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welcome back to the lectures on ah chemistry
with introduction to molecular spectroscopy
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ah this lecture section is ah to help you
with problem solving ah ability and also to
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ah thing about how to organize your third
process when a set of data are given and you
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are ask to find the solution problem solving
is an important part of learning but its not
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the only part ok it is important for you to
listen to lectures its important for you to
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read and understand things discuss them with
your friends and your teachers but then use
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these problems has to give you confidents
that you have understood the material understood
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within codes that material with arrays number
degree of confidents ok
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so in this set of problems are become ten
problems the first problem is ah requiring
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you to identify which molecules have pure
rotational that is microwave spectrum and
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a which molecules have infrared spectrum or
the molecules which have both of them and
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ah you are ask to identify them accordingly
the important point is for a molecule to have
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a pure rotational or a microwave spectrum
the molecules should have a permanent dipole
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moment in its equilibrium geometry or its
disequilibrium configuration for a molecule
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to have infrared spectrum during the vibrational
motion the dipole moment must change the dipole
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movement need not be there to start with but
if in the process of vibrational motion the
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dipose are created and destroyed what you
see is that that kind of a motion leads to
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an oscillation of the electric dipole moment
and then such molecules will have infrared
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spectrum ok
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so obviously molecules which have a permanent
dipole moment will all have an infrared spectrum
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there is no question about that because during
the molecular motion ah some of its during
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at least some of these motion the dipole moment
will change as the molecule vibrates and so
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on so lets take the first example carbon monoxide
carbon monoxide has a strong dipole moment
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therefore it has both microwave and ah i r
spectrum it has c o both microwave and i r
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spectrum ok the next one is c h two c l two
the tetrahedral molecule and this is a tetrahedral
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molecule and it has a permanent dipole moment
due to the c c l bond having electrical dipole
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differently from that of the c h two and therefore
this molecules also has both ah microwave
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spectrum and infrared spectrum carbon dioxide
which is the third one carbon dioxide doesnt
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have a permanent a dipole moment in its equilibrium
geometry or it's a zero point vibrations carbon
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dioxide doesnt have an electric dipole moment
in its equilibrium geometry therefore it does
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not have a microwave spectrum but it has two
motions the bending degrees of freedom
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and the asymmetry stretch in both these normal
mode of vibration the molecule has a changing
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the dipole moment therefore it has the infrared
spectrum for these two degrees of freedom
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the bending mode is w b generate therefore
you have degeneracy involved
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but you have two frequencies one for the bending
mode and the one for the asymmetry stretch
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the molecule ammonium is pyramidal as the
dipole moment therefore it has both microwave
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spectrum and infrared spectrum the molecule
formaldehyde acetaldehyde c h three c h o
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not formaldehyde acetaldehyde yes it has a
dipole moment therefore there are vibrational
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motion were the dipole will change during
the vibration giving rise to both i r and
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microwave spectrum ah methane is a perfect
tetrahedron and in its tetrahedron geometry
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the electric dipole had chancel to each other
and therefore methane the does not have the
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microwave spectrum but methane has vibrational
motions please remember it's a five atom molecules
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therefore it has three n minus six normal
modes which is nine normal modes some of these
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normal modes lead to change in the dipole
moment therefore it has infrared spectrum
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the symmetry stretch of c h four will not
have a dipole moment because all for hydrogen
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atoms vibrate a way from equilibrium and also
going to the such modes are not see but in
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general it has i r spectrum benzene the same
thing in its equilibrium structure does not
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have a dipole moment therefore its not microwave
active so i will write microwave active here
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and this is infrared active here so when it
benzene does not have a microwave spectrum
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but it has many degrees of freedom which lead
to i r spectrum trans dichloro ethane ethylene
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so that is c l c double bond c h c l h this
is given so that there is no equilibrium dipole
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moment
therefore this person have a microwave spectrum
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but it has i r spectrum ozone lets right all
the three oxygens involved please remember
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ozone is an asymmetry top and this oxygen
is differently bonded to the other two oxygen
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therefore asymmetry top has and this molecule
also has dipole moments therefore its microwave
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and i r active i r active ok
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ah acetylene
its quite clear that there is no dipole moment
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for those linier molecule this molecule has
a center of asymmetry therefore its not microwaves
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active that is microwave inactive but it has
vibrational motions therefore its i r active
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hydrogen peroxide is a non planar structure
bending structure
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and therefore it a non planar structure therefore
it has a dipole moment ah therefore its microwave
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active and i r active and the last is propine
c h three c triple bond c h so this has the
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asymmetry groups c h three here therefore
its both microwave active and i r active ok
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so one has to look for the equilibrium geometry
of the molecule and look for the all its possible
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normal modes in determining which of them
