Q. 544.2( 4 Votes )

Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.

Answer :

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.


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