Spectroscopy

Lecture 3

Hydrogen Atom

Rotational Spectroscopy

NC State University

Experimental observation of

hydrogen atom

• Hydrogen atom emission is “quantized”. It

occurs at discrete wavelengths (and therefore

at discrete energies).

• The Balmer series results from four visible

lines at 410 nm, 434 nm, 496 nm and 656 nm.

• The relationship between these lines was

shown to follow the Rydberg relation.

The Solar Spectrum

• There are gaps in the solar emission

called Frauenhofer lines.

• The gaps arise from specific atoms in the sun that absorb radiation.

Atomic spectra

• Atomic spectra consist of series of narrow lines.

• Empirically it has been shown that the

wavenumber of the spectral lines can be fit by

where R is the Rydberg constant, and n1 and n2 are

integers. Note that n2 > n1.

Schrödinger equation for hydrogen:

The kinetic energy operator

The Schrö dinger equation in three dimensions is:

– h2

2 2 + V = E

The operator del-squared is:

The procedure uses a spherical polar coordinate system.

Instead of x, y and z the coordiantes are q, f and r.

2=

2

x2

+ 2

y2

+ 2

z2

The Bohr radius

The quantity a0 = 4pe0h2/me2 is known as the Bohr

radius. The Bohr radius is a0 = 0.529 Å . Since it

emerges from the calculation of the wave functions and

energies of the hydrogen atom it is a fundamental unit.

In so-called atomic units the unit of length is the Bohr

radius. So 1 Å is approximately 2 Bohr radii.

You should do a dimensional (unit) analysis and verify

that a0 has units of length!

Energy levels of hydrogen atom

• The energy levels of the hydrogen atom are

specified by the principal quantum number n:

• All states with the same quantum number n

have the same energy.

• All states of negative energy are bound states,

states of positive energy are unbound and are

part of the continuum.

The Rydberg Constant

• The energy levels calculated using the

Schrö dinger equation permit calculation of the

Rydberg constant.

• One major issue is units. Spectroscopists often

use units of wavenumber or cm-1. At first this

seems odd, but hn = hc/l = hcn where n is the

value of the transition in wavenumbers.

~ ~

in cm-1

The simple form

• Using the Rydberg constant the energy of the

hydrogen atom can be written as:

E = – Rn2

~

~ where R = 109,636 cm-1

Shells and subshells

• All of the orbitals of a given value of n for a shell.

• n = 1, 2, 3, 4 .. correspond to shells K, L, M, N…

• Orbitals with the same value of n and different

values of l form subshells.

• l = 0, 1, 2, ... correspond to subshells s, p, d …

• Using the quantum numbers that emerge from

solution of the Schrö dinger equation the

subshells can be described as orbitals.

Spherical harmonics

These are the spherical harmonics , which

are solutions of the angular Schrodinger equation.

Hydrogen 1 s radial

wavefunction

• The 1s orbital has no

nodes and decays

exponentially.

• R1s = 2(1/a0)3/2e-r/2

r = r/a0

• n = 1 and l = 0 are

the quantum numbers

for this orbital.

2.0

1.5

1.0

0.5R

1,0

1086420r

Y1s

The Radial Distribution in

Hydrogen 2s and 2p orbitals

0.6

0.4

0.2

0.0

Rn,

1086420r

Y2s

Y2p

The Dipole Moment Expansion

The permanent dipole moment of a molecule

oscillates about an equilibrium value as the

molecule vibrates. Thus, the dipole moment

depends on the nuclear coordinate Q.

where is the dipole operator.

Q = 0 +

QQ + ...

Rotational Transitions

Rotational transitions arise from the rotation

of the permanent dipole moment that can

interact with an electromagnetic field in the

microwave region of the spectrum.

Q = 0 +

QQ + ...

z

x

r

q

r sin(q) sin(f)

r cos(q)

y

f

r sin(q) cos(f)

Spherical Polar Coordinates

Rotation in two dimensions

Our first approach is classical.

The angular momentum is Jz = pr.

Using the deBroglie relation p = h/l we also

have a condition for quantization of angular

motion Jz = hr/l.

r

Jz

p m

Classical Rotation

In a circular trajectory Jz = pr and E = Jz2/2I.

I is the moment of inertia.

Mass in a circle I = mr2. Diatomic I = r2

=m1m2

m1 + m2

m

m2

m1

Reduced mass

r r

The 2-D rotational hamiltonian

• The wavelength must be a whole number

fraction of the circumference for the ends to

match after each circuit.

