We have obtained an equation for R(r) which does not contain (θ, φ). So it is true that the solution for a generic potential V(r) can be expressed as the product of an angular part (the known spherical harmonic) and a radial part (which depends on the quantum number l) and is the solution of Equation 5.
Eigenvalues and eigenfunctions of the hydrogen atom
And they depend only on a new quantum number n named the principal quantum number which can take the values 1,2,3,... . However the total eigenfunction also contains the angular part as proposed in Equation 3 which will simply be the spherical harmonics.
The radial part R(r) must depend on the quantum number l, because it is a solution of Equation 7 which contains l. The total eigenfunctions have the following structure (note the indexes/quantum numbers):
For each n the allowed values of the angular momentum quantum number l are l = 0, ... , n-1 and, for each l, the allowed values of the magnetic quantum numbers are m = -l, -l+1, ... , l.
It is often said that l determines the shape of the orbital and m its orientation. The orbitals are also called s, p, d, f, g for l = 0, 1, 2, 3, 4 respectively. So, when an orbital is denoted, for example, as 3d it is meant n=3 and l=2.
The functions R_{nl}(r) are tabulated. You need to remember just their general structure: R_{nl}(r) = (normalization) × (polynomial in r ÷ a) × (decaying exponential ~ exp(-r ÷ a) ).
When we use polar coordinates in 3D the element of volume dxdydz must be substituted by r^{2}sinθdrdθdφ. To integrate a function over all space one has to write:
Remembering that for all space, 0 < r < ∞ ; 0 < θ < π ; 0 < φ < 2π . If the angular and radial part are separable, the triple integral can be solved separately:
The probability density of finding an electron at distance r is called the radial distribution function and it is given by P_{nl}(r) = r^{2} |R_{nl}(r)|^{2} (Eq. 12)
The reason for the extra factor r^{2} can be seen immediately if we express the probability of finding an electron at any angle θ or φ and at a distance between R_{1} and R_{2}:
[The first term is one because spherical harmonics are normalized]
Exercise
Explain why it is possible to write Equation 5 from Equation 4.
Evaluate all the constants in Equation 8 showing that the energy levels of the hydrogen atom are $E_{n}=-{\frac {13.6}{n^{2}}}$ (where the energy is expressed in electronvolts)
What is the ionization energy of the hydrogen atom?
Plot the radial wavefunction and radial distribution function for the H orbitals 1s, 2s, 2p. Indicate if there are nodal planes.
What is the distance where it is most likely to find an electron in the ground state of the hydrogen atom?
Show that the radial equation for the H atom, the He^{1+} ion, and the Li^{2+} ion can be written as $\left\{-{\frac {\hbar ^{2}}{2\mu }}\left({\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}\right)+{\frac {\hbar ^{2}l(l+1)}{2\mu r^{2}}}-{\frac {Ze^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}\right\}R(r)=ER(r)$ where Z=1,2,3 respectively. Atoms with only one electron are hydrogen-like atoms.
How can you write the energy and the wavefunction for all hydrogen-like atoms with any value of Z?
Using your answer to Question 7, calculate the ionization potential of He^{1+}. Find the distance where it is most likely to find the electron in He^{1+}.
Solutions
In this Section it was found $-{\frac {\hbar ^{2}}{2mr^{2}}}\Lambda (\theta ,\phi )Y_{ml}(\theta ,\phi )={\frac {\hbar ^{2}}{2mr^{2}}}l(l+1)Y_{ml}(\theta ,\phi )$$\left\{-{\frac {\hbar ^{2}}{2\mu R^{2}}}\Lambda (\theta ,\phi )\right\}\psi (\theta ,\phi )=E\psi (\theta ,\phi )$
2
The ground state energy of the hydrogen atom is -13.6 eV. The energy with an electron and a proton at infinite distance is zero. The ionization energy is 13.6 eV.
Only 2s has a nodal plane for r=2a.
The general probability is $P_{nl}(r)=r^{2}|R_{nl}(r)|^{2}$. For the 1s orbital it is $P_{\mbox{1s}}(r)=Ar^{2}\exp(-2r/a)$. A is the irrelevant normalization costant. The maximum is found by setting the derivative of P_{1s}(r) = 0. We get $2r\exp(-2r/a)-2{\frac {2r}{a}}\exp(-2r/a)=0\Rightarrow r=a$.
$-{\frac {Ze^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}$ is the interaction energy between a nucleus with charge +Ze and an electron with charge -e.
In the hydrogen-like atom you have the term $-{\frac {Ze^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}$ instead of $-{\frac {e^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}$ as seen in the hydrogen atom. The solution of the hydrogen atom is valid for all hydrogen-like atoms if you substitute e^{2} with Ze^{2} in the eigenvalues and the radial functions. The modified eigenvalues are therefore $E_{n}=-{\frac {\mu Z^{2}e^{4}}{32\pi ^{2}\varepsilon _{0}^{2}\hbar ^{2}n^{2}}}$ and the modified eigenfunctions have a / Z instead of a because $a={\frac {4\mu \varepsilon _{0}\hbar ^{2}}{\mu e^{2}}}$ and $a/Z={\frac {4\mu \varepsilon _{0}\hbar ^{2}}{\mu Ze^{2}}}$
Ionization Potential = 13.6 x Z^{2} = 13.6 x 4 eV.