determine the energy in ev of the photon emitted when an electron jumps down from the n = 3 orbit to the n = 2 orbit of a hydrogen atom.
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In case of hydrogen atom, the energy of the emitted photon can be calculated by the formula
where n= number of states
Here electron jump from n=3 orbit to n=2 orbit
so, initial orbit of electron ni =3
final orbit nf= 2
In our case the energy of photon become, E=E3-E2
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