Answer
See the detailed proof below.
Work Step by Step
Starting with Equation (38.29):
$$ E_n = \frac{1}{2} m \left( \frac{\hbar^2}{m^2 a_B^2 n^2} \right) - \frac{e^2}{4 \pi \epsilon_0 n^2 a_B} $$
Simplifying the first term:
$$ E_n = \frac{\hbar^2}{2 m a_B^2 n^2} - \frac{e^2}{4 \pi \epsilon_0 n^2 a_B} \tag 1$$
Recalling the Bohr radius which is given by
$$ a_B = \frac{4 \pi \epsilon_0 \hbar^2}{m e^2}$$
Substitute this into the first term in (1);
$$ E_n = \frac{\hbar^2}{2 m \left( \dfrac{4 \pi \epsilon_0 \hbar^2}{m e^2} \right) n^2} - \frac{e^2}{4 \pi \epsilon_0 n^2 a_B} $$
Simplifying the first term again;
$$ E_n = \frac{e^2}{8 \pi \epsilon_0 n^2 a_B} - \frac{e^2}{4 \pi \epsilon_0 n^2 a_B} $$
$$E_n=\dfrac{e^2}{4 \pi \epsilon_0 n^2 a_B}\left[\dfrac{1}{2}-1\right]$$
$$E_n=-\dfrac{1}{2}\dfrac{e^2}{4 \pi \epsilon_0 n^2 a_B} $$
This simplifies to:
$$E_n= \dfrac{-e^2}{8 \pi \epsilon_0 n^2 a_B} $$
Manipulating:
$$ \boxed{E_n = -\frac{1}{n^2} \left( \frac{1}{4 \pi \epsilon_0} \frac{e^2}{2 a_B} \right) }$$
