The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level -4.85 eV to -5.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?

E = -13.6 [1/nf2 - 1/ni2]
E = -13.6 [1/32 - 1/22]
E = -13.6 [1/9 - 1/4]
E = -13.6 [-5/36]
E = 1.8889 eV
Since the energy is positive, it is emitted energy.
The energy difference is:
ΔE = Ei - Ef = -4.85 eV - (-5.51 eV) = 0.66 eV
λ = hc/E = (1240 eV·nm) / 0.66 eV = 1878.79 nm
This wavelength belongs to the Paschen series of the Hydrogen spectrum. (Paschen series lies in the infrared region, and its wavelengths range from 820 nm to ∞ nm).