Hydrogen-like Atoms and Multi-electron Atoms

Hydrogenic orbital nodes and superposition

A hydrogen-like orbital \psi_{nlm}(r,\theta,\phi) = R_{nl}(r)\,Y_l^m(\theta,\phi) is an energy eigenstate whose energy depends only on n. Eigenfunctions with different n are orthogonal, and eigenfunctions that differ in l or m are orthogonal as well.

A general state is a linear combination \Psi = \sum_i c_i \psi_i. If \Psi is normalized, \sum_i |c_i|^2 = 1. The probability of obtaining the energy E_n is the sum of |c_i|^2 over all terms that have that n. Only the squared moduli of the coefficients matter, so a complex phase such as -i does not change the probability.

Nodes are points or surfaces where \psi = 0. An orbital with principal quantum number n and azimuthal quantum number l has n-1 nodes in total, divided into two kinds:

• Radial nodes — spherical surfaces where R_{nl}(r) = 0; their number is n-l-1.
• Angular nodes — planes or cones where Y_l^m = 0; their number is l.

For example, a 3d orbital (n=3, l=2) has no radial node and two angular nodes. The 3p radial function has one radial node, at r = 6a_0/Z.

Hydrogen-like ion wavefunctions

A hydrogen-like ion has one electron and nuclear charge Z. Its time-independent Schrödinger equation is a one-body Coulomb problem, which is solved exactly for any Z. Multi-electron systems such as neutral helium do not have this closed form and need approximations.

The 1s wavefunction is \psi_{1s} = \sqrt{Z^3/(\pi a_0^3)}\,e^{-Zr/a_0}, where a_0 is the Bohr radius. The radial probability density is P(r) = 4\pi r^2 |\psi|^2 \propto r^2 e^{-2Zr/a_0}. Setting dP/dr = 0 gives the most probable distance r = a_0/Z. This distance shrinks as Z grows, because the electron is pulled closer to the nucleus.

The energy levels of such an ion are E_n = -\dfrac{Z^2 R}{n^2}, where R is the Rydberg energy with a small reduced-mass correction. Other standard results tested in the same item include the following:

• The zero-point energy of a harmonic oscillator is h\nu/2, from E_n = (n+\tfrac{1}{2})h\nu at n = 0.
• The rigid rotor energy is E_J = \hbar^2 J(J+1)/2I, which varies inversely with the moment of inertia I.
• Under LS coupling, the total angular momentum is the vector sum \mathbf{J} = \mathbf{L} + \mathbf{S}.

Hydrogen-like atom energy levels

The energy of a hydrogen-like ion depends only on Z and n: E_n = -\dfrac{Z^2}{2n^2} in atomic units (hartree). The hartree is E_h = \dfrac{e^2}{4\pi\epsilon_0 a_0}, so \dfrac{e^2}{2\pi\epsilon_0 a_0} = 2E_h. Remember this conversion when a quantity is written with 2\pi\epsilon_0 instead of 4\pi\epsilon_0.

In SI units the energy is E_n = -\dfrac{Z^2 m_e e^4}{8\epsilon_0^2 h^2 n^2}. For example, the ground state of He^+ (Z = 2) has E_1 = -2 hartree.

Point-group symmetry is tested in the same set of items. The symmetry elements of a molecule follow from its point group:

• \text{PCl}_5 (D_{3h}) has three C_2 axes perpendicular to its principal C_3 axis, so a C_2 element is present.
• \text{BF}_3 (D_{3h}) has three \sigma_v planes, the same number that \text{NH}_3 (C_{3v}) has.
• Benzene (D_{6h}) contains one \sigma_h plane and three \sigma_d planes.
• \text{CCl}_4 (T_d) and \text{SF}_6 (O_h) both contain C_2 and C_3 axes.

Angular momentum quantum numbers

For an electron with orbital quantum number l, the orbital angular momentum has magnitude |\mathbf{L}| = \sqrt{l(l+1)}\,\hbar. Its z-component is L_z = m\hbar, with m = -l, \dots, +l. If a given magnitude is stated, solve \sqrt{l(l+1)} for l first.

The angle between \mathbf{L} and the z-axis is smallest when m = l. Then \cos\theta_{\min} = \dfrac{l}{\sqrt{l(l+1)}} = \sqrt{\dfrac{l}{l+1}}. For l = 1 this gives \theta_{\min} = 45^\circ, and the angle approaches 0^\circ only as l becomes large.