The energy levels of the hydrogen atom are given by
[tex]E=-13.6 \frac{1}{n^2} [eV][/tex]
where n is the level number.
For n=3 we have
[tex]E_3 = -13.6 \frac{1}{3^2}=-1.51 eV [/tex]
while for n=1 (ground state) we have
[tex]E_1 = -13.6 \frac{1}{1^2}=-13.6 eV [/tex]
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:
[tex]\Delta E= E_3-E_1 =-1.51 eV - (-13.6 eV)=12.09 eV [/tex]
In joule, this corresponds to
[tex]\Delta E= 12.09 eV \cdot 1.6 \cdot 10^{-19} J/eV = 1.93 \cdot 10^{-18}J[/tex]
