`


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