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)`

Explanation

Area of each plate `A=5xx10^(-4)m^(2)`
Number of photons/sec incident on plate `A=10^(16)xx5xx10^(-4)`
`=5xx10^(12)`
Number of electrons/sec emitted from `A=(5xx10^(12))/(10^(6)) =5xx10^(6)`
(a) Number of electrons emitted in 10 sec
`=5xx10^(6)xx10=5xx10^(7)`
(b) Positive charge acquired by A,
`Q_(1) =5xx10^(7)xx1.6xx10^(-19) =8xx10^(-12)C`
Charge remaining on B,
`Q_(2) =Q_(0) - Q_(1) =(33.7-8)xx10^(-12) =25.7xx10^(-12)C`

p.d. between B and A,
`V=((Q_(2) - Q_(1))//2)/(C=A in_(0)//d) =(8.85xx10^(-12))/(A in_(0)//d)`
`=(d)/(A) =(10^(-2))/(5xx10^(-4)) =20V`
`E=(V)/(d) =(20)/(0.01) =2000 V//m`
(c) Energy of incident photon `=5 eV`
`E =phi + K_(max)`
`5=2 + K_(max) implies K_(max)=2 eV`

After emmission from A, the electrons are accelerated by p.d. 20 V between A and B.
Energy acquired by electron by 20 eV
Total K.E. of electron when it reaches B
`=3 + 20 =23 eV`