Question

A hydrogen atom initially in the n ???? 3 level emits a photon and ends up in the ground state. (a) what is the energy of the emitted photon?

Answer 1

The energy levels of the hydrogen atom are given by

`E=-13.6 (1)/(n^2) [eV]`
where n is the level number.

For n=3 we have

`E_3 = -13.6 (1)/(3^2)=-1.51 eV`
while for n=1 (ground state) we have

`E_1 = -13.6 (1)/(1^2)=-13.6 eV`

The energy of the emitted photon is equal to the energy difference between the two levels in the transition from n=3 to n=1:

`\Delta E= E_3-E_1 =-1.51 eV - (-13.6 eV)=12.09 eV`

In joule, this corresponds to

`\Delta E= 12.09 eV \cdot 1.6 \cdot 10^(-19) J/eV = 1.93 \cdot 10^(-18)J`