Planck, atomic physics, einstein AB coeffecients

Burke, Winter 2003 Last modified 1/2/2003 1:58 AM Lecture 4, Slide # 1Lecture 4: Photons and atoms • Electromagnetic modes in a box • Blackbody radiation; photons, Planck law • Photoelec

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ECE 275B © P.J Burke, Winter 2003 Last modified 1/2/2003 1:58 AM Lecture 4, Slide # 1

Lecture 4: Photons and atoms

• Electromagnetic modes in a box

• Blackbody radiation; photons, Planck law

• Photoelectric effect

• Energy spectrum of hydrogen

• Einstein A/B coefficients

• Three-level laser

• Reading: Ch 7 of Verdeyen

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E ( , , , ) = 3sin( ) ⋅ sin( ) ⋅ cos( ) ω

t i z y

x

E ( , , , ) = 1cos( ) ⋅ sin( ) ⋅ sin( ) ω

t i z y

1 k x + E k y + E k z =

E

( 2 2 2)

22

2

z y

x z

y

L

k k

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ECE 275C © P.J Burke, Winter 2003 Last modified 1/2/2003 1:58 AM Lecture 4, Slide # 3

How much energy is in the box?

1 )

,

Instantaneous energy per unit volume:

Total energy in box:

dV t

r u

2 1

2 2

2

1

E E

E n

n n

So, the amount of energy in the box can have any value.

We will show that this leads to a problem and must be wrong.

The energy in the box must be quantized: these are photons.

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Concrete example of a mode:

3

2 3

2 2

2 1

2 2

2

1 2

1

E E

n n

n

U total = x + y + z + + =

0 ,

, 0

; 1

k n

t i

L

y L

x E

t z y x

, , ,

0 ) , , , ( x y z t =

Ex

0 ) , , , ( x y z t =

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Blackbody radiation

Consider E-M field in thermal equilibrium with

matter at some temperature T

If one is inside a box, do the walls glow?

Yes.

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How is energy in the box related to temperature?

According to the equipartition theorem

from thermodynamics, every mode of the

system has an average energy <U>=(1/2)k B T.

Note: This is already a problem Energy infinite.

What is the energy per frequency, then

we will integrate over frequencies?

There are many modes per unit frequency

Each has energy k B T.

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Modes per frequency

ν ν

ε d kT N ( ) d

2

1 )

• ε(ν)dν is the energy between ν and ν+dν

• (This is the spectrum of the blackbody radiation.)

• N( ν)dν is the number of modes between ν and ν+dν.

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Modes per frequency N ) ( ν d ν

3

2

3 ( 8 )

1 )

ν ν

ε ( ) d = kT ⋅ N ( ) d

3

2 3

8 )

Experiments confirm at low frequencies only.

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Recall Boltzmann factor P( ε):

“The probability for a physical system to be in

a state with energy ε is proportional to .”

Equipartition:

T

kB

e − ε /

(This is fundamentally linked to the concept of temperature.

Take it as an absolute truth for the whole class.)

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In order to get p( ε) to be between 0, 1 we need

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In order to get p( ε) to be between 0, 1 we need

T k

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T k

kT

dE e

1 2

3

23

2

1

E

U total = 0

; 1

T k i

i

B i

ε ε

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“What if….”

2 3

U total = 2 =

3

2 1

n an integer, h “Planck’s constant”

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“What if….”

kT dE

e

dE e

E

T k E

B

2

1 2

1

3

/ 2 1

3

/ 2

1 2

3

23

U total = 2 =

3

2 1

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/

) (

n

T k nh

T k nh n

i

T k i

T

k i

i

i i

B

B i

e

e nh

e

e p

ν

ν ε

ε ε

In HW#2, you will prove:

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Planck at low frequency:

:

T k

h ν << B

1 / k T <<

h ν B

1 x

k h

h e

h

B B

T B k

− +

1 1

/

ν

υ υ

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Planck at high frequency:

:

T k

h ν >> B

Not Equipartition!

T B k h T

B k h

e

h e

h

/ /

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“What if….”

kT dE

e

dE e

E

T k E

B

2

1 2

1

3

/ 2 1

3

/ 2

1 2

3

23

U total = 2 =

3

2 1

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Modes per frequency

ν ν

2

1 )

ν ν

υ ν

π ν

π

υ ν

ν

c e

)

−

= Note: my ε(ν)/L 3 is Verdeyen’s ρ(ν).

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h T

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Intensity:

ν ν

π

υ ν

ν ε ν

c e

h L

d c

d

1

/ )

( )

4 / ( )

=emitted power per unit area

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• We have “discovered” photons.

• However, blacksmiths have known

that hot metal glows red for hundreds of years.

• The arguments are for a “box” but the energy

comes in quanta (photons) for any a.c E-M field

• Note the length of the box did not really matter.

