A beam of electromagnetic radiation intensity `6.4xx10^(-5)W//cm^(2)` is comprised of wavelength ,`lambda`=310 nm.It falls normally on metal (work function 2 eV) of surface are `1 cm^(2)`.If one in `10^(3)` photons ejects an electron ,total number of electrons ejected in 1 s is 10^x [hc=1240 eV-nm ,1 eV=`1.6xx10^(-19)J`],the value of x is....

Options

A. 14
B. 12
C. 11
D. 10

Answer

C

Explanation

P=IA
`=6.4xx10^(-5)xx1`
`=6.4xx10^(-5)W`
`implies`E=hf
`=(hc)/(lambda)`
`=(1240)/(310)=4eV gt 2eV`
`therefore As E gt phi_(0)`, photoelectrons can be emitted.
nE=P
`therefore n=(P)/(E )=(6.4xx10^(-5))/(4xx1.6xx10^(-19))`
`therefore n=10^(14)`
`therefore`No. of electrons emitting out at every second
`(n)/(10^(3))=(10^(14))/(10^(3))`
`10^(x)=10^(11)`
`therefore x=11`