Download Review for Exam 1

4th Set of Notes
Chp. 11 and 14 Notes on Rotational Motion and Hydrogen Atom
Spherical Polar Coordinates (r,
 goes from 0 to 2
 goes from 0 to 
Rotational Motion (Particle on a Sphere) (Rigid Rotor)
Rotor is a dumbbell with masses ma and mb at the ends, held together by a massless bar equal to
the bond length “R”
Solve for the energy levels of this system.
Use the general method as always:
1) Write down the kinetic and potential parts of the Hamiltonian
2) Find solutions to the eigenvalue equation H= E
What would the Hamiltonian be in Cartesian coordinates? Let mass ma have coordinates xa,ya,za,
and mass mb have coordinates xb, yb, and zb.
This proves very difficult if not impossible to solve. Not making use of the symmetry of the
situation!!!
Make reference to the center of mass of the system. Separate into a problem of translation and
rotation which you separate out the translation and rotation parts.
Transform eqn into a mass, a reduced mass , where  = mamb/(ma+mb) rotating at a fixed
distance from the origin of the center of mass. This is essentially then a particle on a
_________.
The Hamiltonian in Schrodingers eqn for the rigid rotor or particle on a sphere then becomes:
two separable eqns. H = H1 + H2
H1 = -h/ 2 /2mA+mB) (..........................................................................................
H2 =-h/ 2 /2 ( 1/r2 /r (r2 /r) + 1/(r2sin) / (sin /) + 1/(r2sin2) 2/ 2)
H1 is the hamiltonian for a free particle translating in space. And H2 is rotation of a particle on a
sphere.
But this is rigid rotor so r is ________________ and /r is equal to _____________.
And R2 = I, where I is called the moment of inertia.
Get:
............-h/ 2 /2 = 1/R2 [1/(sin) / (sin /) + 1/(sin2) 2/ 2)]
Use this SIMPLIFIED HAMILTONIAN to find the allowed rotational energy levels via?
Use trick of plugging in functions which depend only on and . ie. 
Plugging in, and manipulating the eqn. and get
sin21/(sin) / (sin /) + 2IE/h/ 2 ] = -1/ 2/2
If this is to hold for all values of and  then both sides must equal a constant.
Call it 

1. sin21/(sin) / (sin /) + 2IE/h/ 2 ]

2. -1/ 2/2

Looking 1st at number 2, this is a differential eqn whose solution is:
() = Aexp(im), / = A i m exp(im), 22 = A i2 m2 exp(im)
so -1/22-i2 m2 = m2 = 
Thus = m2
If the wavefunction is to be “single-valued” then m = 0,+/-1,+/-2,....
How? Consider Euler’s relationship
For the other part, more manipulations with the substitution that x = cos  are required, but
eventually we reach a differential equation of the form:
(1-x2) 2/x2 - 2x /x + (2IE/h/ 2 - /(1-x2)) = 0 ...........................(C)
(1-x2)  ”- 2x ’ + (l(l+1) - m2/(1-x2)) = 0 .........................................(D)
“ means 2nd derivative with respect to .
‘ means 1st derivative with respect to 

This second equation (D) is the Associated Legendre equation. Solutions to this eqn. exist only
for the conditions that l is a positive integer or zero, and that /m/ < l.
Solutions are the associated Legendre Polynomials
l
m
lm
Comparing (C) and (D) above we find that our eqn for the  dependence of the rigid rotor is just
the same as the Associated Legendre Eqn. with the stipulation that:
........................ 2IE/h/ 2 = l (l+1)
Thus the energy eigenvalues for the rigid rotor are:
..............................E = l (l+1) ............... where (l = 0,1,2,3, ...... ) ............................. (E)
.................... or .... E = J (J+1) .............. where (J = 0,1,2,3, ....... )
Often m is called ml since the value of m depends on the value of l.
In summary the energy eigenfunctions for the Rigid Rotor (the solutions to the wavefunctions
satisfying Schrodingers Eqn) are:
where
and
A is found by?
