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welcome back to the lectures on chemistry
and in this series on the introduction to
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molecular spectroscopy i would like to put
some ah problems sample problems and also
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the method of solving them therefore this
is a video tutorial and i hope it helps you
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to solve the regular tutorial problems of
the course as well as towards the assessment
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in the final exam these are elementary problems
and what i have in this is about ten ah sample
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ah exercises and i shall indicate the method
of solution not necessarily the final answer
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i think that you can calculate but you have
to relate the problem to the concept that
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was presented in the video so there is a question
of thus there is a collection of ten lecture
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ten problems here and i also want to point
out to the readers a very nice site that that
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i came across ah by professor kipeterson of
the university ah state university of wa washington
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washington state university yes and the site
contains a large number of physical chemistry
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course problems which are useful to you so
have a look at them now let me read the first
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problem and then indicate to you what is the
ah solution the first problem asks you to
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find the wave number the frequency and wavelength
of electromagnetic radiation consisting of
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photons each within energy of two electron
moles ok so this is a very straight forward
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ah relation that you have to remember namely
the energy of the photon is related to its
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wave number by the planks constant and the
speed of light times nu bar
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the second one is the energy related to the
frequency in terms of the planks constant
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and the frequency and the third is the energy
related to the wavelength in an inverse fashion
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with the planks constant and the speed of
the light and the wavelength decide/dividing
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dividing it therefore you are asked to calculate
nu bar you are asked to calculate nu and you
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are also asked to calculate nu and you are
also asked to calculate lambda for a given
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energy units are important
important please underline this the unit given
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to you is two electron volt for the energy
and remember planks constant h is in the unit
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of joule second six point six two six times
ten raise to minus thirty four joule second
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these are all si units and the frequency nu
is in hertz or per second and the wave number
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is in the unit of meter inverse since they
are all si units the electron volt should
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obviously be converted to ah joules you recall
that one electron voltage one point six zero
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two times ten to the minus nineteen joules
so if you remember that then everything else
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is known to you h is given c is given therefore
nu bar can be calculated is given h is the
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planks constant so its easy to write its already
written down here as six point six two six
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ah into ten raise to minus thirty four and
the speed of light if you want to write that
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is for this calculation you can use the approximate
value three into ten to the eight meters per
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second
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so everything is given therefore you only
need to calculate the quantity nu so the units
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are i will write one step for cancellation
for hc nu bar as six point six two six times
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ten to the minus thirty four joule second
please write the units always in every such
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problem and c is given as three into ten to
the eight meters per second and you are asked
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to calculate nu bar and the value given here
is one this is ah two eb or one eb this is
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two eb therefore the energy given is two times
one point six zero to into ten to the minus
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nineteen joules per ev please remember when
you write the unit carefully these things
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cancel out ah in the correct manners so if
you write this as to ev and you write this
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as one point six zero two times ten to the
minus nineteen joules per ev see the electron
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volt cancel so what you get is three point
two zero four times ten to the power minus
