Transcript Chapter 6 Quantum Theory of the Hydrogen Atom 6.1 Schrödinger`s

Physics 1
September 27, 2005
Friday, September 30, 4:00 p.m.—special homecoming
colloquium by former student Gavin Buffington. Room 104
Physics. Coffee and donuts at 3:45. Physics 1 students want to
attend.
Three areas will be discussed: (1) modeling laser damage to
human eyes (because you can’t do it experimentally!), (2)
doing research in an undergraduate-only department, and (3)
comments on a post-UMR career.
“Education is that which remains when one has forgotten everything
learned in school. .”—A. Einstein
Chapter 6
Quantum Theory of the Hydrogen Atom
6.1 Schrödinger's Equation
for the Hydrogen Atom
We have discovered a "new" theory—quantum mechanics as
exemplified by Schrödinger's equation.
We tested it on three simple model systems in Chapter 5. We
ought to test it on something a bit more “realistic,” like an
atom.
What’s the simplest atom you can think of?
“Anyone who is not shocked by quantum theory has not understood it.”—
Neils Bohr
The hydrogen atom is the simplest physical system containing
interaction potentials (i.e., not just an isolated particle).
Simple: one proton, one electron, and the electrostatic
(Coulomb) potential that holds them together.
e2
The potential energy in this case is just V = 4πε0r
(the attractive potential between charges of +e and –e,
separated by a distance r).
This is a stationary state potential (no time dependence). We
could just plug it in to Schrödinger’s equation to get
2


ψ
2m
e

+ 2 ( E -  )ψ = 0
2
x
 4πε0r 
…but need to let the symmetry of the problem dictate our
mathematical approach.
2
The spherically symmetric potential “tells” us to use spherical
polar coordinates!
http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html
Now we can re-write the 3D Schrödinger equation
2
2
2
 ψ  ψ  ψ 2m
+ 2 + 2 + 2 (E- V )ψ = 0
2
x
y
z
in three dimensions, and in spherical polar coordinates, as
1   2 ψ 
1
 
ψ 
r
+ 2
 sinθ

2
r r  r  r sinθ θ 
θ 
 e2 
1
 2ψ 2m
+ 2 2
+ 2 ( E - )ψ = 0.
2
r sin θ φ
 4πε0r 
The preceding equation can be solved for the wave function .
It turns out that  is the product of three separate functions.
 = Rnℓ ℓmℓ mℓ ,
with these conditions on the quantum
numbers n, ℓ, and mℓ:
n = 1, 2, 3, ...
ℓ = 0, 1, 2, ..., (n-1)
mℓ = 0, 1, 2, 3, ..., ℓ
On the next slide are the possible hydrogen electron wave
functions for n=1, 2, and 3 (n=1 is the ground state).
Let’s look at the ground state wave function:
1s =
1
r/a0
e
a3/2
0
This function contains all that is “knowable” about the ground
state electron in hydrogen.
The radial part is the only “interesting” part.
R1s =
2 r/a0
e
3/2
a0
Note that  for the ground state electron is a function of r
only. You will learn in quantum mechanics that * evaluated
for some state gives the probability of finding the system in
that state.
In spherical polar coordinates, P(r)=R*R r2 dr gives the
probability of finding the electron within some small dr in
space centered at r.
Use the 1s radial wave function and Mathcad to show that the
1s electron in hydrogen is most likely to be found at r=a0.
Hints: set a0=1 for simplicity. You could use calculus to find
where the radial probability function is maximum. In Mathcad
it is easier to plot the probability function and see where the
maximum value occurs.
The quantum mechanical equivalent of the average value is
the expectation value, given by
G =  GdV
The expectation value of the radial part of the electron’s wave
function is


r =  R rR r dr =  Rr 3dr
0

2
0
Use Mathcad to show that the expectation value for the 1s
electron’s radial coordinate is 1.5a0 (i.e., 1½ times as far from
the nucleus as its most likely coordinate).
Hints: set a0=1 for simplicity. Use the “integral function,”
obtained by entering the “&” symbol in Mathcad. Mathcad
complains if you try to integrate from 0 to . If you integrate
from 0 to 50 (meaning 0a0 to 50a0 if you set a0=1), you have
gone “far enough out in r” to approach infinity.
Does it make sense that the average electron position is
different than its most likely position?
Absolutely! You may have to wait until you take Modern
Physics to see why.