The normalized product of  and  is called te Spherical Harmonics (see Table 12.3). Should
be able to write out the wavefunctions (eigenfunctions) such as 2,1
for every value of ..l.. there are ___________________ values of ml.
Also since E = El = ......................................................... where l = 0,1,2,3,... and I = mr2
What can be said about 1,0, 1,1, 1,-1?
So the degeneracy of the lth state is?
Associated with ROTATION is ANGULAR MOMENTUM.
Angular momentum, J, is a vector quantity, meaning it has both direction and magnitude.
Classically the magnitude of the J, is related to Erot by
Erot = J2/(2I), where I is the moment of inertia or Erot = L2/2I ...........................(F)
Comparing Eqns. (E) and (F) leads to the conclusion that the magnitude of the total angular
momentum
.................................. L2 = l(l+1)h/ 2 ............. and in fact L2 rot = l(l+1)h/ 2 rot
The angular momentum vector quantity is given by J = r x p ................. where r and p are the
radial
direction of the particle and the linear momentum as defined in the fig. above.
In general for vectors a and b, the cross product a x b is given by:
The magnitude of the yth component is given by: (axbz - azby).
The magnitude of the xth component is given by (aybz - byaz)
So for angular momentum L = r x p.
In cartesian coordinates the magnitude of the components in the x, y, and z, directions become:
lx =
ly =
lz =
In terms of quantum mechanics, this becomes
Also putting it in terms of the symmetry of the problem (spherical polar coordinates) gives:
lx
ly
lz
See if the eigenfunctions to the rigid rotor will also give eigenvalues for the zth component of the
angular momentum operator.
Also lz  = lz 










Find that Lz = hm
/ l .................................................. where ml = 0, +/-1, +/-2, .....
The zth component of the angular momentum can only take on discrete values:
2l+1values.
Since ml is restricted by l.
Thus the projection of L on the z axis is allowed to be only certain discrete (quantized)
values). The plane of rotation of the particle can only be oriented in certain discrete
ways.
Quantum Mechanics says that a rotating body may not take up any arbitrary orientation
with respect to some axis, only certain discrete orientations are allowed
Furthermore: by uncertainty principle, if lz is known, lx and y can not be precisely known.
A better model is a cone model where lx and ly are not known.
So far angular momentum due to orbit of a particle. (Orbit of electron about nucleus!)
Also have spin angular momentum - motion of electron about its own axis - internal
angular momentum of the electron is called its ________________.
Use quantum number s intead of l, and ms instead of ml, for zth component.
Magnitude of spin angular momentum S is:
Magnitude of the zth component of the sz is:
For an electron only one value of s is allowed:......... s = ½
electron i s aspin ½ particle. Thus the magnitude of the angular momentum is:
neutrons and protons are also spin ½ particles, more massive, but spin slower.
½ integral spin particles are called ______________
integral spin particles are called ________________
Matter is an assembly of fermions held together by forces conveyed by bosons.
Elementary particls are all fermions.
Electronic Energy: Ideal Model is the Hydrogen Atom
Hydrogenic Atoms - Atoms which have a single electron
H, He+, Li2+, etc.
Spectra of H emission revealed line spectra
Lyman
Balmer
Paschen
We can solve Schrodingers equation for this case if we consider the hydrogen atom as a
system of two interacting point particles.
What is potential? Coulombic attraction?
Charge of nucleus is +Ze,
Charge of electron is -e
Z is atomic number
So potential energy is:
V = -Ze2/r
or V = -Ze2 /(4or) (in SI units)
r = distance between the electron and the nucleus
Write Schrodingers Eqn.
Cartesian x1, y1, z1 coordinates of nucleus, m1 is the mass of the nucleus
............... x2, y2, z2 coordinates of electrons
-h/2/2m1 (....................................................) + -h/2/2m2 (......................................................)
+ V
ET 
V= -ze2/ [(x2-x1)2 + (y2-y1)2 + (z2-z1)2]1/2
Since interested in motion of the electron about the nucleus, try to separate this into two
eqns., one representing the translational motion of the molecule as a whole, ......and
the other - involves the relative motion of the nucleus and the electron
Introduce new variables x, y, z, which are the Cartesian coordinates of the center of mass
system,
and r, and , and the polar coordinates of the relative motions.