• The condition 2pr = Ml combined with the

deBroglie relation leads to a quantized

expression,Jz = Mh where M is a quantum

number for rotation in two dimensions.

• The hamiltonian is:

Energy level spacing

Energy levels

Energy Differences

of DJ = ± 1

Quantization of rotational motion:

solution of the f equation

The corresponding wavefunctions are:

with the constraint that:

Since the energy is constrained to values Jz2/2I we

find that

The wavefunctions of a rigid rotor are

called spherical harmonics

The solutions to the q and f equation (angular part) are the spherical harmonics Y(q,f )= Q(q)F(f) Separation of variables using the functions Q(q) and F(f) allows solution of the rotational wave equation.

– h2

2I1

sin2q

2Y

2+ 1

sinq

qsinqY

q= EY

We can obtain a q and f equation from the above equation.

Rotational Wavefunctions

J = 0 J = 1 J = 2

These are the spherical harmonics YJM, which

are solutions of the angular Schrodinger equation.

The form of the spherical harmonics

Y00 = 1

4p

Y10 = 3

4pcosq

Y1±1 = 3

8psinqe±if

Including normalization the spherical harmonics are

Y2

0= 5

16p3cos

2q – 1

Y2

±1= 15

8psinqcosqe±if

Y2

2= 15

32psin

2qe±2if

The form commonly used to represent p and d orbitals are linear combinations of these functions

Euler relation

e±if = cosq ±isinq

sinq e if – e–if

2i

cosq e if + e–if

2

Linear combinations are formed using the Euler relation

Projection along the z-axis is usually taken using z = rcosq. Projection in the x,y plane is taken using x = rsinqcosf and y = rsinqsinf

Solutions to the 3-D rotational

hamiltonian • There are two quantum numbers

J is the total angular momentum quantum number

M is the z-component of the angular momentum

• The spherical harmonics called YJM are functions whose probability |YJM|2 has the well known shape of the s, p and d orbitals etc.

J = 0 is s , M = 0

J = 1 is p , M = -1, 0 , 1

J = 2 is d , M = -2 , -1, 0 , 1, 2

J = 3 is f , M = -3 , -2 , -1, 0 , 1, 2, 3

etc.

The degeneracy of the solutions

• The solutions form a set of 2J + 1 functions at each

energy (the energies are

• A set of levels that are equal in energy is called a

degenerate set.

J = 0

J = 1

J = 2

J = 3

Orthogonality of wavefunctions

• The rotational wavefunctions can be represented as the product of sines and cosines.

• Ignoring normalization we have:

• s 1

• p cosq, sinqcosf, sinqsinf

• d 1/2(3cos2q - 1), cos2qcos2f , cos2qsin2f ,

cosqsinqcosf , cosqsinqsinf

• The differential angular element is sinqdqdf/4p over

• the limits q = 0 to p and f = 0 to 2p.

• The angular wavefunctions are orthogonal.

Orthogonality of wavefunctions

• For the theta (q) integrals we can use the substitution

• x = cosq and dx = sinqdq

• For example, for s and p-type rotational wave functions,

for the theta integral we have

• Using these substitutions we can turn all of these integrals

into polynomials.

The moment of inertia The kinetic energy of a rotating body is 1/2Iw2.

The moment of inertia is given by:

The rigid rotor approximation assumes that

molecules do not distort under rotation. The types

or rotor are (with moments Ia , Ib , Ic)

- Spherical: Three equal moments (CH4, SF6)

(Note: No dipole moment)

- Symmetric: Two equal moments (NH3, CH3CN)

- Linear: One moment (CO2, HCl, HCN)

(Note: Dipole moment depends on asymmetry)

- Asymmetric: Three unequal moments (H2O)

I = miri2

i = 1

Polyatomic Molecules

• There are 3N total degrees of freedom in a

molecule that contains N atoms.

• There are three translational degrees of freedom.

These correspond to motion of the center of mass

of the molecule.

• In a linear molecule there are two rotational

degrees of freedom. In a non-linear molecule there

are 3 rotational degrees of freedom.

• The remaining degrees of freedom are vibrational.

Spectroscopy of atomic hydrogen

• Spectra reported in wavenumbers (cm-1)

• Rydberg fit all of the series of hydrogen spectra

with a single equation,

• Absorption or emission of a photon of frequency

n occurs in resonance with an energy change,

DE = hn (Bohr frequency condition).

• Solutions of Schrö dinger equation result in

further selection rules.

n = R 1n1

2 – 1n2

2

Spectroscopic transitions

• A transition requires a transfer from one state

with its quantum numbers (n1, l1, m1) to another

state (n2, l2, m2).

