(See attached file for full problem description)
Note on problem solving.
A handy way to calculate photon energies and wavelengths is as follows: We know
E = hc/lambda and  = hc/E
for photons. The quantity hc may be expressed as 1240 eV-nm. Thus, a 2.0-eV photon as a wavelength
lambda = (1240 eV-nm)/(2.0 eV) = 620 nm
A 1.4-nm x-ray has an energy of
E = (1240 eV-nm)/(1.4 nm) = 886 eV
This avoids the messy and cumbersome use of exponents when using Joules.
1. (A) Looking at Figure 42-1, find the energy of a photon emitted from an n = 2 to n = 1 transition, expressing the answer in eV. Now find the wavelength. Is this photon in the infrared, the visible, or the ultraviolet part of the spectrum? ( 122 nm )
(B) Photons emitted from the Balmer series of hydrogen are due to transitions of n values 3 to 2, 4 to 2, 5 to 2, and so on. Draw arrows on the diagram below, representing the first three transitions of this series.
This solution provides step by step calculations and explanations for finding the energy of a photon and drawing transitions from a Balmer series.