Then for the relative motion coordinates
x = x2 - x1 = r sincos
y = y2 - y1 = r sinsin
z=
Have two parts which depend only on x, y, and z and another which depends on r, and
Each
part must be equal to a constant or in this case zero.
Resulting Eqns are:
................................................................................................. = 0 ...........................(A)
and
............................................................................................. = 0 ................................... (B)
Eqn A is the eqn for the motion of a ______________ ____________________, the
translation of the ______________ _____________ in space.
Eqn,. B is identical with the wave eqn of a single partile of mass  under the influence of
a potential function V(r,Actually, the potential depends only on ______ as it is
centrosymmetric.
Try separating Eqn. B Further: ...................(r,) = R(r)()
Introducing this into eqn. B and dividing by Rmultiplying by r2sin gives:
sin2R d/dr(r2dR/dr) + 1/d2/d2 + sin/ d/d(sin d/d) + 2/h/2 (E-V(r)) r2sin2
=0
Note that this is starting to separate into three parts: one depending only on r, another
depending only on , and one on . Now there is a part depending only on  so as
before:
1/2 d2/d2 = must be equal to a constant, and looks like RIGID ROTOR so = -ml2
substituting in this -ml2 for the  dependence and dividing by sin2 gives:
1/R d/dr
(..................................................................................................................................
................................................ = 0.
Now the 2nd and third parts of this equation are independent of r and the remaining part is
independent of  so again equate each to a constant. Set part = -  and r part = +.
Also then multiplying the part by  and the r part by R/r2 then it all falls out
wonderfully as:
1/sin d/d(sin d/d) - ml/sin2  +  = 0 ..................................... (C)
1/r2 d/dr(r2dR/dr) - /r2R + 2h/2(E-V(r))R = 0 .................................... (D)
Solved the first part before
Found that = l (l+l) and was the Associated Legendre Polynomial. Also Remember
that the  dependence was  = (1/2)½ exp(iml)
and the product Yl ml which is known as the _______________ ________________
which are listed in TABLE ______.
restrictions are: l = 0, 1, 2, …
ml = -l, -l+1, …..., +l
So just have the Solution of the R Eqn. to contend with. Note we haven’t put in V(r) yet
so this is applicable to any system of two particles which interact with one another in a
way expressible by V(r).
Put in V(r) = -Ze2/r (in cgs) or -Ze2/(40 r) (in SI)
Consider E to be negative - bound states; the total energy is insufficient to ionize
Now the R eqn. is easier to solve with the following substitutions:
a2 = - 2m/h/2 E,...... l = m Ze2/(h/2 ), ......  = 2r, .......... and S() = R(r)
then the wave eqn. becomes:
1/2 d/d(2 dS/d) + [-1/4 - l(l+ 1)/2 + ] S = 0 ...........0 <  < infinity
Recognize that this is differential eqn. whose solution are the Associated Laguerre
polynomials.
Rn l is the Associated ______________ _______________ Note that the functions
depend on the value of the quantum numbers n and l.
The quantum number n has the restriction such that n = 1, 2, 3, 4, ...........
And the quantum number l has the restriction such that l = 0, 1, 2, .........n-1. NOTE that
the value of l is restricted by the value of n. This is how the Associated Laguerre
Polynomials work.
The total wavefunction then is:
n,l,ml = Rn l Yl m
Also remember that ml is restricted by the values of l such that ml = -l, ..., -1, 0, 1, .....+l
and there are 2l + l values of ml for every value of l.
Whydrogen are the energy eigenfunctions and are the ONE ELECTRON OR
HYDROGENIC ATOMIC ORBITALS
An orbital is a one electron eigenfunction. Why is it called one-electron?
Atomic Orbital - one electron energy wavefunction describing the distribution of an
electron in an atom. How does it describe the distribution?
n l m = Rn l Yl m is the form of all hydrogenic wavefunctions.
