Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.
Electronic transitions in the hydrogen atom are given by the Rydberg’s equation:
Where 
is the Rydberg constant.
And ni and nf indicates the initial and final energy levels (n being the principal quantum number).
• For the given problem, ni implies to 3 and nf implies to 2
Hence,
= 
=
× 0. 1389
= - 3.02802 × 10-19 J/H atom.
Here, energy is released or emitted.
When it comes to the frequency of the photon emitted, the energy would be taken in terms magnitude only,
, where, h= 6.626 × 10-34 J S, Planck’s constant and 
is the frequency of the radiation (energy) emitted (photon) due to the transition of the electron.
Hence, 
= 
=4.569 × 1014 S-1.