04 quantum theory

8/15/2019 04 Quantum Theory

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Quantum Theory of Matter


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2 2

20

1 v4

e

e mF

r r πε = =

The Classical Atomic ModelConsider an atom as a planetary system.The Newton’s 2 nd Law force of attraction onthe electron by the nucleus is:

where v is the tangential velocity of theelectron:

The total energy is then:

0

v4

e

mr πε =

221 1

2 20

v4

eK m

r πε = =

This is negative, sothe system is bound,which is good.

This model fails since accelerating charges radiate and loose energy.


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The Bohr Model of the Hydrogen Atom

Bohr’s general assumptions:

1. Stationary states , in which orbiting

electrons do not radiate energy, exist inatoms and have well-defined energies,

E n. Transitions can occur between them,yielding light of energy:

E = E n − E n’

= hν 2. Classical laws of physics do not applyto transitions between stationary states,but they do apply elsewhere.

3. The angular momentum of the n th state is:where n is called the Principal Quantum Number.

hn

n = 1

n = 3

n = 2

Angularmomentum is

quantized!


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Consequences of the Bohr Model

The angular momentum is:

hnr m L == v

0v 4

e

mr πε =

mr n / v h=

04

2 2 2

2 2

n e

m r mr πε =

hBut: So:

Solving for r n: 2

0nr n a=

So the velocity is:

00

4 22a me

πε ≡

hwhere:

a 0 is called the Bohr radius. It’s the diameter of the Hydrogenatom (in its lowest-energy, or “ground,” state).

a 0


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The HydrogenAtom Energies

So the energies of the stationarystates are:

where E 0 = 13.6 eV .

r

e E

0

2

8πε

−−=

04 2 2

n 2

nr

me

πε =

h

Use the classicalresult for the

energy:

and:

E n = − E 0 /n 2or:


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Transitionsin the

Hydrogen

AtomThe atom will remain in

the excited state for ashort time beforeemitting a photon and

returning to a lowerstationary state.


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Limitations of theBohr Model

Works only for single-electron (“hydrogenic”) atoms.

Could not account for the intensities or the fine structure ofthe spectral lines (for example, in magnetic fields).

Could not explain the binding of atoms into molecules.

Failures:

The Bohr model was a greatstep in the new quantumtheory, but it had its limitations.


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Quantum Mechanics and theSchrödinger Wave Equation

The Schrödinger wave equation in its time-dependent form for aparticle of energy E moving in a potential V in one dimension is:

where i is the square root of -1.

The Schrodinger Equation is THE fundamentalequation of Quantum Mechanics.

where V = V ( x,t )


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Normalization and ProbabilityThe probability P ( x) dx of a particle being between x and x + dx isgiven in the equation

The probability of the particle being between x1 and x2 is given by

The wave function must also be normalized so that the probabilityof the particle being somewhere on the x axis is 1.


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Application of the Schrödinger Equation tothe Hydrogen AtomThe potential energy of the electron-proton system is electrostatic:

The three-dimensional time-independent Schrödinger Equation hasthe solution (in spherical coordinates):


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Quantum Numbers

The four quantum numbers:n : Principal quantum number

ℓ: Orbital angular momentum quantum numberm ℓ: Magnetic (azimuthal) quantum numberm s: magnetic spin quantum number

The restrictions for the quantum numbers:n = 1, 2, 3, 4, . . .

ℓ = 0, 1, 2, 3, . . . , n − 1m ℓ = − ℓ, − ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓm s = ±½.

The energy levels for H are:


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Probability Distribution Functions

The probability density for the hydrogen atom for three differentelectron states.


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Multi-electron atoms

When more than one electron is involved, the potential and thewave function are functions of more than one position:

1 2( , ,..., ) N V V r r r = r r r 1 2( , ,..., , ) N r r r t Ψ = Ψ r r r

Solving the Schrodinger Equation in this case can be very hard.

But we can approximate the solution as the product of single-particle wave functions:

1 2 1 1 2 2( , ,..., , ) ( , ) ( , ) ( , ) N N N r r r t r t r t r t Ψ = Ψ Ψ Ψr r r r r r

L

And it turns out that we’ll be able to approximate each Ψ i with aHydrogen wave function.


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Pauli Exclusion Principle

To understand atomic spectroscopic data, Pauli proposed hisexclusion principle:

No two electrons in an atom may have the same set ofquantum numbers ( n , ℓ , m ℓ , m s )....

It applies to all particles of half-integer spin, which are called fermions ,and particles in the nucleus are also fermions.

The periodic table can be understood by two rules:

The electrons in an atom tend to occupy the lowest energylevels available to them.The Pauli exclusion principle.


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Atomic Structure

Hydrogen : (n , ℓ, m ℓ, m s ) = (1, 0, 0, ±½) in ground state.In the absence of a magnetic field, the state m s = ½ is degenerate with

the m s = −½ state.

Helium : (1, 0, 0, ½) for the first electron.(1, 0, 0, −½) for the second electron.

Electrons have anti-aligned ( m s = +½ and m s = −½) spins.

The principle quantum number also has letter codes.n = 1 2 3 4...Letter = K L M N…

n = shells (eg: K shell, L shell, etc.)n ℓ = subshells (e.g.: 1 s , 2 p , 3 d )

Electrons for H and He atomsare in the K shell.

H: 1 s

He: 1 s 2


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Atomic StructureHow many electrons may be in each subshell?