The three quantum #s n, l, and ml define which hydrogenic orbital or eigenstate the
electron is occupying.
l and ml came from the _____________ solution.
Remember l describes the magnitude of the angular momentum by
L = l l + 1)½ h/; so the orbital angular momentum of the electron is given by this.
Also the zth component of the angular momentum of the electron is given by:
lz = ml h/
What about the ENEGY OF THE HYDROGEN ATOM? The energy eigenvalues are
determined in the solution of the differential eqn. above and the energy is:
En = -hcR/n2; ............... hCR = 2e4/(32202 h/2) (in SI)
**** Note that the energy depends only on the quantum number n: n = 1,2,3, ......... for
hydrogenic atoms.
Orbital Energy Diagram (depends only on n)
Summarizing then:
The eigenfucntions (eigenstates, one electron atomic orbitals) are defined by the quantum
#s
n - principal quantum # - tells about the size and energy of the electron in the atom
......... allowed values of n are n =
l - azimuthal or angular momentum quantum # -tells about the ____________of the
__________ _________.
..........allowed values of l are l =
ml - zth component of the ______________ __________ or also called the magnetic
quantum #; ...... since it tells about the magnitude of the ____ _________ of the
___________ __________. ...... allowed values are ml =
 n, l, ml is the product of angular spherical harmonics and radial associated Laguerre
Polymials.
The electron also has an intrinsic momentum, and INTRINSIC angular momentum - spin
angular momentum.
Full specification of the electronic state of a hydrogenic atom then requires the use of
another .........quantum #. Since the spin angular momentum quantum number s is fixed
at ½, use ms .........instead.
ms - specifies the orientation of the spin angular momentum
....... Allowed values for ms are ms =
SO  n l ml ms is the
Definitions:
1) All orbitals of a given n are said to be in the same _________ or ________. Those
orbitals in the same shell have the same _______________ and similar ___________.
n = 1, 2, 3, ...
......K, L, M, N, ......
2) All atomic orbitals with the same value of n, but different l values are said to be in
different
_____________ or _____________ of a given shell.
l = 0, 1, 2, 3, 4, ......... n-1
......s, p, d, f, g,
Letters run alphabetically with the omission of j.
For n =2, l = 1, or 21 this is called the __________ subshell or sublevel.
3) ml = -l, ........, 0, ............ +l
For the case above, 21, the allowed values of ml are ? ........................ and 21 turns into:
when all three of the values of ml are considered. Thus when ml is included we are
looking at the
2p orbitals. An electron occupying the 2p orbital is called a 2p electron.
Since l can go from 0 to n-1, there are ____ subshells of a shell.
# of orbitals is 2l + 1 so for n = 1
l = _____ and ml = ________________ so there are ____ orbitals.
Example: Case when n = 3?
Overall, the number of orbitals in a shell with principal quantum number n is ______.
Thus in a hydrogenic atom each shell is n2 __________________.
Remember we also have spin angular momentum in addition to the orbital angular
momentum
Talked about this before. It is the motion of the electron about its own axis - __________
angular momentum
Use quantum number s instead of l, and ms instead of ml, for the zth component.
Magnitude of spin angular momentum, S, is:
and the magnitude of the zth component, sz is:
for a single electron only one value of s is allowed
s=1/2
and ms = ?
In each orbital can only have at most 2 __, one with ms = ½ and other with ms = -1/2
ENERGY DIAGRAM
The orbital occupied in the ground state is ______________________
or _____________________________ which is the 1s level or state.
Writing out the wavefunction Table
Since wavefunction is largest at r=0, expect probabiltiy to be largest there as well!
Decays exponentially form maximum at r=0.
The ground state of the hydrogen atom n=1, ml = 0 has the electron closest to the nucleus.
The successive excited
states where n=2, n=3, ……. show the electron to be spread out over more space away
from the nucleus.
.
Way to represent the probability density of electron is 2 shown by density of shading
generated by computer.
Fig
The probability of finding the electron at a distance r from the nucleus is proportional to
the volume of a spherical shell of radius r around the nucleus. The volume is directly
proportional to r2 .
