Answer
The photon is part of the Paschen series, and is emitted when an electron falls from the n= 4 state down to the n = 3 state in hydrogen.
Work Step by Step
Use the relationship for photons, E = pc, to find the energy.
$$E=pc$$
$$=(3.53\times10^{-28}kg \cdot m/s)(3.00\times 10^8 m/s)(\frac{eV}{1.60\times10^{-19}J})$$
$$=0.662 eV$$
From Figure 27–29, we see that all Lyman series photons have energy greater than 10.2 eV. All Balmer series photons have energy greater than 1.9 eV. This photon is not part of those series.
The 0.662 eV photon is part of the Paschen series, and is emitted when an electron falls from n= 4 down to n = 3.
To verify that, calculate the energy of a photon falling from the n = 4 state down to the n = 3 state in hydrogen.
$$\Delta E=E_4-E_3=-(13.6eV)(\frac{1}{4^2}-\frac{1}{3^2})\approx0.661eV$$
