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A Rydberg hydrogen atom makes a downward transition to the n equals 225 state. If the photon emitted has energy 9 32 μeV what was the original state?
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∙ 14y ago
The energy difference between two energy states of a Rydberg hydrogen atom is given by the formula: ΔE = 13.6 eV * (1/n_initial^2 - 1/n_final^2), where n_initial and n_final are the initial and final quantum numbers, respectively. Given that the energy of the photon emitted is 9.32 μeV, we convert this to electron volts (eV) and use it to calculate n_initial. Substituting the values, we find that the original state was n equals 229.
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