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electrons that are liberated

from electrode A by the

• Electrons bound to metal by work function W

• If one photon is absorbed, energy of electron after being liberated is h ν−W=h(ν−ν c )

• eV 0 is the “stopping potential = h( ν-ν c )

• Slope of V 0 vs ν is h/e

(higher intensity)

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• Maxwell’s equations are still valid.

• However, the energy of any E/M wave is quantized:

ε = n h ν

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Energy spectrum of hydrogen: Emission

Atoms excited by electrical discharge.

Lines are seen at well defined wavelengths.

For example:

6 , 5 , 4 , 3

1 4

λ

m=1 Lyman series; m=2 Balmer series, etc.

Others are seen:

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Energy spectrum of hydrogen: Absorption

Certain wavelengths are strongly absorbed These are the Lyman series and at elevated temperatures the Balmer series.

What does it mean? For photons, ε = n h ν

Hydrogen energy levels are quantized.

initial final

2 2

hcR

m n

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Energy spectrum of hydrogen:

n=1

n=2

n=3 n=inf.

Lyman

Balmer

2

1 eV

6

13

n

ε

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Energy spectrum of all atoms:

n=1

n=2

n=3 n=inf.

Similar to Hydrogen, but electron-electron interactions make them different.

Pauli exclusion principle:

No two electrons can occupy the same quantum state at the same time.

But the n=1 energy level has more than one quantum state.

1s↑ 1s↓

n=2 has more states:

2s↑ 2s↓ 2p ↑(m=1) 2p ↑(m=0) 2p ↑(m=-1) 2p ↓(m=1) 2p ↓(m=0) 2p ↓(m=-1)

And so on…

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Hydrogen

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Helium

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Lithium

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Berylium

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Boron

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Carbon

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Nitrogen

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Oxygen

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Flourine

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

Neon

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n=1

n=2

n=3 n=inf.

2s↑ 2s↓ 2p↑ (m=1) 2p↑ (m=0) 2p↑(m=-1) 2p↓(m=1) 2p↓(m=0) 2p ↓(m=-1)

1s↑ 1s↓

And so on for all the elements…

Important: In reality, all the n=2 states do not have the same energy!

Also, there are selection rules: only transitions between certain

classes of states are allowed.

In a course on atomic physics, you would calculate and learn all the levels.

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Einstein A/B coeffecients

• Two-level atom (“two-level-onium”)

• Equilibrium occupation at temperature T

• Spontaneous emission rate

• Stimulated emission rate

• Absorption rate

• Relationship between all three rates

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Two-level atom (“two-level-onium”)

∆ε

ε 2

ε 1

Consider an ensemble of them.

Let N 1 be the # of atoms in state 1.

Let N 2 be the # of atoms in state 2.

Recall Boltzmann factor:

a state with energy ε is proportional to .” e − ε / kBT

T

kB

e N

N 2 / 1 = − ∆ ε /

⇒

Total

N N

Total

T k

B

e N

N e

e N

N

/

1 /

/ 2

=

⇒

In thermal equilibrium, always more in state 1 than state 2.

This will mean later that we can’t make a laser from a system

in thermal equilibrium Need a pump.

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First photon “stimulates”

emission of second photon

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Add ‘em up:

dt

dN AN

N B

N

B dt

2 2

21 1

12

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Add ‘em up:

dt

dN AN

N B

N

B dt

2 2

21 1

12

In thermal equilibrium, the average number N 2 and N 1 stay the same.

T k Total

T k

B

e N

N e

e N

N

/

1 /

/ 2

=

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Add ‘em up:

dt

dN AN

N B

N

B dt

2 2

21 1

12

T k Total

T k

B

e N

N e

e N

N

/

1 /

/ 2

=

ν ν

π

υ ν

ν

c e

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Add ‘em up:

dt

dN AN

N B

N

B dt

2 2

21 1

12

T k Total

T k

B

e N

N e

e N

N

/

1 /

/ 2

=

ν ν

π

υ ν

ν

c e

)

−

=

In thermal equilibrium, we know ρ(ν) in a box!

Einstein showed that this can only be true if:

21 3

3 21

12

8

and

c

h A

B

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Can we get gain?

I N B h

I abs = − ⋅ ⋅ ∆

∆ . υ 12 1 ( ν )

z c

I N B h

I stim em = ⋅ ⋅ ∆

∆ . . υ 21 2 ( ν )

) ( ) ( 2 1

21 ν

c

h z

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Can we get gain?

two-level-onium

∆z

) ( ) ( 2 1

21 ν

c

h z

Need N 2 >N 1 for gain

If we put all atoms into excited state and pass the wave through a few times, they will eventually all end up half excited and half ground, and we will no longer have gain (This can be shown rigorously.)

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Then back to step one

Tiêu đề	Planck, Atomic Physics, Einstein AB Coefficients
Trường học	University of Electronic Science and Technology of China
Chuyên ngành	Atomic Physics
Thể loại	Lecture
Năm xuất bản	2003
Thành phố	Chengdu

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Số trang	55
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