Answer
See answers.
Work Step by Step
Equation 27-15b says that for hydrogen, the energy of a level is $E_n=\frac{-13.6eV}{n^2}$.
a. This transition is an absorption, because the final state energy is higher. The energy of the photon involved is the higher energy minus the lower energy.
$$E=-13.6eV(\frac{1}{3^2}-\frac{1}{1^2})=12.1eV$$
b. This transition is an emission, because the initial state energy is higher. The energy of the photon involved is the higher energy minus the lower energy.
$$E=-13.6eV(\frac{1}{6^2}-\frac{1}{2^2})=3.0eV$$
c. This transition is an absorption, because the final state energy is higher. The energy of the photon involved is the higher energy minus the lower energy.
$$E=-13.6eV(\frac{1}{5^2}-\frac{1}{4^2})=0.31eV$$
The photon involved in the transition from n = 1 to n’ = 3 has the largest energy.
