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 n_i and n_f indicates the initial and final energy levels (n being the principal quantum number).

• For the given problem, n_i implies to 3 and n_f 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 × 10^{14 S-1}.