or i r active and which of them are microwave
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active and so on
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so lets go to the second problem this is an
interesting elementary though problem on the
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determination of the moment of inertia of
water molecule about its symmetry axis the
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symmetry axis of course passes through oxygen
and ah it bisect the h o h bonds so the o
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h bond distance is given to you as ninety
five point seven picometers and the h o h
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bond angle is also given as one naught four
point five degrees so the moment of the inertia
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if you recall for water molecule if you have
to calculate about the symmetry axis this
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is the symmetry axis c two symmetry axis ok
and since on the c two axis oxygen resides
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oxygen doesnt contribute to the m i r i square
term r i is zero for oxygen because on the
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axis however the hydrogens are or this distance
perpendicular from the ah axis which the perpendicular
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distance that you have to take into account
and therefore you have a mass of hydrogen
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and since this angle is given has one naught
four point five degrees the half angle is
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fifty two point two five degrees and the bond
length is give as ninety five point seven
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picometers
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therefore its easy for you to calculate this
distance as r i as ninety five point seven
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so this is a sign fifty two point two five
ok that c r i therefore two hydrogen atoms
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contribute to the overall moment moment of
the inertia for this molecule therefore this
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is ah this value and since its a symmetry
axis that we are talking about hydrogen atoms
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on this side is also r one and r two are both
equal therefore what you have is two m h times
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ninety five point seven into ten raise two
minus twelve meter square times sin square
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fifty two point two five and this is kilogram
that you have to use therefore you will have
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the moment of the inertia i in terms of kilogram
meters square i think the numbers can be calculated
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by you for let me not write that here ok
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lets go to the next problem the next problem
is about to the fundamental vibrational frequency
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of hydrogen molecule and how it is related
to the vibrational frequency of its isotopic
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subsequent the h d in which one hydrogen item
is substituted by the deuterium and d two
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in which both the atoms are replaced by deuterium
and one assumes that in these three species
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the chemical nature namely the force constant
ah which contribute the bond stunts of these
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three molecules do not change by much its
in that limit how are these vibrational frequency
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is related to each other this is mean so very
simple problem to look at because you know
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vibrational frequency for problem three vibrational
frequency is related to the force constant
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and the last of the molecule that is the center
of mass the sorry ah the reduced mass mu
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this is the expression therefore if you are
writing mu h two is mu of h two and k of h
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two doesnt change from k of h d which is now
obviously related to the mu of h d by the
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same relation mu h d and mu of d two is given
by i by two pi square root of k mu d two therefore
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if you know this number has four thousand
four hundred and one point two centimeter
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inverse then you can calculate where should
the nu h d should be by taking the ratio of
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this mainly u h d divided divided by mu h
two will turn out to be one by because the
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case will cancel the two pis will cancel you
will have one by square root or let me write
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to the the final form the square root of this
h d therefore its mu of h two divided by mu
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of h d where inversely proportional to the
square therefore the ratio is done
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and mu of h two is easy to calculate if you
assume h two to be the h atomic mass as one
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point zero zero eight a m u if you multiply
that the one point six six one into ten raise
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to minus twenty seven kilograms that's the
amount for one a m u that's how that's the
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mass of a m u associated with that then h
two is the reduced mass is nothing but one
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point zero zero eight one point zero zero
eight divided by two point zero one six times
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one point six six one times ten to the minus
twenty seven kilograms this is the mu of h
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two and likewise the mu of h d if you want
to do that please remember the relation m
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one m two by m one plus m two ok so for mu
of h d the d is i believe ah two point zero
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one four so h is one point zero zero eight
times one point six six one times ten to the
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minus twenty seven kilograms and the d is
two point zero one four times one point six
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six one six six one times ten to the minus
twenty seven kilograms therefore the mu of
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h d can be calculated as one point zero zero
eight times two point zero one four divided
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by three point ah zero two two the sum of
the two times one point six six one into ten
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raise to minus twenty seven
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so this is the mu of h d so you can see that
if you substitute mu of the h d the mu of
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h d is slightly more than the mu of the h
two and therefore the square root of this
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number tells you that the nu of h d is a little
less than the nu of h two ah now it's a same
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thing that you have to do for = d two for
d two you would use two point zero one four
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times two point zero one four divided by four
point zero two eight times ten to the minus
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twenty seven into six point six six one one
point six six one kilograms therefore the
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mu of d two is given by this number then you
can calculate mu of h two sorry the nu of
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h two the frequency h two divided by the frequency
of d two is the square root of the mu of d
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two divided by the mu of h two ok
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so these are simple manipulations of the harmonic
oscillator frequencies and also a isotopic
