Two metallic plate `A` and `B` , each of area `5 xx 10^(-4)m^(2)`, are placed parallel to each at a separation of `1 cm` plate `B` carries a positive charge of `33.7 xx 10^(-12) C` A monocharonatic beam of light , with photoes 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 fall on it per square meter per second. Also assume that all the emitted photoelectrons are collected by plate `B` and the work function of plate `A` remain constant at the value `2 eV` Determine
(a) the number of photoelectrons emitted up to `i = 10s,`
(b) the magnitude of the electron field between the plate `A` and `B` at `i = 10 s, and`
(c ) the kinetic energy of the most energotic photoelectrons emitted at `i = 10 s ` whenit reaches plate `B`
Negilect the time taken by the photoelectrons to reach plate `B` Take `epsilon_(0) = 8.85 xx 10^(-12)C^(2)N- m^(2)`

Answer

A

Explanation

Area of plates,`= xx10^(-4) m^2`
Distance between the plates, d = 1 cm = 10^(-2) m`
(a) Number of photoelectrons emitted upto t = 10 s are
`n = ((Number of photons falling on unit area in unit time )xx(areaxxtime))/(10^6)`
`=(1)/(10^6) [(10)^16xx(5xx10^(-4))xx(10)]`
`5.0xx10^7`
(b) At time t =10 s,
Charge on plate A, `q_A = + ne (1.0xx10^7)(1.6xx10^(-19)`
`=8.0xx10^(-12) C`
`:. Electric field between the plates, E = ((q_B - q_A)/(2Aepsilon_0))`
or `E = ((25.7- 8.0xx10 ^(-12) /(2xx(5xx10^(-4)(8.85xx10^(-12))`
`= 2xx10^3 N//C`
(c ) Energy of most energetic photoelectrons at plate 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 plate B
`= (20+3) eV = 23 eV`