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nineteen joules and thats equal to six point
six two six times ten to the minus thirty
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four joule second multiplied by three times
ten to the eight meters per second the second
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cancels off and these are equated and therefore
you see that nu bar is going to be in terms
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of meter inverse namely it is ah one point
sorry this is ah three point two zero four
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into ten raise to minus nineteen divided by
six point six two six into three into ten
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to the minus twenty six and the whole unit
is one by meter ok so dimensions and units
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are extremely important therefore in the similar
manner you can calculate nu you can calculate
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hc by lambda ok lets look at problem two ok
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in the second problem what is given are three
energy levels in the order a less than b less
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than c meaning the energy state a is like
this b is like this and c is like this and
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this is the increasing order of increasing
energy so this is you call it as a b c the
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energy difference between a and c is one electron
volt this delta e is given ok and that delta
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e is one electron volt and the wavelength
of right required light required for resonant
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transition from a to b is nine hundred nano
meters this lambda is given let me write it
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as delta e one and write this as delta e two
then energy conservation tells you that the
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delta e should be equal to delta e one plus
delta e two and you are given the wavelength
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for this transition as nine hundred nano meters
and you are asked to calculate what the wavelength
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of light required for resonant transition
from b to c that is this transition remember
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that delta e is inversely proportional to
the wavelength therefore this is delta e and
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this one is delta e one is given as nine hundred
nano meters therefore if you write hc by lambda
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one plus hc by lambda two and this is given
as one electron volt therefore what you have
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is one point six zero two times ten to the
minus nineteen joules on this side and thats
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equal to hc times one by nine hundred nano
meters plus one by lambda two ok nano meters
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therefore recall the units carefully you have
one point six zero two times ten to the minus
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nineteen joules and this is given as the sum
of two energies six point six two six times
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ten to the minus thirty four times three times
ten to the eight this is joules second and
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this is meters per second and the joule and
joule will go away the second and second will
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go away and what you have is one by nine hundred
nano meters which is nine hundred into ten
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to the minus nine meter plus one by lambda
two ok so once you take care of the units
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properly you see how everything cancels out
correctly and you can see that lambda two
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will be given in terms of meters ok this is
something easy to calculate please do that
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completely numerical exercise
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third problem these are all very elementary
problems the purpose of a discussing these
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problems in greater detail is to tell you
the methodology of solving them and also algebraically
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how you have to pay attention to details even
if they are trivial because at the end of
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it the devil is in the details please understand
that therefore the solution of any problem
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you have to pay attention to all the details
so here the third problem tells you the usual
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frequency range for vibrational spectroscopy
is four hundred to four thousand centimeter
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inverse ok this is in wave numbers and you
are given a typical vibrational stretching
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line at two thousand centimeter inverse and
you are all asked to calculate the energy
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corresponding to this two thousand centimeters
inverse in two units one as a joule per molecule
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and the other is the as a joule per mole ah
a typical vibration refers to the change in
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energy in the single molecule two thousand
centimeter inverse means the energy required
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to raise the energy levels of a single molecule
therefore this nu bar which corresponds to
