Spherical harmonics are still useful in the presence of V(r)[edit | edit source]
In the past Lesson we wrote the Hamiltonian for a system of two particles with a potential energy that depends only on their distance:
We have described the solution in the particular case where r is constant (r = R):
We can hope that the solution in the general case has the following structure:
i.e. it is a product of a function of only the coordinate r times the spherical harmonics, eigenfunctions of the rigid rotor.
We are searching for the solution of:
which can be rewritten as (see Exercises to see why):
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[edit | edit source]
The potential energy between an electron and a proton is
So the radial equation to be solved is
The eigenvalues (given without proof) are:
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 hydrogen atom radial function[edit | edit source]
The functions Rnl(r) are tabulated. You need to remember just their general structure: Rnl(r) = (normalization) × (polynomial in r ÷ a) × (decaying exponential ~ exp(-r ÷ a) ).
A few examples are given below:
- 1s: Rn=1,l=0(r) = 2a-3/2exp(-r/a)
- 2s: Rn=2,l=0(r) = (2a)-3/2(2 - r/a)exp(-r/2a)
- 2p: Rn=2,l=1(r) = 3-1/2(2a)-3/2(r/a)exp(-r/2a)
where a ≈ 0.52 Å.
These are easy to plot considering that the number of nodes is n-l and that they tend to be more extended for larger values of n.
Multiple integrals involving polar coordinates[edit | edit source]
When we have to integrate a function of x,y,z over all space, we write a triple integral in this way:
When we use polar coordinates in 3D the element of volume dxdydz must be substituted by r2sinθdrdθdφ. 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:
To practice with these integrals, we will check that the 1s wavefunction of the hydrogen atom is normalized.
The wavefunction is normalized if the integral of |Ψ1s|² = 1. In this particular case
It is easy to show that both terms in the square parentheses are equal to 1.
[In the last integral we substitute r/a for x]
Radial distribution function[edit | edit source]
The probability density of finding an electron at distance r is called the radial distribution function and it is given by Pnl(r) = r2 |Rnl(r)|2 (Eq. 12)
The reason for the extra factor r2 can be seen immediately if we express the probability of finding an electron at any angle θ or φ and at a distance between R1 and R2:
[The first term is one because spherical harmonics are normalized]
- 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 (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 He1+ ion, and the Li2+ ion can be written as 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 He1+. Find the distance where it is most likely to find the electron in He1+.
- In this Section it was found
- 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 . For the 1s orbital it is . A is the irrelevant normalization costant. The maximum is found by setting the derivative of P1s(r) = 0. We get .
- 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 instead of as seen in the hydrogen atom. The solution of the hydrogen atom is valid for all hydrogen-like atoms if you substitute e2 with Ze2 in the eigenvalues and the radial functions. The modified eigenvalues are therefore and the modified eigenfunctions have a / Z instead of a because and
- Ionization Potential = 13.6 x Z2 = 13.6 x 4 eV.
Next: Lesson 8 - Operators and Measurements