Asked by Marty
an electron in a mercury atom jumps from level a to level g by absorbing a single photon. determine the energy of the photon in joules.
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Answered by
Anonymous
Ei-Eo=delta E 10.38-2.48=7.9eV
7.9ev* (1.60*10^-19J)= 1.264* 10^-18 joule
7.9ev* (1.60*10^-19J)= 1.264* 10^-18 joule
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