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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.... |
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Answer» 14 `=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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