4.6: Energy Levels of Hydrogen Atom

Consider a hydrogen atom, for which the potential takes the specific form

$R(r)$ satisfies Equation $\ref{395}$, which can be written $\ref{397}$

Here, $m_e$ ) and the proton (of mass $\mu$ rotating about a fixed point. Let us write the product $P(r)$. The above equation transforms to

$\mu$ moving in the effective potential $a= \sqrt{\frac{-\hbar^2}{2\,\mu \,E}},$ $\ref{400}$

and $P(r) = f(y) \exp(-y).$

$\ref{401}$

Here, it is assumed that the energy eigenvalue $E$ is negative. Equation $\ref{398}$ transforms to

$f(y) = \sum_{n} c_n\, y^{\,n}.$ $\ref{403}$

Substituting this solution into Equation $\ref{402}$, we obtain

$\ref{404}$

Equating the coefficients of $c_n\,[n\,(n-1) - l\,(l+1)] = c_{n-1} \left [2\,(n-1) - \frac{2\,\mu\, e^2\, a}{4\pi\, \epsilon_0\, \hbar^2}\right].$

$\ref{405}$

Now, the power law series $\ref{403}$ must terminate at small $n$, otherwise $y\rightarrow 0$. This is only possible if $c_{n_{\rm min}}\,y^{\,n_{\rm min}}$. There are two possibilities: $n_{\rm min} = l+1$. The former predicts unphysical behavior of the wavefunction at $n_{\rm min} = l+1$. Note that for an $l>0$ state there is zero probability of finding the electron at the nucleus (i.e., $r=0$, except when $r\rightarrow 0$ if $y$, the ratio of successive terms in the series $\ref{403}$ is

$\sum_n \frac{(2\,y)^{\,n}}{n!},$ $\ref{407}$

which converges to $f(y)\rightarrow \exp(2\,y)$ as $y\rightarrow \infty$ . It follows from Equation $\ref{401}$ that as . This does not correspond to physically acceptable behavior of the wavefunction, since $n$. According to the recursion relation $\ref{405}$, this is only possible if

$c_n\, y^{\,n}$. It follows from Equation $\ref{400}$ that the energy eigenvalues are quantized, and can only take the values $E_0 = - \frac{\mu\, e^4}{32\pi^2\,\epsilon_0^{\,2}\, \hbar^2} = - 13.6\,{\rm eV}$ $\ref{410}$

is the ground state energy. Here, $l$, otherwise there would be no terms in the series $\ref{403}$.

The properly normalized wavefunction of a hydrogen atom is written

$R_{n\,l}(r) = {\cal R}_{n\,l}(r/a),$ $\ref{412}$

and

$a_0 =\frac{4\pi\, \epsilon_0\,\hbar^2}{\mu \,e^2} = 5.3\times 10^{-11}\,\,\,{\rm meters}$ $\ref{414}$

is the Bohr radius, and $\left[\frac{1}{x^2} \frac{d}{dx}\, x^2 \,\frac{d}{dx}-\frac{l\,(l+1)}{x^2} + \frac{2\,n}{x} - 1\right] {\cal R}_{n\,l} = 0$

$\ref{415}$

that is consistent with the normalization constraint

$Y_{l\,m}$ are spherical harmonics. The restrictions on the quantum numbers are $n$ is a positive integer, $m$ an integer.

The ground state of hydrogen corresponds to $l=0$ and $n=2$. The other quantum numbers are allowed to take the values $m=0$ or $m=-1, 0, 1$. Thus, there are $l$, despite the fact that $1/r$ Coulomb potential.

In addition to the quantized negative energy states of the hydrogen atom, which we have just found, there is also a continuum of unbound positive energy states.

Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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