Calculate the ground-state energy of the hydrogen atom using SI units and convert the result to electronvolts.
The $\mathrm{H}$ atom ground-state energy is given by (6.94) with $n=1$ and $Z=1$ as $E=-\mu e^4 / 8 h^2 \varepsilon_0^2$. Use of (6.105) for $\mu$ gives
$$
\begin{gathered}
E=-\frac{0.9994557\left(9.109383 \times 10^{-31} \mathrm{~kg}\right)\left(1.6021766 \times 10^{-19} \mathrm{C}\right)^4}{8\left(6.626070 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)^2\left(8.8541878 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}-\mathrm{m}^2\right)^2} \frac{Z^2}{n^2} \\
E=-\left(2.178686 \times 10^{-18} \mathrm{~J}\right)\left(Z^2 / n^2\right)\left[(1 \mathrm{eV}) /\left(1.6021766 \times 10^{-19} \mathrm{~J}\right)\right]
\end{gathered}
$$
$$
E=-(13.598 \mathrm{eV})\left(Z^2 / n^2\right)=-13.598 \mathrm{eV}
$$
a number worth remembering. The minimum energy needed to ionize a ground-state hydrogen atom is $13.598 \mathrm{eV}$.
