Question

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`.
A. `6.25xx10^(8)`
B. `1.25xx10^(9)`
C. `1.25xx10^(6)`
D. `6.25xx10^(11)`

Answer 1

Correct Answer - D
(d) : (d) : Incident energy `E=10.6eV=10.6xx(1.6xx10^(19))J=16.96xx10^(19)J`
`"Given":("Energy incident")/("area"xx"time")=2W//m^(2)`
`:.("Number of incident photons")/("area"xx"time")=(2)/(16.96xx10^(-19))`
`=1.18xx10^(18)`
`:.("Incident photons")/("time")=(1.18xx10^(8))xx"area"`
`=1.18xx10^(18)xx(1.0xx10^(-4))=1.18xx10^(14)`
`:.("Number of photoelectrons")/("time")=((0.53)/(100))xx(1.18xx10^(14))`
or `n=6.25xx10^(11)`