Answer
See answers.
Work Step by Step
Equation 27-15b says that for hydrogen, the energy of a level is $E_n=\frac{-13.6eV}{n^2}$. Combine this with equation 27–10. $E_1$, $E_2$, and $E_3$ are shown in the textbook. We will also use $E_4=-0.85eV$ and $E_5=-0.54eV$.
a. The second Balmer line is the transition from n = 4 to n = 2.
$$\lambda=\frac{hc}{E_4-E_2}=\frac{1240eV\cdot nm}{-0.85eV-(-3.40eV)}=486nm$$
b. The second Lyman line is the transition from n = 3 to n = 1.
$$\lambda=\frac{hc}{E_4-E_2}=\frac{1240eV\cdot nm}{-1.51eV-(-13.6eV)}=103nm$$
c. The third Balmer line is the transition from n = 5 to n = 2.
$$\lambda=\frac{hc}{E_4-E_2}=\frac{1240eV\cdot nm}{-0.54eV-(-3.40eV)}=434nm$$