Two metalic plates A and b has the same area of `5xx10^(-4)m^(2)` are placed parallel to each other at a distance of 1 cm. Plate B carrins a positive charge of `33.7xx10^(-12)C`. A monochromatic beam of light, with photons of energy 5eV each, starts falling on plate A at t=0 so that `10^(16)` photons fell on it per square metre per second. Assume that one photoelectron is being emitted for every `10^(6)` incident photons. Assuming that all the emitted photoelectrons are collected by plate b and th work function of plat A remains constant at the value 2 eV, the numbeer of photoelectrons emitted up to t=10 s is `nxx10^(6)`. Find the value of n.
Negleect the time taken by the photoelectron to reach plate B.
(Take `epsilon_(0)=8.85xx10^(-12)C^(2)/N-m^(2)`)

Answer

`50.00`

Explanation

Area of plates `A=5xx10^(-4)m^(2)`
Distance between the plates `d=1cm=10^(-2)m`
upto=10s number of photoelectrons emitted are
`n=(("number of photons falling in unit area in unit time")xx("area"xx"time"))/(10^(6))`
`=1/(10^(6))[(10)^(16)xx(5xx10^(-4))xx(10)]`
`=00xx10^(5)`