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a transition between say some vibrational
state due to the stretching line having a
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v one and v one plus one this corresponds
to about two thousand centimeters inverse
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this is centimeter inverse si units are meter
inverse therefore first of all you need to
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make sure that nu bar is expressed in the
right unit namely since it is two thousand
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by centimeter its essentially two thousand
into hundred centimeter per meter so that
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it is also centimeter here that cancels off
ok so what you have is two into ten raise
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to five meter inverse so this is the nu bar
given to you and you are asked to calculate
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the j joules per molecule and thats given
as hc nu bar if everything else is expressed
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in the appropriate si units therefore you
do the same thing six point six two six times
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ten to the minus thirty four joule second
times three into ten to the eight meters per
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second into two times ten to the five per
meter so you see that goes off then the meter
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the second goes off that and what is left
over is joules and this is per molecule molecule
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and what is the value for per mole whatever
that you get energy in joules per molecule
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since there are avagadro number of molecules
in one mol
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namely six point zero two two times ten to
the twenty three per mole you can calculate
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the energy that you have done earlier this
number you multiply this number by the avagadro
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number then you get the energy per mol delta
e joules per molecule into six point zero
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two two times ten to the twenty three molecules
per mol so you have that gives you joules
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per mol ok so dimensionally you see things
ah look at these things carefully the next
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problem talks about the relative populations
of molecules in a two level system this is
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a model he level one is called the excited
state and the level zero is called the ground
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state and n one corresponds to the number
of molecules in the excited state and the
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number n zero corresponds to the number of
molecules in the ground state therefore you
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are asked to calculate the population ratio
subject to transition that would occur due
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to electromagnetic radiation of three different
types therefore what you have in this problem
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is three different sets of energies two level
system this is one system e zero e one with
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n zero n one this is system one corresponding
to a transition of five hundred angstroms
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the transition energy requires a wavelength
of five hundred angstroms this is another
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system i am not worried about the scale but
i am simply writing this as system two corresponding
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to a transition of thirty centimeter inverse
this is system three except that here frequency
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is given as ten to the seventeen hertz again
the temperature you will find out is an irrelevant
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data for this problem
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you are given three different units for the
energies and you are asked to calculate n
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one by n zero please remember he boltzmann
population
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formula for thermal equilibrium
if the two energy levels are thermally at
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equilibrium with each other the molecules
going up and down due to the absorption of
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energy so on then its ah immediately you should
remember that the number of molecule which
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are in any particular energy state e is proportional
to this ratio to this factor exponential minus
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e by kb t if the energy level is non degenerate
that is its its only one ah wave one function
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one state for one energy this is the formula
and therefore if you are looking at to the
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umber in the energy state e one verses the
number n the energy state e zero then you
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have to write to this as since proportionality
constant is the same for both of them the
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ratio is exponential minus e one by kb t with
exponential minus e zero by kb t where kb
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s the boltzmann constant so fundamental constant
and t is a temperature in kelvin in kelvin
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ok therefore he answer is exponential minus
