1.

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....

Answer»

14
12
11
10

Solution :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`


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