A photon of 300nm is absorbed by a gas and then reemitted two photons. one of the reemitted photon has wavelength 496 nm, then the other photon has
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Energy absorbed for photon = hc/lambda
Ephoton−absorbed=[6.626×10^−34×3.0×10^8] / −9] / [300×10^−9]
=6.626×10^19 J
Energy re-emitted out for photon = hc/lambda
Ere−emitted=[6.626×10^−34×3.0×10^8] / [496×10^−9]
=4.0×10^−19J
E absorbed = E for first emitted photon + E for second emitted photon
Therefore E for second emitted photon = 2.626 * 10^-19 J
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