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

Options

A. `10^(10)` and 5 eV
B. `10^(12)` and 4 eV
C. `10^(11)` and 5 eV
D. `10^(14)` and 10 eV

Answer

C

Explanation

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)`