Question

when a beam of `10.6 eV` photons of intensity `2.0 W//m^(2)` falls on a platinum surface of area `1.0 xx 10^(4) m^(2)` and work function `5.6 eV , 0.53 %` of the incidentphotons eject photoelectrons find the number of photoelectrons emited per second and their minimum energies (in eV)Take `1 eV= 1.6 xx 10^(-19) J`

Answer 1

Correct Answer - A::B
No of photons // sec
`= (Energy incident on platimum surface per sec and )/(Energy of one photon )`
No , of photons incident per second
`= (2 xx 10 xx 10^(-4))/(10.6 xx 1.6 xx 10^(-19)) = 1.18 xx 10^(14)`
As `0.53% ` of incident photon can eject photoelectron
:. no of photoelectron eject per second
`= 1.18 xx 10^(14) xx (0.53)/(100) = 6.25 xx 10^(11)`
`minimum energy = 0eV`
`Maximum energy = (10.6 - 5.6 = 5 eV`