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concepts that vibraional and rotation frequencies
do depend on the isotopic mass of the individual
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ok then the forth problem is on the morse
potential and you recall the morse potential
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expression which is as v of r the potential
energy and its given by the factor d which
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is the dissociation constant times one minus
e two the minus alpha r minus r e r e is the
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equilibrium distance r is the distance during
the vibration therefore this the potential
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term that would you use to solve for the morse
oscillator the kinetic energy of the morse
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oscillator is already given by p square by
two m and this is v r and therefore you use
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this to solve the potential energy
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now the question is show that in the limit
of small displacement that is an r minus r
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e is very very small that this is the same
as the harmonic oscillator approximation and
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identify the harmonic frequency also show
that the asymptotic value is d what are the
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interpretations for the parameter for the
morse potential ok so first of all v of r
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is d times one minus e to the r minus r e
times alpha whole square ok so suppose for
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r minus r e times alpha much less than one
is easy for us to expand this by writing d
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one minus please remember exponential of minus
a x when a x is much less than one is one
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minus a x plus a square x square by two minus
and so on we are looking at small values this
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therefore we stop with that so what we have
is d is equal to one minus r minus r e alpha
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and also there is a one ok plus one because
the exponential has one minus that so if we
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stop with that term just the first term and
take the square of this term which is already
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a very small one we don't even go to the second
term then what you see is it is d into r minus
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r e whole square and please remember this
is the potential energy form and for the harmonic
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oscillator the potential energy form half
k x square or in the rotation of r minus r
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e it is half k r minus r e whole square ok
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therefore you see that the ah alpha there
is an alphas here therefore this is an alpha
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square ok so d times r minus r e square times
alpha square is approximately the potential
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v of r for small values of r about r e r e
so what you have is d alpha square is equal
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to half k and therefore alpha is given as
square root of k by two d ok so this is the
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morse oscillator parameter and please remember
the morse oscillator parameter has a dimension
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by one the length and k as the dimensions
mass into ah this is a force constant therefore
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it is mass t to the minus two and this is
energy which is mass into length square into
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t to the minus two therefore the square root
of this is one by length ok
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so the interpretation of alpha is that if
this k the force constant is basically two
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d alpha square ok now the next question is
what is the harmonic fre frequency please
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remember the harmonic frequency is one by
two pi into square root of k by m k by mu
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and for the morse oscillator we just now found
that k is given by k is given by two d alpha
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square therefore the morse oscillator frequency
in harmonic limit will be one by two pi square
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root of two d sorry this is alpha square not
two d alpha square therefore two d alpha square
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by mu ok so the morse parameters d and alpha
and the reduced mass of the molecule or you
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use to define the ah harmonic frequency in
the limit of small approximation
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and second thing is second question in that
is show that the asymptotic value of v r is
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d now try to plot v r as a function of r minus
r e ok its d one minus e to the minus alpha
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r minus r e whole square ok at r is equal
to r e the exponential has zero exponent and
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therefore its one and therefore one minus
one is zero so the v of r is zero this is
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v of r ok therefore v r is zero at r is equal
to r e ok for all values of r greater than
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or e this number is smaller than one and therefore
this number is positive one minus e to the
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minus alpha r minus r e equal whole square
is positive and since this ah keeps on becoming
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i mean this become smaller and smaller what
happens is this keeps on increasing until
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it reaches the maximum value of a d then r
minus r e so large that this exponent is almost
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zero because exponential of minus alpha x
and if x is very very large then the exponential
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of minus alpha x goes zero therefore you see
that the v r reaches an asymptotic values
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and this the asymptote
therefore the d is the value for infinitely
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large ah or e or r compare to r e and when
r is less than r e please remember this is
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negative and therefore the exponential is
a positive ah it has a positive exponent therefore
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the exponential increases and this is the
square so one minus this is negative but this
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square of the that still increasing therefore
you see this increase very very quick very
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very steeply this is the shape of the morse
potentially in which the d is the dissociation
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energy from equilibrium
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you cannot measure d you can only measure
the dissociation energy from the zero point
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energy experimentally zero point energy will
be the first v is equal zero state and that
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will still have some energy namely half h
nu if you remember that the nu being the harmonic
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frequency and therefore this ah number in
morse oscillator called is d e dissociation
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energy from equilibrium but the experimental
values are d naught which is this number from
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zero point energy that's d naught ok so this
is how interpretation for the most potentials
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are given for their parameters
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so let me stop here for a movement and we
continue with the remaining five or six problems
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in the next part of this video