e one by e e one minus e naught by kb t ok
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if you go back to the problem you see one
data point is missing ah here t is equal to
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twenty seven is important we need that data
t is equal to twenty seven or its three hundred
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kelvin ok
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so what you have is kb is known kb is one
point three eight times ten to the minus twenty
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three joules per kelvin therefore kb t means
that t should be in kelvin so twenty seven
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degrees essentially means two seventy three
plus twenty seven kelvin which is three hundred
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kelvin kelvin so kb is known t is known and
delta e is given by the corresponding either
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the lambda or the nu bar or the nu please
remember for this the delta e is hc by lambda
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and for this the delta e is hc nu bar and
for this the delta e is h nu so there are
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three different units used and therefore in
each one of them you can calculate the only
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thing you have to remember is this angstrom
and angstrom refers to ten raise to say if
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it is five hundred angstrom what it means
is that five hundred angstrom is five hundred
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into ten to the minus eight centimeters or
its five hundred into ten to the minus ten
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meters you have to use si units therefore
convert everything into meters seconds and
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joules if energy is given in ah electron volts
convert that into joules if the energy is
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given in calories convert that into joules
and then its easy to use the si unit and calculate
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the ratio exponential minus e one by e naught
e one minus e naught is that delta e corresponding
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to this number so the lambda is five hundred
here multiplied by ten to the minus ten to
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give you the corresponding unit in meters
and this is thirty centimeter inverse therefore
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you must multiply by hundred to give you meter
inverse and this is in the right unit hertz
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therefore you multiply by h so you can calculate
delta e for this delta e for this and delta
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e for this you are three different ah models
three different what are called the pairs
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of energy levels therefore you get three answers
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the next problem this is something i had also
mentioned in my lecture show that for a diatomic
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molecule with two atoms of mass m one and
m two the moment of inertia i for an axis
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passing through the center mass of the molecule
and perpendicular to the bond axis is given
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by the formula i is equal to mu r square ok
now whats your picture picture is you have
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a diatomic molecule with two different masses
m one and m two the center of mass is closer
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to the heavier nucleus and you are asked to
calculate the moment of inertia about an axis
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which is perpendicular to the bond axis this
is the bond axis and this is the center of
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mass point the origin k for this system the
center of mass ok and if you know that the
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radius vector of this atom from this origin
is r one and the radius vector of the center
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of this atom from the origin is r two hen
the inter atomic distance or let me write
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to the proper vector unit the inter atomic
distance r is given that one direction is
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positive suppose this is the positive x axis
then r two is positive but r is on the opposite
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sign therefore r will be r two minus r one
vectorally and what is the moment of inertia
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the moment of inertia is m one r one square
plus m two r two square because r one and
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r two are the perpendicular distances of the
two masses m one and m two from the axis about
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which we want to calculate the center of mass
so m one r one square and m two r two square
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this is what is to be rewritten as mu r square
where mu is the reduced mass m one m two by
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m one plus m two ok
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and the radius vector the distance vector
r is related to the two vectors r two and
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r one by this choice but also please remember
the definition of the center of mass means
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the following that m one vector r one plus
m two vector r two that should be zero because
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the center of mass is also the origin therefore
