Two metallic plates A and B each of area `5xx10^(-4) m^2` , are placed parallel to each other at a separation of 1 cm . Plate B carries a positive charge of `33.7xx10^(-12) C`. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that `10^16` photons fall on it per square meter per second. Assume that one photoelectron is emitted for every `10^6` incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work - function of plate A remains constant at the value 2 eV. Determine (a) the number of photoelectrons emitted up to t = 10 s.
(b) the magnitude of the electric field between the plate A and B at t = 10s and
(c ) the kinetic energy of the most energetic photelectron emitted at t=10s when it reaches plate B. Neglect the time taken by the photoelectron to reach plate B. `(Take, epsilon_0 = 8.85xx10^(-12) c^2 | N-m^2).`

Answer

(i) `5 xx 10^(2)` (ii) `2 xx 10^(3) N//C` (iii) `23 eV`

Explanation

Area of plates `A=5xx10^(-4) m^(2)`
distance between the plates `d=1 cm=10^(-2) m`
(i) Number of photoelectrons emitted upto `t=10 s` are
(No. of photons falling)
`n=("in unit area in unit time")/10^(6) xx("area"xx"time")`
`=1/10^(6) [(10)^(16)xx(5xx10^(-4))xx(10)]=5.0xx10^(7)`
(ii) At time `t=10s` charge on plate A,
`q_(A)=+n e=(5.0xx10^(7))(1.6xx10^(-19))=8.0xx10^(-12) C`
and charge on plates B,
`q_(B)=(33.7xx10^(-12)-8.0xx10^(-12))=25.7xx10^(-12) C`
`:.` Electric field between the plates `E=((q_(B)-q_(A)))/(2A in_(0))`
`implies E=((25.7-8.0)xx10^(-12))/(2xx(5xx10^(-4))(8.85xx10^(-12)))=2xx10^(3) N/C`
(iii) Energy of photoelectrons at plates A
`=E-W=(5-2) eV=3eV`
Increase in energy of photoelectrons
=(eEd) joule =(Ed) eV
`=(2xx10^(3))(10^(-2))eV=20 eV`
Energy of photoelectrons at plates B
`=(20+3)eV=23 eV`