Principles of Chemical Science_Hydrogen Atom Wave functions (Orbitals) - Lec6
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Course
Chemistry
Institution
Chemistry
Lecture 6: Hydrogen Atom Wave functions (Orbitals)
1. Wavefunctions (Orbitals) for the Hydrogen Atom
2. Shape and Size of S and P Orbitals
3. Electron Spin and the Pauli Exclusion Principle
5.111 Lecture Summary #6 Monday, September 15, 2014
Readings for today: Section 1.9 – Atomic Orbitals. Section 1.10 – Electron Spin, Section 1.11 –
The Electronic Structure of Hydrogen. (Same sections in 4th ed.)
Read for Lecture #7: Section 1.12 – Orbital Energies (of many-electron atoms), Section 1.13 –
The Building-Up Principle. (Same sections in 4th and 5th ed.)
Topics: I. Wavefunctions (Orbitals) for the Hydrogen Atom
II. Shape and Size of S and P Orbitals
III. Electron Spin and the Pauli Exclusion Principle
I. WAVEFUNCTIONS (ORBITALS) FOR THE HYDROGEN ATOM
Solving the Schrödinger Equation provides values for En and Ψ(r,θ,φ).
A total of 3 quantum numbers are needed to describe a wavefunction in 3D.
1. n ≡ principal quantum number
n =
determines binding energy (energy level or shell)
2. l ≡ angular momentum quantum number
l =
l is related to n, determines angular momentum, describes subshell, shape of orbital
largest value of l = n – 1
3. m ≡ magnetic quantum number
m=
m is related to l, determines behavior in magnetic field, describes the specific orbital
To describe an orbital, we need to use all three quantum numbers:
Ψnlm(r,θ,φ)
The wavefunction describing the ground state is .
Using the terminology of chemists:
The Ψ100 orbital is instead called the orbital.
n designates the shell or energy level (1,2,3…)
l designates the subshell (shape of orbital) (s, p, d, f…)
m designates orbital orientation (specific orbital) (px, py, pz…)
=0⇒ orbital = 1 ⇒ orbital = 2 ⇒ orbital = 3 ⇒ orbital
for = 1: m = 0 is pz orbital, m = ±1 are the px and py orbitals
1
, State label wavefunction orbital H atom En H atom En[J]
n=1
=0 ψ100 –2.18 × 10–18J
m=0
n=2
=0 ψ200 -5.45 × 10–19J
m=0
n=2
=1 ψ211 -5.45 × 10–19J
m = +1
n=2
=1 210 ψ210 –RH/22 -5.45 × 10–19J
m=0
n= 2
=1 21-1 ψ21-1 –RH/22 -5.45 × 10–19J
m = -1
What is the corresponding orbital for a 5,1,0 state?
For a hydrogen atom, orbitals with the same n
value have the same energy: E = -RH/n2.
≡ having the same energy
For any principle quantum number, n, there are
degenerate orbitals in hydrogen (or any other 1 electron atom).
IN THEIR OWN WORDS
MIT graduate student Benjamin Ofori-Okai discusses how energy levels relate to
research in nanoscale MRI (magnetic resonance imaging), a technique that allows
3-D imaging of biological molecules, such as proteins, and viruses.
Image from "Behind the Scenes at MIT”. The Drennan Education Laboratory. Licensed
under a Creative Commons Attribution-NonCommercial-ShareAlike License. 2
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