if its a pure vibration without the molecule
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moving anywhere then the pure vibration is
essentially means of the center of the mass
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remains where it is that means this condition
is satisfied for any vibrational amplitude
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or for m one r one plus m two r two is zero
and vectorally and the inter nuclear distance
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vector r is the difference r two minus r one
because of the choice of the directions with
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this you have to do a simple substitution
and get the answer let me write that now ok
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so what you do is given the fact that r is
r two minus r one think thats what i have
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written and so r two minus r one yes thats
right r two is equal to r plus r one the condition
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that m one r one plus m two r two is equal
to zero gives you also r two is equal to minus
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m one r one by m two we substitute that you
get the result namely r two is this value
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r plus r one is that therefore from these
two you can get the expression r is equal
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to minus m one r one by m two minus r one
ok r two minus r one therefore thats what
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you have so this is the definition of r using
one vector r one
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now i is m one r one square plus m two r two
square so let us replace r two by whatever
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we have namely m one r one square plus m two
times r two square which is m one square by
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m two square times r one square these are
scalars therefore the minus sign goes away
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and so what is left over is you have m one
if you take the common factor m one m two
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square r one square plus m two m one square
r one square divided by m two square ok so
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this you have to substitute the r one you
have to substitute into this if you do this
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substitution of r one as given here using
r ok what is r one r is minus m one plus m
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00:25:57,600 --> 00:26:12,070
two by m two times r one from this equation
therefore r one is m two r with the minus
199
00:26:12,070 --> 00:26:18,880
sign divided by m one plus m two ok so if
you substitute r one square here you will
200
00:26:18,880 --> 00:26:29,910
get immediately the expression that i is equal
to m one m two by m one plus m two times r
201
00:26:29,910 --> 00:26:44,660
square ok please remember this is nothing
other than m one plus m two by m two times
202
00:26:44,660 --> 00:26:54,310
r one square so substitute this and you get
the moment of inertia i as this value ok this
203
00:26:54,310 --> 00:26:59,990
is the standard expression its a very simple
algebra but that be ah vector notation has
204
00:26:59,990 --> 00:27:07,750
to be taken into account and you must know
what is meant by the center of mass this is
205
00:27:07,750 --> 00:27:11,670
used in rotational spectroscopy in diatomic
molecular systems
206
00:27:11,670 --> 00:27:20,870
now the next problem h in this problem we
are asked to calculate the force constant
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00:27:20,870 --> 00:27:36,100
in the unit of newton per meter for the hydrogen
bromide molecule given that the fundamental
208
00:27:36,100 --> 00:27:45,260
the fundamental frequency is given as for
car car for hbr it is two six five zero centimeter
209
00:27:45,260 --> 00:27:51,040
inverse and also calculate the same for carbon
monoxide that is the force constant given
210
00:27:51,040 --> 00:27:57,820
that the frequency is given as twenty one
seventy centimeter inverse and you are also
211
00:27:57,820 --> 00:28:04,070
asked to calculate on the strengths of the
bonds based on your result ok first of all
212
00:28:04,070 --> 00:28:17,350
remember from the elementary vibrational spectroscopy
that the fundamental frequency nu in hertz
213
00:28:17,350 --> 00:28:41,450
is given as one by two pi square root of k
by mu mu is the reduced mass of diatomic
214
00:28:41,450 --> 00:28:54,940
k is the force constant remember the unit
newton that is basically kilogram meter per
215
00:28:54,940 --> 00:29:10,080
second square force into distance is energy
so given that newton meter inverse is kilogram
216
00:29:10,080 --> 00:29:18,980
per second square thats the force constant
k the unit ok the reduced mass mu should be
217
00:29:18,980 --> 00:29:27,650
in kilograms therefore nu will be square root
of second to the minus two and so it is hertz
218
00:29:27,650 --> 00:29:35,770
ok so this is hertz you are given nu bar as
twenty six fifty centimeter inverse therefore
219
00:29:35,770 --> 00:29:44,270
whats the relation between nu and nu bar please
remember this is hertz and i think you will
220
00:29:44,270 --> 00:29:54,830
remember this as per meter and it is meter
per second which is the speed of light gives
221
00:29:54,830 --> 00:30:05,070
you per second so the relation is nu hertz
is c nu bar and nu bar is in centimeter inverse
222
00:30:05,070 --> 00:30:15,200
since you are using si units convert that
as c u bar into hundred ah you will get the
223
00:30:15,200 --> 00:30:19,250
hertz ok so these are the things you have
to remember
224
00:30:19,250 --> 00:30:31,280
what about the reduced mass for h br if we
use the atomic masses simply as one and eighty
225
00:30:31,280 --> 00:30:36,450
and that worried about the details you can
