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 sqare mater for every `10^(6)` incident 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 photoelectron 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, B, C

Explanation

Number of electron falling on the metal plate `A`
` = 10^(16) xx (5 xx 10^(-4))`

`:. `number of photoelectrons emitted from metal plate `A` upto `10` sec is
`n_(e) = (( 5xx 10^(4)) xx 10^(16))/(10^(6)) xx 10 = 5 xx 10^(7)`
change on plate `B` at `t = 10 sec`
` Q_(0) = 33.7 xx 10^(-12) - 5 xx 10^(7) xx 1.6 xx 10^(-19) = 25.7 xx 10^(-12) C `
`also Q_(0) = 8 xx 10^(-12) C `
` E = (sigma_(B))/(2s_(0)) - (sigma_(A))/(2s_(0)) = (1)/(2As_(0)) (Q_(B) -Q_(A))`
` = ((17.7 xx 10^(-12))/(5 xx 10^(-4) xx 8.85 xx 10^(-12)) = 2000 N//C`
`K.E of most energtic particles
` = (hc - phi) + e(Ed) = 23 eV`