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Radial Solution of Hydrogen Lesson


Shodor > CSERD > Resources > Activities > Radial Solution of Hydrogen Lesson

  Lesson  •  Materials  •  Lesson Plan


Lesson - Radial Solution, Hydrogen Wavefunction

Schrodinger's equation applied to a single electron Hydrogen atom takes the following form:


\begin{displaymath}
\mathbf{H} \Psi(\vec{r}) \equiv
\left[ -\frac{\hbar^2}{2m}\Delta+V(\vec{r})\right]
\Psi(\vec{r}) = E \Psi(\vec{r})
\end{displaymath}

Written in terms of angular and radial portions, Schrodinger's equation is given by


\begin{displaymath}
\mathbf{H} \Psi(\vec{r}) \equiv
\left[ -\frac{p_r^2}{2m}+\fr...
...l}^2}{2mr^2}+V(\vec{r})\right]
\Psi(\vec{r}) = E \Psi(\vec{r})
\end{displaymath}

where


\begin{displaymath}
\mathbf{l} \equiv -i \hbar (r \times \nabla)
\end{displaymath}

The eigenfunctions of the angular momentum operator are a set of complex functions called the spherical harmonics $Y_l^m(\theta, \phi)$, and they have the eigenvalues


\begin{displaymath}
\mathbf{l}^2 \Psi = l(l+1) \Psi
\end{displaymath}

Using separation of variables, it is assumed that since the potential depends only on r, the solution may be written as


\begin{displaymath}
\Psi(\vec{r}) = Y_l^m(\theta,\phi) \frac{y_l(r)}{r}
\end{displaymath}

where $y_l(r)/r$ is the solution to the radial portion of the equation.

Leaving the form of the spherical harmonics for a different discussion, it is this radial solution that is now considered.

The primary force between the electron and the proton in Hydrogen is the Coulomb force, which describes the electric force between two point charges. The electron is not observed to be in the nucleus, where strong and weak nuclear forces would dominate, nor are the particles massive enough for gravitational forces to dominate. This gives us a potential of $V(r) = e^2/r$.

Using a suitable separation of variables, $x = 2 \sqrt{E}$ and $\nu = - \frac{m e^2}{\sqrt{E}\hbar^2}$


\begin{displaymath}
\mathbf{H} \Psi(\vec{r}) \equiv
\left[ \frac{d^2}{dx^2}-\fra...
...rac{\nu}{x}-\frac{1}{4}\right]
\Psi(\vec{r}) = E \Psi(\vec{r})
\end{displaymath}

While Schrodinger's original equation was in three dimensions, it has now been reduced to a single dimension. What's more, while the equation as a whole required a solution using complex numbers, the radial term can be written in terms of real numbers.

But what does this represent? The wave function given by Schrodinger's equation has been linked to the probability distribution of the electron. The actual probability density is given by


\begin{displaymath}
P(\vec{r}) = \Psi^*\Psi
\end{displaymath}

Now, consider a sphere of radius r centered on the nucleus of the hydrogen atom. If you increase the radius of the sphere, you also increase the surface area of the sphere. So, if you were to ignore for a second the dependence of $\Psi$ on $\theta$ and $\phi$, and just add up the contribution due to $y_l(r)$, you would get that the probability of finding the electron between $r-\Delta r$ and $r+\Delta r$ is


\begin{displaymath}
P(r-\Delta r, r+\Delta r)
\int_{r-\Delta r}^{r + \Delta r}{
\frac{y_l(r^{\prime})^2}{r^{\prime 2}}
r^{\prime 2}dr^{\prime}
}
\end{displaymath}

We can then interpret $y_l(r)$ as the square root of the radial probability density of the electron in the Hydrogen atom. Where $y_l(r)$ is small, we are less likely to find the electron. Where $y_l(r)$ is large, we are more likely to find the electron.

The equation for $y_l(r)$ can be solved with the usual techniques, but like many problems in physics has many solutions that will not satisfy the boundary conditions. We want the specific eigenfunctions and eigenvalues that will satisfy the boundary conditions, which are that as you get very far away from the atom it should be less likely to find the electron and that as you get very close to the center of the atom it should be hard to find the electron.

You can try building your own model of this, but be careful how you set up your solution to the radial equation. There are solutions to the equation that do not satisfy the boundary conditions, and in fact get very large as you get far away from the atom, so if you integrate the problem from the inside out, numerically it is likely that you will "jump" to these solutions which do not satisfy the boundary conditions.

It is customary to integrate from the outside in, which lets you rule out this instability in the numerical solution and force the outer boundary condition. Forcing the outer boundary condition does not do anything about the inner boundary condition, however, and it is up to you to find out what values of $\nu$ and $l$ will allow for a solution that satisfies your inner boundary condition.

You may also use a model that has been built for you. Using this model, determine the values of $\nu$ and $l$ for which a solution is possible.

What do you notice about the allowable values of $l$ and $\nu$?

What happens to the solution for values which do not match the boundary conditions?

Given the possible values for $\nu$, what does that tell you about the allowed energies of the electron in the Hydrogen atom (remember that $\nu$ was defined by $\nu = - \frac{m e^2}{\sqrt{E}\hbar^2}$)?

Observations of the energy levels of hydrogen show that the energy given off as a Hydrogen electron transitions from state n to state m is given by


\begin{displaymath}
E = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)
\end{displaymath}

Does this make sense given the solutions that you have found?


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