actually put the exact number but the process
226
00:30:36,450 --> 00:30:43,750
is one and eighty then the reduced mass s
eighty into one divided by eighty one m one
227
00:30:43,750 --> 00:30:51,660
plus m two and this is has to be divided by
six point zero two two into ten raise to twenty
228
00:30:51,660 --> 00:31:00,920
three to get the mass of one molecule in grams
therefore you have to multiply this by ten
229
00:31:00,920 --> 00:31:09,430
raise to minus three eighty one into six point
zero two two times ten to the twenty three
230
00:31:09,430 --> 00:31:21,100
so that would give you kilo grams ok so given
then nu is equal to two pi nu is equal to
231
00:31:21,100 --> 00:31:32,820
square root of k by mu you get immediately
k is equal to four pi square c nu bar because
232
00:31:32,820 --> 00:31:42,700
thats nu times mu c is known nu bar is known
in ah meter inverse mu is known therefore
233
00:31:42,700 --> 00:31:51,070
you calculate k k you calculate for h br k
you calculate for co using the reduced mass
234
00:31:51,070 --> 00:31:59,500
for co this expression namely twelve carbon
sixteen twelve into sixteen divided by twenty
235
00:31:59,500 --> 00:32:05,530
eight into six point zero two two times ten
to the twenty three and also multiply by ten
236
00:32:05,530 --> 00:32:11,690
to the minus three to give you the kilograms
therefore for carbon monoxide you know the
237
00:32:11,690 --> 00:32:16,820
mu for carbon monoxide you are given the nu
bar in met centimeter inverse convert that
238
00:32:16,820 --> 00:32:22,910
into meter inverse put the c in meters per
second and then you see that you can calculate
239
00:32:22,910 --> 00:32:31,510
k find out which is greater than which the
larger the force constant the stronger the
240
00:32:31,510 --> 00:32:46,130
bond larger the k the stronger the bond therefore
you can find out which of the k the carbon
241
00:32:46,130 --> 00:32:52,050
monoxide or h br is larger and you will see
this all the numerical results a numbers will
242
00:32:52,050 --> 00:32:58,010
be there in your websites along with the problems
but this is how i would like to pay attention
243
00:32:58,010 --> 00:33:00,250
to the details
244
00:33:00,250 --> 00:33:08,690
now the next problem is something very similar
for hydro hydrogen chlorine calculate the
245
00:33:08,690 --> 00:33:15,280
maximum amplitude of the lowest vibrational
level can the molecule be approximated as
246
00:33:15,280 --> 00:33:21,010
a simple harmonic oscillator and what are
given you are given the force constant and
247
00:33:21,010 --> 00:33:31,700
since the molecule is given as hcl you also
know that the ah reduced mass can be calculated
248
00:33:31,700 --> 00:33:36,420
what is meant by the maximum amplitude of
the lowest vibrational level what it means
249
00:33:36,420 --> 00:33:48,650
is that if the entire energy of the harmonic
oscillator is vibrational potential energy
250
00:33:48,650 --> 00:33:59,230
and its equal to half k x maximum square please
remember if you do a classical model if this
251
00:33:59,230 --> 00:34:05,610
is the energy these are called the turning
points in the classical model corresponding
252
00:34:05,610 --> 00:34:13,730
to the maximum amplitude of vibration where
the potential energy is equal to the total
253
00:34:13,730 --> 00:34:22,290
energy right in the middle that is of the
equilibrium position the molecule has no potential
254
00:34:22,290 --> 00:34:28,419
energy because we assume that the potential
energy is zero at equilibrium distance therefore
255
00:34:28,419 --> 00:34:34,669
the entire energy is kinetic energy therefoer
the molecule has the maximum vibrational velocity
256
00:34:34,669 --> 00:34:52,200
so you have two conditions that at the turning
points half k x maximum square sorry this
257
00:34:52,200 --> 00:35:11,950
is x maximum square x maximum square thats
equal to e and in the middle half m v max
258
00:35:11,950 --> 00:35:19,549
square is equal to e so if you are ask to
calculate maximum vibrational amplitude where
259
00:35:19,549 --> 00:35:24,210
it it means essentially the potential energy
is all the total energy and since you are
260
00:35:24,210 --> 00:35:29,980
given the value of k and you are given the
value of e you can calculate x max square
261
00:35:29,980 --> 00:35:36,230
pay attention to the units if in a similar
problem if somebody ask you what s the maximum
262
00:35:36,230 --> 00:35:42,369
vibrational velocity what it means is that
assuming that the molecule has no potential
263
00:35:42,369 --> 00:35:48,570
energy during that vibration meaning is right
at the equilibrium during the vibration whats
264
00:35:48,570 --> 00:35:53,119
the speed of the atom this is a classical
model so either way you can calculate k is
265
00:35:53,119 --> 00:36:09,410
given as kilogram per second square therefore
you can calculate e the equilibrium bond length
266
00:36:09,410 --> 00:36:19,310
is given and in order to calculate e you also
need the vibrational energy in the ground
267
00:36:19,310 --> 00:36:35,900
state therefore you need as an input vibrational
energy in the ground state that can be calculated
268
00:36:35,900 --> 00:36:45,059
as remember the frequency in nu is one by
two pi square root of k by mu since k is given
269
00:36:45,059 --> 00:36:53,549
and mu can be calculated for hcl you know
what the nu is and once nu is known in hertz
270
00:36:53,549 --> 00:37:04,630
the energy e is h nu ok therefore the energy
is known and immediately you can write that
271
00:37:04,630 --> 00:37:19,349
this energy is all half kx maximum square
giving you x maximum is two e by k square
272
00:37:19,349 --> 00:37:32,690
root and if k is one by sorry if k is a known
e has been calculated therefore x max can
273
00:37:32,690 --> 00:37:33,690
be calculated
274
00:37:33,690 --> 00:37:43,750
now why are they given the data equilibrium
bond distance equilibrium bond distance is
275
00:37:43,750 --> 00:37:53,460
one twenty seven point four pico meters this
is r equilibrium is one twenty seven point
276
00:37:53,460 --> 00:38:00,130
four pico meters meaning it is one twenty
seven point four into ten raise to minus twelve
