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Transcript of Lecture 3 Hydrogen Atom Rotational Spectroscopyfranzen/public_html/CH331/lecture/Lecture_3.pdf ·...
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Spectroscopy
Lecture 3
Hydrogen Atom
Rotational Spectroscopy
NC State University
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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.
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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.
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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.
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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
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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!
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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.
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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
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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
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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.
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Spherical harmonics
These are the spherical harmonics , which
are solutions of the angular Schrodinger equation.
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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
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The Radial Distribution in
Hydrogen 2s and 2p orbitals
0.6
0.4
0.2
0.0
Rn,
1086420r
Y2s
Y2p
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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 + ...
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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 + ...
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z
x
r
q
r sin(q) sin(f)
r cos(q)
y
f
r sin(q) cos(f)
Spherical Polar Coordinates
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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
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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
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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:
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Energy level spacing
Energy levels
Energy Differences
of DJ = ± 1
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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
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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.
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Rotational Wavefunctions
J = 0 J = 1 J = 2
These are the spherical harmonics YJM, which
are solutions of the angular Schrodinger equation.
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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
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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
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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.
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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
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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.
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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.
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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
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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.
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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
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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.
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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
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A propagating wave of electromagnetic radiation
of wavelength l has an oscillating electric dipole, E
(magnetic dipole not shown)
l
E
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The oscillating electric dipole, E, can induce an oscillating dipole
in a molecule as the radiation
passes through the sample
l
E
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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
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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
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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
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Rotational spectroscopy
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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
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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.
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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
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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
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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
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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
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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.
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Energy level spacing
Energy levels
Energy Differences
of DJ = ± 1
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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:
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A pure rotational spectrum
A pure rotational spectrum is observed in the
microwave range of electromagnetic spectrum.
2B ~
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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.