(Solved by Humans)-A researcher observes hydrogen emitting photons of energy 0.661
Question
A researcher observes hydrogen emitting photons of energy 0.661 eV .?What are the quantum numbers of the two states involved in the transition that emits these photons?
Energy emission of hydrogen atom for different quantum state transitions are as under:
The formula for energy emission is
E = 0.661(1/m^2 - 1/n^2), n > m
m = 1, n = 2 => E = 0.49575 eV
m = 2,...
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