View
33
Download
0
Category
Tags:
Preview:
DESCRIPTION
Physics 3313 - Lecture 13. Monday March 23, 2009 Dr. Andrew Brandt. Loose ends from Ch. 4 Nuclear Motion+Lasers QM Particle in a box Finite Potential Well. Nuclear Motion. In Bohr atom, we have implicitly assumed nucleus is fixed, since we only considered electron KE. - PowerPoint PPT Presentation
Citation preview
3313 Andrew Brandt 1
Physics 3313 - Lecture 13
3/23/2009
Monday March 23, 2009Dr. Andrew Brandt
1. Loose ends from Ch. 4 Nuclear Motion+Lasers2. QM Particle in a box3. Finite Potential Well
3313 Andrew Brandt 2
Nuclear Motion• In Bohr atom, we have implicitly assumed nucleus
is fixed, since we only considered electron KE. • Since nucleus does not (quite) have infinite mass, it
will have motion if the total momentum of the atom is zero
• pe+pN=0 so pe=-pN with KEN= p2/2M
• where the reduced mass is defined as
• For M=m get but in the case of Hydrogen so • What about heavier atom?
3/23/2009
22 2 2
2 2 2 2e
e e
p M mp p pKEM m Mm
e
e
m Mm M
2000p eM m
2
/ 2M MM M
220002000
ee
e e
mm
m m
3313 Andrew Brandt 3
Electron Transition Example• An electron makes a transition from n=2 state to n=1, find , of emitted
photon
3/23/2009
2 2
1 1 1
f i
Rn n
1 1 31
4 4RR
7 1
4 121.53 1.097 10
nmx m
cv
3313 Andrew Brandt 4
Laser• Light Amplification by Stimulated Emission of Radiation• laser light is monochromatic (one color), coherent (all in phase), can be
very intense, small divergence (shine laser on mirror left on moon) [I knew I forgot something]
3/23/2009
3313 Andrew Brandt 53/23/2009
Three Level Laser
3313 Andrew Brandt 6
Particle in a Box Again• http://user.mc.net/~buckeroo/PODB.html
• Solutions:
• Would have cos also, but boundary condition at x=0 implies coefficient =0• Use boundary condition at x=L gives
• which is equivalent to classical expression
• Final wave function
3/23/2009
2mE L n
2 2 2
22nnEmL
3313 Andrew Brandt 7
Particle in a Box Example• What is the probability that a particle in a box is
between 0.45L and 0.55L for n=1? n=2?• What is the classical answer? • 10% since this is 1/10 of the length of the box
• with
• Integrating gives
• for n=1 P=0.198 (about twice expectation), while for n=2 P=0.0065
• Does this make sense?
3/23/2009
2
1
2 22 sinx
nx
n xP dx dxL L
2 1sin 1 cos 22
2
1
.55
.45
1 2sin2
x
x
x n xPL n L
Probability
Wave Function
3313 Andrew Brandt 8
Particle in a 3-D Box• Need 3-D Schrodinger Equation:
• Factorizes into product of 3 1-D equations so
• Note 3-fold degeneracy for one dimension in n=2 state
• For rectangular box
3/23/2009
2
2 , ,2h U x y z Em
1 2 3
2 22 2 21 2 322n n nE n n n
mL
112 121 211E E E
1 2 3
22 22 231 2
2 2 21 2 32n n n
nn nEm L L L
3313 Andrew Brandt 9
Finite Potential Well• Classically if E<U than particle bounces off sides, but quantum mechanically, particle can penetrate into regions I and III• For I rewrite as
• With
• Solutions are and
• What about in the box? Since U=0 with (this is similar to infinite potential well, aka particle in
box)
3/23/2009
2
2 2
2 0m E Ux
2
2ma U E
ax axI Ce De ax ax
III Fe Ge
ikx ikxII e Be
2
2mEk
22
2 0ax
2
2 2
2 0mEx
3313 Andrew Brandt 10
Finite Well BC• At x=- =0 so for D must be 0 ,so
• Similarly at x=+ =0 so for F must be 0 and
• Finally what about boundary conditions for ? Is it 0 at x=0?
• Nope at x=0
• And at x=L • But so too many unknowns! Should we quit?• Need other constraints. Derivatives must also be continuous (match
slopes)
• After some math again get specific energy levels, but wavelengths a little longer than infinite well and from De Broglie, this means momentum and thus energy is smaller
3/23/2009
ax axI Ce De
ax axIII Fe Ge
axIII Ge
ikx ikxII e Be
II
I IIC aL
II III Ge
axI Ce
Recommended