When a beam of 10.6 eV photons of intensity `2.0 W //m^2` falls
on a platinum surface of area `1.0xx10^(-4) m^2` and work function 5.6 eV, 0.53% of
the incident photons eject photoelectrons. Find the number of photoelectrons
emitted per second and their minimum and maximum energy (in eV). Take
`1 eV = 1.6xx 10^(-19) J`.

Answer

A, B

Explanation

Number of photoelectrons emitted per second
`=((instensity)(Area)/(Energy of each photon) xx(0x53)/(100)`
`=(2.0) (1.0xx10^(-4)xx(10.6)xx1.6xx10^(-19) xx(0.53)/(100)`
`=6.25xx10^11`
Minimum kinetic energy of photoelectrons,
`K_(min) = 0`
and maximum kinetic energy is, `K_(max = E-W = (10,6 - 5.6)eV`
= 5.0 eV.