277
00:38:00,130 --> 00:38:09,230
meters ok f you calculate x max and if the
x max is very large compared to the equilibrium
278
00:38:09,230 --> 00:38:17,660
bond distance then what it means is that the
molecule at that lower energy level its vibrational
279
00:38:17,660 --> 00:38:30,029
velocity is such that the harmonic model is
meaningful if however the maximum bond length
280
00:38:30,029 --> 00:38:36,819
is very far away from the equilibrium then
you have to think about that ok whether the
281
00:38:36,819 --> 00:38:44,569
vibrational the unharmonocity ah the harmonic
model is appropriate to that of the equilibrium
282
00:38:44,569 --> 00:38:50,900
bond distance therefore at this energy if
the maximum amplitude is very close to the
283
00:38:50,900 --> 00:38:58,380
bond distance then what it means is that the
vibrational ah amplitude is not very different
284
00:38:58,380 --> 00:39:05,309
from is very small and therefore the equilibrium
bond distance and the fully stretched vibrational
285
00:39:05,309 --> 00:39:15,310
molecule the two bonds are very close to each
other for example if this is that re given
286
00:39:15,310 --> 00:39:23,759
by one twenty seven point four and corresponding
to this energy this number is a fraction of
287
00:39:23,759 --> 00:39:28,510
one twenty seven point four if it is less
than say ah ten percent or five percent of
288
00:39:28,510 --> 00:39:35,509
one twenty seven point four then it make sense
that the vibration is harmonic because the
289
00:39:35,509 --> 00:39:40,680
amplitude of vibration is very small if its
a very large number hen the amplitude is very
290
00:39:40,680 --> 00:39:45,359
large if the amplitude is very large then
the harmonic model breaks down therefore when
291
00:39:45,359 --> 00:39:51,900
you calculate x max you will know how close
it is to the equilibrium bond distance and
292
00:39:51,900 --> 00:39:57,470
that will give you the picture whether the
molecular vibration that energy is nearly
293
00:39:57,470 --> 00:39:59,759
harmonic or not ok
294
00:39:59,759 --> 00:40:11,349
so you look at the numbers given in the website
for solutions the next problem talks about
295
00:40:11,349 --> 00:40:18,150
the vibrational frequency dependence on the
masses of the redu reduced masses of the atoms
296
00:40:18,150 --> 00:40:28,869
and here the oh stretching vibration corresponding
to the reduced mass species oh is about three
297
00:40:28,869 --> 00:40:36,279
thousand six hundred centimeters so if you
write to the nu as one by two pi times k by
298
00:40:36,279 --> 00:40:44,749
mu this mu corresponds to mu of oh assuming
that the force constant of the bond to remain
299
00:40:44,749 --> 00:40:54,609
the same due to isotopic substitution of h
by b so let us call this as nu one corresponding
300
00:40:54,609 --> 00:41:04,759
to oh we are asked to find out the expected
shift in the stretching frequency so if this
301
00:41:04,759 --> 00:41:20,119
is nu one its one by two pi k one by mu oh
you are told that the force constant is the
302
00:41:20,119 --> 00:41:25,430
same therefore if there is any difference
in the vibrational stretching frequency it
303
00:41:25,430 --> 00:41:35,789
is due to the fact that it is k one by mu
od the reduced mass of od and oh are different
304
00:41:35,789 --> 00:41:45,160
please remember oh is sixteen into one by
seventeen times six point ah zero two two
305
00:41:45,160 --> 00:41:52,460
times ten to the twenty six if you want to
write this in kilograms ok whereas the mu
306
00:41:52,460 --> 00:42:01,099
of od if you remember is sixteen into two
by eighteen sixteen plus two times six point
307
00:42:01,099 --> 00:42:10,720
zero two two into ten to the twenty six kilogram
therefore the mu of od and the mu of oh are
308
00:42:10,720 --> 00:42:16,230
quite different and since it is a isotopic
substitution if the force constant remains
309
00:42:16,230 --> 00:42:23,989
the same the bond strength is very much the
same then what is the shift in nu one ah due
310
00:42:23,989 --> 00:42:34,730
to the substitution therefore you are only
calculating the ratio nu one by nu two as
311
00:42:34,730 --> 00:42:47,650
one by two pi square root of k by mu oh divided
by one by two pi square root of k by mu od
312
00:42:47,650 --> 00:42:57,200
and obviously after canceling what you see
is mu od by mu oh so you can see the ratio
313
00:42:57,200 --> 00:43:03,489
nu one and nu two and therefore ah if nu two
is substantially different from nu one you
314
00:43:03,489 --> 00:43:09,230
you can see that this number od is bigger
than the number mu oh and therefore the nu
315
00:43:09,230 --> 00:43:14,730
one for oh is more than that for nu two the
vibrational frequency is reduced when you
316
00:43:14,730 --> 00:43:19,720
substitute with the heavier mass and because
its inversely proportional to reduced mass
317
00:43:19,720 --> 00:43:24,150
and you can see immediately that nu two will
be different whats the difference between
318
00:43:24,150 --> 00:43:36,609
nu one and nu two thats the shift in the vibrational
frequency is the shift so substitute the numbers
319
00:43:36,609 --> 00:43:40,789
here and you will immediately get the answer
320
00:43:40,789 --> 00:43:59,520
ok the next problem is on the beer lambert
law a certain solution in cell absorbs ten
321
00:43:59,520 --> 00:44:05,880
percent of incident light what fraction of
the incident light will be absorbed in a cell
322
00:44:05,880 --> 00:44:14,900
five times longer i think i changed that number
from ten to ah eighty percent just to get
323
00:44:14,900 --> 00:44:21,400
some logarithmic ratios reasonable ok eighty
percent of incident light so recall the beer
324
00:44:21,400 --> 00:44:32,309
lambert law that the absorbents which is the
logarithm of incident intensity of light to
325
00:44:32,309 --> 00:44:41,150
the intensity of light that is emitted thats
the molar extension coefficient epsilon times
326
00:44:41,150 --> 00:44:50,119
concentration of the substance times the length
of the cell now when you say a cell absorbs
327
00:44:50,119 --> 00:44:58,710
eighty percent of incident light what it means
