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

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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

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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

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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]

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Three Level Laser

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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

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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

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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

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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

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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