The Minimum Energy of a Hydrogen Atom in Quantum Mechanics
Theorem
The minimum energy of a hydrogen atom is as follows.
$$ E_{min}=-\frac{1}{2}mc^2\alpha^{2} $$
Here $m$ is the mass of the hydrogen atom, $c$ is the speed of light, and $\alpha$ is the fine structure constant.
Explanation
Here $\alpha$ is the fine structure constant defined by $\alpha = \dfrac{e^2}{\hbar c}$, and its value is $\alpha\simeq\dfrac{1}{137}$.
Proof
The energy $E$ of a hydrogen atom is
$$ \begin{align*} E &= \frac{p^2}{2m}-\frac{e^2}{r} \\ &= \frac{1}{2m}\frac{{\hbar}^2}{r^2}-\frac{e^2}{r} \end{align*} $$
By the uncertainty principle, $$pr \simeq \hbar$$ To find where $E$ is at its minimum, $$ \left. \frac{\partial E}{\partial r} \right|_{r=r_{0}}=0 -\frac{{\hbar}^2}{m{r_{0}}^3}+\frac{e^2}{{r_{0}}^2}=0 -\frac{{\hbar}^2}{m}+e^2r_{0}=0 r_{0}=\frac{{\hbar}^2}{me^2} $$
$$ \begin{align*} \implies E_{min} &= \frac{{\hbar}^2}{2m}\frac{m^2e^4}{{\hbar}^4}-\frac{me^4}{{\hbar}^2} \\ &= \frac{me^4}{2{\hbar}^2}-\frac{me^4}{{\hbar}^2} \\ &= -\frac{1}{2}\frac{me^4}{{\hbar}^2} \\ &= -\frac{1}{2}\frac{me^4{\color{blue}{c^2}}}{{\hbar}^2{\color{blue}{c^2}}} \\ &= -\frac{1}{2}mc^2\alpha ^2 \end{align*} $$
Therefore, the minimum energy of a hydrogen atom is
$$ E_{min}=-\frac{1}{2}mc^2\alpha ^2 $$
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