Recall: ℓ = 0 1 2 3 4 5 …letter = s p d f g h …

ℓ = 0, ( s state) can have two electrons.ℓ = 1, ( p state) can have six electrons, and so on.

The lower ℓ values have more elliptical orbits thanthe higher ℓ values.

Electrons with higher ℓ values are moreshielded from the nuclear charge.

Electrons with higher ℓ values lie higher inenergy than those with lower ℓ values.

4s fills before 3 d .

Total

For each m ℓ: two values of m s 2

For eachℓ: (2

ℓ+ 1) values of m ℓ 2(2

ℓ+ 1)


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ThePeriodic

Table


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Motion in atoms and molecules

Electrons vibrate in their motion around nucleiHigh frequency: ~10 14 - 10 17 cycles per second.

Nuclei in molecules vibratewith respect to each other

Intermediate frequency:~10 11 - 10 13 cycles per second.

Nuclei in molecules rotateLow frequency: ~10 9 - 10 10 cycles per second.

Rotational and vibrational levels are also quantized.


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Vibrational States

A vibrational energy mode can also be excited.Thermal excitation of a vibrational mode can occur.

It is also possible to stimulate vibrations in molecules with light.

Assume that the two atoms arepoint masses connected by amassless spring with simpleharmonic motion.


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The energy levels are those ofa quantum-mechanical oscillator.

The frequency of atwo-particle oscillator is:

where the reduced mass is µ = m 1m 2 / ( m 1 + m 2) and the spring constantis κ .If it’s a purely ionic bond, we can compute κ by assuming that the forceholding the masses together is Coulomb.

Vibrational States

and


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Rotational States

Molecular spectroscopy :

We can learn about molecules by

studying how molecules absorb,emit, and scatter light.

A diatomic molecule may be thought of as two atoms held togetherwith a massless, rigid rod ( rigid rotator model ).

In a purely rotational system, the kinetic energy is expressed interms of the angular momentum L and rotational inertia I .


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Rotational States

L is quantized.

The energy levels are

E rot varies only as a function of the quantumnumber ℓ .


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Vibration and Rotation Combined

It’s possible to excite rotational and vibrational modes simultaneously.Total energy of simple vibration-rotation system:

Vibrational energies are spaced at regular intervals.

Transition from ℓ + 1 to ℓ :Photons will haveenergies atregular intervals:


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∆ E increaseslinearly with ℓ.

VibrationandRotation

Combined


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In what energy levels do molecules reside?Boltzmann population factors

N i is thenumberdensity ofmolecules instate i (i.e.,the number

of moleculesper cm 3).

T is thetemperature,and k B isBoltzmann’sconstant.

[ ]exp / i i B N E k T −

E n e r g y

Population density

N 1

N 3

N 2

E 3

E 1

E 2


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Blackbody radiation

Blackbody radiation is emitted from a hot body. It's anything but black!

The name comes from the assumption that the body absorbs at everyfrequency and hence would look black at low temperature.

It results from a combination of spontaneous emission, stimulatedemission, and absorption occurring in a medium at a giventemperature.

It assumes that thebox is filled with manydifferent molecules

that that, together,have transitions(absorptions) at everywavelength.


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Spectral Density DistributionThe problem is to find the energy density of blackbody radiation as a function offrequency. The distribution function:

ρ νd ν : energy density in the frequency range d ν to ν + d ν

ρ ν depends only on frequency and temperature. It does NOT depend on theshape of the cavity. For simplicity, we will consider a rectangular cavity withperfectly conducting (reflective) walls.

The field inside mustsatisfy the wave equation:

With boundary conditionon each wall:


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Standing Wave Cavity ModesSeparating the space and time dependences and using our previous timedependence exp(-i ω t).

Which is the Helmholtzequation.

The solution is (you’ll verify in a HW problem)

The indices l, m and n are positive integers.

These are the supportedfrequencies or modes of the cavity.

c n is the speed of light in the cavity.


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Number of Resonant ModesLet’s now find number of modes with frequencies from 0 to ν or wavevectormagnitudes from 0 to 2πν /c.

k-space representation: allowed values of k-vectors are vectors fromorigin to the nodal points of this lattice.

Number of allowed modes:

There are two possible E-field directions for a given k-vector.

Total cavity volume


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Density of Resonant Modes

Number of modes per unit volume, per unit frequency:

As a result, the energy density becomes:where is the average energy in each mode.

According to Boltzman statistics:

Rayleigh-Jeans formula

This formula is a catastrophe! It tells us that the density increases

rapidy for low frequencies. Also, the total energy is infinite.


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Planck’s HypothesisThe UV catastrophe was solved by Planck’s hypothesis that the light energy isquantized. Hence, the energy of a mode could only be E=nh ν .

With this assumption, the average energy becomes:

Hence:

This result agrees perfectly with the experiments!

Planck’s hypothesis also implies the existence of photons,the quanta of light.


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Plancks formula resolves the UV

catastrophe

(You will get to plot these in a HW problem)


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Wien's Displacement Law: Blackbody peakwavelength scales as 1/Temperature.Writing the blackbody

spectrum vs. wavelength:

[ ]{ }

2 58 / ( )

exp / 1 B

hc I

hc k T λ

π λ λ

λ

=−

(You will prove this in another HW problem)


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We can tell how hot a star is by itsblackbody emission spectrum.


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04 quantum theory

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