absabsemit2(0.80).4ARRrπ=Given that 62abs2.00 10 mA−=×and 3.00 m,r=withabs4.000 photons/s,R=we find the rate at which photon is emitted to be ()228emitabsabs44(3.00m)4.000 photons/s2.83 10 photons/s(0.80)(0.80)(2.00 10 m )rAππ−===××. Since the energy of each emitted photon is ph1240 eV nm2.48 eV500nmhcEλ⋅=, the power output of source is
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