Download Physical Chemistry 20130517 week 7 Friday May 17 2013

Physical Chemistry 20130517 week 7 Friday May 17, 2013 page 1
tunneling:
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scanning tunneling microscope (STM), important for analyzing surfaces, electrons tunnel from
tip to surface
electron tunneling
proton tunneling
hydrogen atoms tunneling
α particle tunneling
Tunneling is a quantum phenomenon. It has no equivalent in classical physics.
A piezoelectric substance converts pressure to voltage or voltage to pressure. Voltage makes it expand
and contract.
2 particle rigid rotor
V = potential energy = 0
One mass is m1, the other mass is m2, the distance from the center of m1 to the center of m2 is r. The
book calls r d instead. These particles can’t vibrate.
dV=-Fdr=0
since the radius doesn’t change
energy = KE + rotational energy
Erot =J(J+1)
ℏ2
2I
(no potential energy)
I is moment of inertia
J=0,1,2,3…
μ=reduced mass=
I=μr 2 for diatomic molecule
m1 m2
m1 +m2
energy levels
J=3
J=2
J=1
J=0
E=12
E=6
E=2
E=0
The E in that table is really
Notice the spacing between energy levels isn’t constant.
polar coordinates
E
to simplify.
ℏ2
( 2I )
Consider a graph with mass m1 at the origin and mass m2 somewhere else. The distance between the
two masses is r. The angle between the z axis and the radius r is θ. Project r down onto the xy plane
gives a vector with length rsinθ. The angle between the x axis and the vector rsinθ is φ. Given that, we
know:
x=rsinθcosφ
yrsinθsinφ
z=rcosθ
ψrot = ΘθΦ(φ) where Θ and Φ are functions
The formula doesn’t include r since r doesn’t change.
Θ depends on J and MJ.
Φ depends on MJ.
MJ is the component of J along the z axis.
J = 0,1,2,3,…
For J = 2 MJ can be -2, -1, 0, 1, 2.
For a given J there are 2J+1 values for MJ.
For J=1:
MJ=+1
ψrot=ψ11
MJ=0
Ψrot=ψ10
MJ=-1
ψrot=ψ1-1
Therefore each energy level J has 2N+1 degeneracy.
See approximation methods handout.
For n=1: E=
E(trial) = ∫φ*Ĥφdτ ≥ Egs
n2 h2
h2
h2
=
=.125
8ma2 8ma2
ma2
Egs is ground state energy
Ĥ is the time independent Hamiltonian
Both φ and φ* are already normalized. If they weren’t normalized, use this:
∫ φ*Ĥφdτ
≥Egs
∫ φ*φdτ
Suppose we approximate ψ(x) using x(x-a). We want to show that it fits the boundary conditions for a
particle in a box:
lim x(x-a) =0
x→a
and
̂ ψ=x(x-a)
ψH
lim x(x-a)=0
x→0
-h2 d2
x(x-a)
8π2 m dx 2
In this case the complex conjugate and the function are the same since they’re real.
a
̂ ψdx=
∫ ψH
0
a
a
a
-h2
-h2
-h2 a3 a3
2
∫
x(x-a)dx
=
x
dx
-a
∫
xdx
=
(∫
)
( - )
4π2 m 0
4π2 m 0
4π2 m 3 2
0
=
-h2 -a3
h 2 a3
=
4π2 m 6 24π2 m
That’s not normalized, so take the integral of the bottom too.
a
a
a
a
a
∫ ψ2 dx = ∫ x 2 (x-a)2 dx = ∫ x 4 dx +a2 ∫ x 2 dx -2a ∫ x 3 dx =
0
0
0
0
0
a5 a5 a 5 a5
+ - =
5 3 2 30
h 2 a3
(
) 30h2 a3
5h2
.127h2
24π2 m
E' = a
=
=
=
=
ma2
a5
24π2 ma5 4π2 ma2
∫0 ψψdx
( )
30
a
∫0 ψĤψdx
% error=
.127-.125
*100%=1.6% error (high by quantum mechanics standards)
.125
Next we’ll try approximating ψ with x2(x-a)2