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welcome back to the video for solving problems
in elementary spectroscopy this continues
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from what i just did for four problems in
the video tutorial two and the we shall start
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with next problem namely problem five n which
we are asked to do the following determine
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the fraction of molecules ah in the v is equal
to one two and three state relative to the
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ground state v is equal to zero for the molecule
bromine monoflouride the i think the last
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some type of graphical things here we frequency
of the ah transition is six six nine point
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seven centimeter inverse ok so what the problem
say is at t is equal to thousand kelvin at
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t is equal to thousand kelvin and when nu
is given by six six nine point seven centimeter
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inverse ah for the molecule bromine monoflouride
ok you are asked to calculate the number of
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molecules in the v is equal to one relative
to the number of molecules in the v is equal
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to zero state ok this is very elementary problem
in the sense it looks at the boltzmann population
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at equilibrium and since vibrational states
for this single degree of freedom or non degenerate
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vibrational states are non degenerate
the ratio is given by the simple exponential
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factor namely exponential minus e v is equal
to one by kb t divided by exponential minus
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e v is equal to zero by kb t and you know
the difference between v is equal to one and
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v is equal to zero the energy the answer is
h nu ok and nu is given as sixty nine therefore
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hc nu bar therefore the delta e e to the minus
delta e by kb t for this ration n is equal
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to v is equal to one to n at v is equal to
zero this delta e is given by hc nu bar ok
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and the nu bar is given as six sixty nine
point ah seven centimeter inverse therefore
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what you have to do is to ensure that we use
the right units namely its given as six sixty
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nine point seven into ten to the two meter
inverse times six point six two six into ten
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raise to minus thirty four this is h this
is nu bar times three into ten to the eight
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meters per second and this is joules and the
meter inverse and so what you see is the delta
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e is given as ah ah joule second ok joule
second times that therefore the answer is
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given in terms of joules so this can be calculated
and kb t is ah one point three eight times
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ten to the minus twenty three joules per kelvin
and its calculated at thousand kelvin so that
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is given as joules ok
therefore delta e by kb t is a numerical exercise
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ok for you have to do that and take the exponential
of minus that in order to calculate the ratios
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of n between the two different levels and
the only thing that you have to do for two
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and three is that for two the energy difference
between e zero and e two is two h nu and for
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ah three it is three h nu therefore you will
see that there are relative populations in
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the first second and third excited state drastically
come down to very very small values thats
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the purpose of this exercise to show that
in vibrational ah spectroscopy the number
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of molecules in the higher energy levels is
much less than the number of molecules in
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the ground state and this number s less if
the frequency is even more ok that was the
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idea for this problem ok therefore you can
calculate that ok then next problem is formal
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ah process is basically asking you to write
all the simple harmonic oscillator wave functions
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for n equal to zero to n equal to three is
a text book exercise and most important for
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this is determine why the selection rule for
vibrational spectrum of a harmonic oscillator
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contains only one line with the selection
rule delta v is equal to plus minus one thats
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important ok problem seven and you are asked
to write the harmonic oscillator functions
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so text book exercise please remember that
the psi n is a normalization constant which
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is a function of n and then it is e to the
minus alpha x square by two times hn of root
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alpha x ok
therefore what is alpha alpha is equal to
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square root of k mu by h bar square so k is
the force constant mu is the reduced mass
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and those are the only two parameters of this
problem therefore this is the vibrational
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wave function and h zero is one h one of root
alpha x is two root alpha x h two of root
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alpha x is four alpha x square minus two and
the third h three root alpha x is eight alpha
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root alpha x cube minus twelve root alpha
x please remember this makes the whole expression
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dimensionless because the alpha has a dimension
of one by length and therefore you see that
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the harmonic oscillator x is the length parameter
its distance from the equilibrium ah for vibration
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and therefore this all quantity polynomials
are dimensionless so these are the corresponding
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forms and each one of them you have to multiply
by to the minus alpha x square by two and
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also put down the general formula for n n
these are available in text books
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but the second part of the problem is important
namely why the selection rule delta nu is
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equal to plus minus one ok ok this is a mathematical
statement that the selection rule for transition
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between e n with the wave functions psi n
and e lets write n double prime n prime psi
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n prime n double prime the selection rule
for vibrational transition is due to the dipole
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moment matrix element which connects these
two states in fact the transition probability
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this is the statement you have to take it
from me transition probability for ah transition
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between n prime and n double prime n prime
and n double prime is given by the absolute
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square of the integral from minus infinity
to plus infinity psi n double prime psi n
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prime star mu which let me not write ah yeah
mu is he dipole moment here dipole moment
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ok not the reduced mass mu dipole moment psi
