Calculate the frequency of the photon emitted by a hydrogen atom making a transition from the $n=4$ to the $n=3$ state. Compare your result with the frequency of revolution for the electron in these two Bohr orbits.
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Step 1
Given: Energy at $n=4$ state, $E_4 = -13.6 \text{ eV}$ Energy at $n=3$ state, $E_3 = -13.6 \text{ eV}$ Energy of the photon emitted, $E_{\text{photon}} = E_4 - E_3 = -13.6 \text{ eV} - (-13.6 \text{ eV}) = 0.6611 \text{ eV}$ Show more…
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