THUS the probability of finding the electron at a distance r is proportional to r2 2 . Plots
of this versus r/ao are given where ao is called the Bohr radius (radius of Bohr’s n=1
orbital = 0.0529A.
Fig
A second way is a boundary surface. Surface whose shape and size represent capturing
about 90% of the electron probability. For 1s, the surface is a ______________
Fig
Example:
For the 1s orbital of the hydrogen atom, calculate the probability of finding the electron
within a distance ao from the nucleus.
Radial nodes
# of radial nodes = # of places where the radial wavefunction passes through zero or
where the probability is equal to zero. How many radial nodes are there???
See figure
1s has _________ nodes. 2s has ________ nodes. etc.
These can be solved for by setting the radial part of the wavefunction to zero.
R10
The wavefunction or more precisely, *d gives the probability of finding an electron
in any region. For 1s * is proportional to exp(-2r/a0)
Also talk about probability of finding electron on spherical shell of thickness dr ar radius
r. Interested in spherical shell of volume 4r2dr which is d
so the Probability = 4r2 2 dr
Since r2 increases with radius from zero at the nucleus and 2 decreases to 0 at infinity.
P on the other hand is zero both at _______________ and at________________.
Pmax (how do you get it) gives the most probable radius.
So far only considered S one-electron atomic orbitals p orbitals are different
p has nonzero angular momentum L = l(l+1)½ h/
l is equal to______ for electron occupying orbitals in p subshell.
All orbitals with l > 0 have zero amplitude at the nucleus so no probability of finding
electron there.
Fact that electron in p subshells have nonzero angular momentum changes the shape!
Classically there is a centrifugal force that flings the electrons away from the nucleus.
2p subshell
There are three 2p orbitals ml distinguishes
Angular Nodes
x,y plane is a nodal plane of the orbital
The 2pz orbital is 2 1 0
2px, and 2py orbitals are real linear combinations of the _____ and ________
d subshell and d orbitals
n = 3 .......... l is 0,1,or ____
In the shell there is one 3s, three 3ps, and five 3d orbitals
have
Spectroscopic Transitions
For a transition - electron changes its eigenstate or orbital. Goes from state with
quantums
ni, li, ml .... to a state with nf lf, ml
E = Enf - Eni = Ephoton = h = h c/
Can electron move to any old atomic orbital?
Angular momentum must be conserved. Since energy given by a photon or taken away
by a photon. The angular momentum must change to compensate. Go from l=2 to l=0
subshells?
What is the angular momentum of a photon?
Give Selection Rules:
Grotrian Diagram
Many Electron Atoms
For the hydrogenic atoms there was one electron surrounding the nucleus with an
effective charge Z.
For many-electron atoms must consider not only the interaction between the nucleus and
electrons, but also the interaction of the electrons with themselves
What would Schrodinger’s equation look like
H=
-h/2/2me
where ri is the distance for the ith electron from the nucleus
ri j is the distance between the ith and jth electrons
Z is the atomic number .... and ...... me is the mass of the electron
Consider He, Li+ Be2+ etc. Z = 2,3,4 etc
H
If we omit the term which is the interaction between the electrons (ri j) terms, this eqn. is
separable into N three-dimensional equations (one for each electron) (CALLED THE
ORBITAL APPROX)
To this degree of approx. the wavefunction for the atom may be built up out of singleelectron wavefunctions - That is a solution to the equation for the many electron atom
with
Omission of electron interaction terms,


And Z -> Zeff
where the individual orbitals resemble the hydrogenic orbitals, but with the nuclear
charge modified to account for the “screening effect”?
Within the orbital approx. the electronic structure can be expressed by reporting its
atomic electronic configuration which is a list of the occupied hydrogenic atomic
orbitals.
Remember in General Chemistry we did this. Used Aufbau Principle, or Building-up
Principle
(Add electrons in succession into atomic hydrogenic orbitals)
Need also the Pauli Exclusion Principle
No more than two electrons may occupy any given orbital, and if two do occupy one
orbital, their spins must be paired.