n double prime dx
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therefore mu is essentially mu naught times
ah x and this x keeps changing therefore during
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vibration the dipole is created and changed
destroyed and so are increased and decreased
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therefore this x is the operator the position
operator and the mu naught is a ah is the
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constant per for the permanent dipole moment
position operator ok therefore the dipole
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keeps changing from mu naught to mu naught
times x during the process of vibration and
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the integral that you needed to calculate
is psi n prime of x from minus infinity to
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plus infinity star mu naught times x and then
you have psi n double prime x dx ok this is
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the this is what is call the transition probability
amplitude and the absolute square of a of
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this is the transition probability ah per
unit time whatever units that you wanted to
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use but the absolute square is the one which
is connected with the actual process of transition
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between vibrational levels and therefore it
is important for this bit what is called the
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matrix elements of the dipole moment operator
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between the two states x n prime and n prime
ok n double prime and n prime so the calculation
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essentially boils down therefore to writing
down in the case of vibrational spectroscopy
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the psi n are given by the exponential minus
alpha x square
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so let us write that minus infinity to plus
infinity the normalization constant associated
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with psi n prime which is en n prime and to
the function is e to the minus alpha x square
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by two times h n prime root alpha x this is
psi including this his is psi n prime ok and
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then you have x times the mu not and then
you have e to the minus alpha x square by
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two h n double prime root alpha x dx so it
turns out to be from minus infinity to plus
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infinity proportional to let me drop the constants
out there is also n double prime for the wave
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function associated with this state ok thats
the normalization concept for that state so
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this is the integral that needs to be calculated
its called the transition moment integral
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transition moment integral
now without any further explanation i will
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just make the statement that this integral
will be non zero only if when you write that
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like that the alpha x square by two the alpha
x square by two add up and then this integral
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will be non zero only if n prime and n double
prime differ by plus or minus one because
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it involves x if it involves x square then
the dipole moment is now being considered
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as the square of the amplitude for vibration
that i mean those kind of things do happen
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when you talk about the unharmonic states
of the molecule and so on the dipole moment
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in the lowest level of approximation is simply
proportional to the length and therefore what
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you see is the length x itself and this is
harmonic oscillator model and therefore the
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harmonic oscillator model this integral is
non zero only if n double prime is n prime
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plus or minus one because of the nature of
the hermite polynomial please understand the
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hermite polynomials are odd and even and so
what you see here as that the integral becomes
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even only when n prime and n double prime
and x these three products give you an even
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function but that argument should be very
carefully stated because this does not mean
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n and n prime can differ by three now doesnt
work this integral is still zero it cannot
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differ by other than one and therefore its
a property of the wave functions which results
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with the transition moments being connected
only in the vibrational energy levels if i
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have to write the potential energy and then
simply write the vibrational energies as something
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like that n equal to zero n equal to one n
equal to two and so on n equal to three the
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only transition that possible that can be
measured experimentally are this because of
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the nature of the wave functions therefore
the transition moment integral needed to be
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introduced in order for you to understand
this problem ok this will not happen k will
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not happen so vibrationally there is only
one line for a harmonic oscillator corresponding
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to this frequency ok they are all the same
therefore there is only one line nu bar ok
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thats the important point next problem ok
i miss problem six
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so let us go back to problem six ok problem
six is on a morse oscillator model ok please
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remember the morse oscillator model the energies
are given as h omega e h bar omega e v plus
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half minus ok here i am using omega e in terms
of wave numbers therefore let me write to
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this as h omega e and then minus h omega e
xe v plus half whole square this is the morse
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this is the expression for the morse oscillator
energies ok ah if you are writing this in
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terms of centimeter inverse lets do one more
thing lets get rid of this and lets write
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this as sum nu e as omega e xe minus omega
e times e plus half minus omega e xe times
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v plus half whole square therefore all are
in wave number units ok
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now the problem that is given to you is br
f let me now read the problem the problem
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given to you is that the vibrational frequency
omega e and unharmonocity constant omega ex
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e have been given from have informed from
experiments to be about centimeter inverse
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to be six six nine point seven and three point
eight six nine centimeter inverse and you
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are asked to calculate the first four morse
oscillator energy levels for those model molecule
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and also the transition frequencies for the
transitions v is equal to zero to two and
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v is equal to one to three ok
so lets look at that so the nu corresponding
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to the vibrational quantum number v is omega
e times v plus half minus omega e xe this
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is the unharmonocity v plus half whole square
