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)`

Options

A. `2 xx 10^(3) N//C`
B. `10^(3) N//C`
C. `5 xx 10^(3) N//C`
D. Zero

Answer

A

Explanation

Number of photoelectrons emitted up to ` t = 10 sec` are :
` n = (("Number of photons per unit area per unit time ") xx ("Area" xx "Time"))/(10^(6))`
`= (1)/(10^(6)) [ (10)^(16) xx ( 5 xx 10^(-4)) xx (10)] = 5 xx 10^(7)`
At time `t = 10 sec`
Charge on plate `A , q_(A) = + ne = 5 xx 107 xx 1.6 xx 10^(-19)`
` = 8 xx 10^(-12) C = 8 pC`
and charge on plate `B , q_(B) = 33.7 - 8 = 25.7 pc`
Electric field between the plates
` E = ((qB - qA))/( 2 epsilon_(0) A) = (( 25.7 - 8) xx 10^(-12))/(2 xx 8.85 xx 10^(-12) xx 5 xx 10^(-4))`
`= 2 xx 10^(3) (N)/(C )`