No spin angular momentum when the spin is paired. The spin of one electron cancels the
spin of the other.
Also another way of stating Pauli, no two electrons in an atom may have the same
________
__________ ____________.
For the hydrogen atom then - with one electron- that electron goes into the 1s orbital
n = 1 l = ............., ml
Atomic Electronic Configuration is:
1s1
What about Li?
In building up - build up based on energy of electronic orbital? Do electrons in a 2p and
2s have the same energy?
Shielding Effects
Many-electron atoms experience Coulombic repulsion forma all the other electrons
present. Other electrons “effectively” reduce the attraction of the outer electrons to the
nucleus.
Zeff = Z -  where  is the shielding constant
Zeff is different for p and s electrons! Why?
Figure
“s” electrons have higher probability of being closer to the nucleus. - s experiences less
shielding since it penetrates further. Principal Quantum # muost important generally!
Combining penetration and shielding shows that energies of subshells with same
principal quantum #s is s,p,d
Many Electron vs Hydrogenic
Consider Li again (Building up)
Aufbau principle says that the order of occupation is
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 6p etc.
Get this order from the periodic table - where are the main exceptions?
Abbreviated Notation [Noble Gas Symbol] rest of config
Example: Write the Atomic Electronic Configs for the following many-electron atoms
C
P
Fe
Cr
Cr2+
Ag
O2Valence Electrons come off lst when forming cation.
Valence electrons in the outermost shell of an atom int its ground state - responsible for
chemical bonds.
Core Electrons - non valence electrons - not as reactive.
What about electron orbitals?
Electronic configuration gives an idea about the arrangement of electrons in?
Atomic electronic orbital diagram gives more information!
There is more information as the arrangement of the electrons in their orbitals - Again
based on Energy!
Hunds Rule: An atom in its ground state (lowest energy state) adopts a configuration with
the greatest # of unpaired electrons.
Spin Correlation - electrons with same spin stay well apart from each other - SO less
repulsion
Repulsive term is smaller, it makes positeve contribution to the electronic energy - so
energy is more negative when spins are parallel.
Consider atomic orbital diagram for Carbon .... C
What about N
Periodicity
1) Size
Size increase going up or down a group (family, column)
size decrease going left to right across the period (row)
2) Ionization Energy
lst Ionization Energy - energy required to remove and electron
Na -> Na+ + 1eTrend
See Fig 9.21
Overhead (kink in curve at Be and N Why?
2nd Ionization energy?
Does this orbital approximation work well enough for all predictions?
Can we get better approximations to the true wavefunctions - orbitals?
Well remember the problem is the term
V=
ri j is the separation of electrons i j
this term is difficult to deal with (can’t deal with it analytically) must use approcimations
Hartree Fock Procedure
If we know what the rough structure of the atom is, (Use orbital approx as the start - that
is approximate by hydrogenic orbitals) then a Schrodinger eqn can be written for this
electron by ascribing a potenetial energy to it arises from the nuclear attraction and now
the average repulsion
from other electrons in their approximate orbitals.
For instance take AluminumL
orbital approximation suggests configuration of
with hydrogenic atomic orbitals
consider 3p .......... Solve for 3p in approx Schrodinger
with numerical methods where Vee depends on the wavefunction of the other electrons and is the average term.
Once 3p is solved move on to 3s now with Vee modified with the new 3p This is solved
giving a new value for 3s. Move on to 2p, 2s etc. and iterate ............
These solutions are said to be self-consistent.
The Electron-Electron interaction causes several terms (with different energies) to arise
from a given atomic electronic interaction. This means that the Atomic Electronic
Configurations lack some information.
So in addition to Atomic Electronic Configurations, USE TERM SYMBOLS
1) Electron - Electron Interaction
.... .A) Configuration interaction - spins of electrons may interact and cause splittings
..... B) Spin Orbit Coupling (interaction) between the spin and orbital angular momenta of
an .......... electron.
...
......... ml & ms values tell about the relative orientation of individual angular momenta w
respect to each other.
Different Orientations of the spin and orbital angular momenta correspond to different
couplings and different overall energies