• Not all transitions are possible: there are

selection rules, Dl = 1, m = 0, 1

• These rules demand conservation of angular

momentum. Since a photon carries an intrinsic

angular momentum of 1.

The interaction of electromagnetic

radiation with a transition moment

The electromagnetic wave has an angular

momentum of 1. Therefore, an atom or

molecule must have a change of 1 in its

orbital angular momentum to conserve this

quantity. This can be seen for hydrogen atom:

l = 0 l = 1

Electric vector

of radiation

A propagating wave of electromagnetic radiation

of wavelength l has an oscillating electric dipole, E

(magnetic dipole not shown)

l

E

The oscillating electric dipole, E, can induce an oscillating dipole

in a molecule as the radiation

passes through the sample

l

E

The oscillating electric dipole, E, can induce an oscillating dipole

in a molecule as the radiation passes through the sample

l

The type of induced oscillating dipole depends on l.

If l corresponds to an electronic energy gap, then radiation will be

absorbed, and a molecular electronic transition will result

n=0

n=1

∆E = hc/l

The oscillating electric dipole, E, can induce an oscillating dipole

in a molecule as the radiation passes through the sample

l

If l corresponds to a electronic energy gap, then radiation will be

absorbed, and an electron will be promoted to an unfilled MO

HOMO

LUMO

∆E = hc/l

The oscillating electric dipole, E, can induce an oscillating dipole

in a molecule as the radiation passes through the sample

l

The type of induced oscillating dipole depends on l.

If l corresponds to a vibrational energy gap, then radiation will

be absorbed, and a molecular vibrational transition will result

v=0

v=1

∆E = hc/l

O

R

R

O

R

R

Rotational spectroscopy

Rotational Transitions

• Electromagnetic radiation can interact with a molecule to

change the rotational state.

• Typical rotational transitions occur in the microwave

region of the electromagnetic spectrum.

• There is a selection rule that states that the quantum

number can change only by + or - 1 for an allowed

rotational transition (DJ = 1).

J = 0

J = 1

J = 2

Rotational Transitions

• Treat the electromagnetic wave as having a polarization

along x, y, or z.

• The transition integral is not zero in this case since the z-

polarized transition is matched to the pz rotational orbital.

The total wave function in

calculation of transition moment

The total wave function can be factored into

an electronic, a vibrational and a rotational

wave function.

= elvYJM

Mrot = v* YJM el

* eldel vYJMdnuc

= v* YJM0vYJMdnuc

= v* vdQ YJM0YJMsinqdqdf

The rotational transition moment

The transition moment is a dipolar term that

connects two states. Here we are considering

rotational states that have quantum numbers

J, M in the initial state and J’,M’ in the final

State

The electronic integral gives 0, the permanent

dipole moment. The vibrational wave functions

are normalized and the integral is 1.

Mrot = 0 YJMYJMsinqdqdf

Interaction with radiation

For example, if an oscillating electromagnetic

field enters as E0cos(wt)cos(q) such that

hw is equal to a rotational energy level

difference, the interaction of the electric field

with the transition moment is

Mrot = 0 YJ+1,M

*cos q YJMsin q dqdf

0

p

0

2p

Mrot= E0

Interaction with radiation

The choice of cos(q) means that we consider

z-polarized microwave light. In general

we could consider x- or y-polarized as well.

x sin(q)cos(f)

y sin(q)sin(f)

z cos(q)

0 = Xi + Y j + Zk

0 = 0 sinqcosfi + sinqsinf j +cosqk

Pure rotational spectra

• A pure rotational spectrum is obtained by

microwave absorption.

• The range in wavenumbers is from 0-200 cm-1.

• Rotational selection rules dictate that the

change in quantum number must be

DJ = ± 1

• A molecule must possess a ground state dipole

moment in order to have a pure rotational

spectrum.

Energy level spacing

Energy levels

Energy Differences

of DJ = ± 1

The rotational constant The spacing of rotational levels in spectra is given

by DE = EJ+1- EJ according to the selection rule

The line spacing is proportional to the rotational

constant B

In units of wavenumbers (cm-1) this is:

A pure rotational spectrum

A pure rotational spectrum is observed in the

microwave range of electromagnetic spectrum.

2B ~

Key Points • The vibrational energy levels are given by:

The energy levels have a degeneracy of 2J + 1

• The wave functions are the spherical harmonics.

Transition energies are given by:

• B is the rotational constant

• Rotational spectra consist of a series lines

separated by 2B.