The frequency of the emitted photon is given by:
[tex]f= \frac{c}{\lambda} [/tex]
where
[tex]c=3 \cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda=1879 nm=1.879 \cdot 10^{-6} m[/tex]
So we find
[tex]f= \frac{3 \cdot 10^8 m/s}{1.879 \cdot 10^{-6} m}=1.60 \cdot 10^{14} Hz [/tex]
And the energy of the emitted photon is given by
[tex]E=hf[/tex]
where [tex]h=6.6 \cdot 10^{-34} Js[/tex] is the Planck constant
f is the frequency
Substituting, we find
[tex]E=(6.6 \cdot 10^{-34} Js)(1.60 \cdot 10^{14} Hz)=1.05 \cdot 10^{-19} J[/tex]
If we convert this into electronvolts:
[tex]E= \frac{1.05 \cdot 10^{-19} J }{1.6 \cdot 10^{-19} J/eV}=0.66 eV [/tex]
This is the amount of energy lost by the electron in the orbit transition. If we have a look at the values of the orbit energy in the hydrogen atom:
https://i.stack.imgur.com/6raba.jpg
We can see that this difference of energy corresponds to the transition between the orbit n=4 (-0.85 eV) to the orbit n=3 (-1.51 eV):
[tex]\Delta E= -0.85 eV -(-1.51 eV)= 0.66 eV[/tex]