is that obviously when i naught and i are
328
00:44:58,710 --> 00:45:08,710
looked at i is twenty percent of i naught
because the rest is absorbed so when eighty
329
00:45:08,710 --> 00:45:13,019
percent is absorbed what you means is twenty
percent is transit transmitted therefore i
330
00:45:13,019 --> 00:45:23,980
naught by i is five so you know log five is
sum epsilon for that substance times its concentration
331
00:45:23,980 --> 00:45:30,380
times the length of the cell now you are asked
to calculate what is the corresponding log
332
00:45:30,380 --> 00:45:38,530
i naught prime by i prime sorry if the concentration
of the cell remains the same concentration
333
00:45:38,530 --> 00:45:42,329
of the substance remains the same then the
length becomes the five times the length of
334
00:45:42,329 --> 00:45:48,799
the cell a cell five times longer so you can
calculate this number by taking the ratio
335
00:45:48,799 --> 00:45:54,160
because this goes away this goes away this
in the ratio when you divide the length also
336
00:45:54,160 --> 00:46:03,109
goes away so what you have is log five by
a prime due to the five times longer length
337
00:46:03,109 --> 00:46:11,650
is one by five therefore you can calculate
a prime as five times log five log five is
338
00:46:11,650 --> 00:46:18,700
point six nine nine zero so you have five
times point six nine nine zero thats a prime
339
00:46:18,700 --> 00:46:30,400
but what is a prime a prime is log i naught
by i therefore from this taking the antilogarithm
340
00:46:30,400 --> 00:46:36,609
of this value you can calculate i naught and
the ratio i naught by i so if the ratio for
341
00:46:36,609 --> 00:46:46,369
example is three percent if what it means
is that if sorry the ratio is not three percent
342
00:46:46,369 --> 00:46:55,910
if the ratio is thirty three point three what
it means is that three percent of i naught
343
00:46:55,910 --> 00:47:01,440
gets transmitted the remaining ninety seven
percent gets absorbed therefore based on the
344
00:47:01,440 --> 00:47:11,390
antilogarithm you can calculate i naught by
i as anti log a prime and from this you can
345
00:47:11,390 --> 00:47:23,200
get the value of i as the fraction some of
i naught so this number will also be there
346
00:47:23,200 --> 00:47:28,099
in the text
347
00:47:28,099 --> 00:47:37,519
the last problem is again on beer lamberts
law and it talks about another compound methanol
348
00:47:37,519 --> 00:47:44,029
absorbing at a particular frequency particular
ah light wavelength and it has some absorbance
349
00:47:44,029 --> 00:47:54,720
of one fifty dm cube per mol per centimeter
so absorbance is a given by epsilon into concentration
350
00:47:54,720 --> 00:48:09,930
length l epsilon is given as one fifty dm
cube per mol and the concentration of the
351
00:48:09,930 --> 00:48:22,369
cell is given as point o one mol per dm cube
and the length of the cell this is c and the
352
00:48:22,369 --> 00:48:29,630
length of the cell is one centimeter there
is also here per centimeter so dimensionally
353
00:48:29,630 --> 00:48:35,799
absorbance is has no no dimensions case that
cancels are and the dm cube mol inverse is
354
00:48:35,799 --> 00:48:42,839
a ah the inverse of the mol dm cube so what
you have is one point five ok absorbance is
355
00:48:42,839 --> 00:48:48,550
one point five and what is the percentage
of light transmitted this is equal to log
356
00:48:48,550 --> 00:48:59,880
t naught by i and in this case if i look at
the numbers i think it is about thirty one
357
00:48:59,880 --> 00:49:09,499
point that is i naught by i is a log of one
point five its about thirty one point seven
358
00:49:09,499 --> 00:49:17,440
so what it means is that ah i is thirty one
point seven times less than i naught so the
359
00:49:17,440 --> 00:49:23,670
percentage of light that comes out is roughly
three point one five percent because thats
360
00:49:23,670 --> 00:49:34,979
what it is i naught by thirty one point seven
will be i and thats three point one five percent
361
00:49:34,979 --> 00:49:43,559
ok of the original i naught so these are simple
ways of manipulating ah elementary exercise
362
00:49:43,559 --> 00:49:51,069
of course you have to learn more and more
about the computations of small parameter
363
00:49:51,069 --> 00:49:57,119
as well as the numerical parameters that you
need to calculate in solving these problems
364
00:49:57,119 --> 00:50:07,400
and let me share with you the last slide of
the ppt which contains these problems in text
365
00:50:07,400 --> 00:50:21,849
form for you let me go to the presentation
last slide is quite important most important
366
00:50:21,849 --> 00:50:28,529
whatever you do you have to enjoy your learning
and if there are problems of similar nature
367
00:50:28,529 --> 00:50:32,970
that you want to solve in this course by looking
at text books and other slides please do send
368
00:50:32,970 --> 00:50:39,780
and share them ah with the ah other students
in the website but please remember even though
369
00:50:39,780 --> 00:50:45,019
i have given you a tutorial please remember
learning is only reinforced through these
370
00:50:45,019 --> 00:50:49,619
problem solving methods its not the other
way around you dont learn by solving problems
371
00:50:49,619 --> 00:50:57,680
your learning is through understanding a concept
thinking about it and discussing that problems
372
00:50:57,680 --> 00:51:04,140
ensure that the way you have understood that
is reinforced by solving them right therefore
373
00:51:04,140 --> 00:51:11,509
please remember learning is only reinforced
through problem solving it is important that
374
00:51:11,509 --> 00:51:17,079
problem solving is necessary its not a sufficient
exercise to understand a subject therefore
375
00:51:17,079 --> 00:51:23,509
please read more please listen to the lecture
please find your own problems and online learning
376
00:51:23,509 --> 00:51:28,359
through this course or by any other course
on your own is actually by self motivation
377
00:51:28,359 --> 00:51:31,859
and while i continue to solve similar problems
i wish you all the best
378
00:51:31,859 --> 00:51:32,359
thank you