so when v is equal to zero nu zero is half
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omega e minus one by four omega e xe nu one
is three by two omega e minus this is three
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by two square therefore nine by four omega
e xe and like wise you are asked to do that
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for four levels isnt it so nu two is five
by two omega e minus twenty five by four omega
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e xe and the last one omega three is seven
by two omega e minus forty nine by four omega
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e xe because the second term is the square
v plus half square so everything is a square
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here so this is these are the expressions
for the four energy levels of the morse oscillator
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ok now you are given omega e and omega e xe
so this is given as six six nine so if you
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write nu zero it is six six nine point seven
by two centimeter inverse and the other is
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one by four minus one by four times three
point eight six nine centimeter inverse so
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you know that therefore in a similar way you
can calculate nu one nu two nu three
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now what is important is the next step which
is v is equal to zero to v is equal to two
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so what is delta nu e zero to two if you look
at that its five by two omega e minus one
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by two omega e thats the first term five by
two omega first term minus one by two this
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is between two and zero and the next one is
minus twenty five by four omega e xe minus
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twenty five by four omega e xe minus one by
four omega e xe and so what you have is two
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omega e minus six omega e xe so this is the
second frequency and likewise delta nu e from
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one to three is seven by two omega e minus
one to three is one is three by two omega
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e minus seven by two corresponds to forty
nine by four omega e xe minus omega one corresponds
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to one by no ah nine by fourth three by two
whole square therefore nine by four omega
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e xe so what you have is two this uis also
two omega e xe minus forty by four so it is
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ten ten omega e xe what do we gain from this
process of calculating thats important is
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not just the problem what is important from
this problem is that if you look at it he
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morse oscillator model which has a potential
energy r v of r if you remember has something
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of this kind we saw in another problem if
you look at it the lowest energy level is
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half omega e minus one by four omega e xe
the next energy level if you look at it s
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three by two omega e minus nine by four omega
e xe the difference between the two is two
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omega e xe the unharmonocity term is two omega
e xe
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but if you go to the next level the levels
are closing in thats a purpose of doing this
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exercise the levels are closing in that is
successive energy differences are smaller
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because you remember this omega e xe term
is negative to any omega e term because you
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see any near by energy level say v is equal
to zero v is equal to one if you look at the
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transition between two nearby levels you see
this is zero to one is omega e minus two omega
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e xe but from one to two if you look at that
v is equal to two it is again omega e but
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one corresponds to nine by four and two corresponds
to ah twenty five by four thats is five by
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two whole square therefore the difference
between the two is four omega e xe so the
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second energy level difference v is equal
to one to two is smaller than the first energy
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level difference therefore the morse oscillator
energy levels actually they converge and so
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what you see is that there are many lines
that you can see in the vibration spectrum
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because not every one of them is not equal
to any every other thing they are different
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and also in morse oscillator this possible
for you to have multiple transitions v is
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equal to zero to two v is equal to one to
three and so on therefore you see morse oscillator
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gives you a real model of what is called the
vibrational spectrum a spectrum not one line
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thats the purpose for doing this exercise
lets go to problem eight now problem eight
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talks about molecule boron trifluoride and
it asks you to calculate the three principle
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moments of inertia and identify the time and
it also calculates ask us to calculate all
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the energy levels of this molecule for the
rotational quantum number j up to and including
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three ok first of all this is a symmetric
top bf three so is actually a an equilateral
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triangle that you can draw b the three florents
and these are the bond distances between the
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three fluorines they are identical ok this
is what you have
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now you are asked to calculate all the three
moments of inertia is a very simple exercise
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so what are these axes so we can choose one
axis let me use a different color let me choose
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one axis as the axis that passes through one
of the bonds ok the center of mass is right
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way the boron is because of the symmetry of
the system and therefore the second axis which
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is perpendicular to that is that axis again
passing through the boron and the third axis
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is perpendicular to the plane of this ah screen
ok thats perfect thats a third axis so you
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have what is call the ix iy and iz axis ok
now is easy to see that with respect to these
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two axis this axis as well as this axis the
moment of inertia that you calculate are identical
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lets do that ok the bf bond distance is given
please remember these angles are one twenty
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degrees therefore you can see that for if
we calculated for this axis we call it as
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the yeah we will call it as the x y and z
will be up the other so lets call this as
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a x axis one fluorine is on the axis and therefore
it doesnt contribute to the moment of inertia
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mu i mu one r one square call it as one thats
zero r is zero the other two fluorine atoms
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if you think about that this is sixty degrees
this is sixty degrees and this is the fluorine
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boron bond distance so if you write this as
r b f this distance as r b f then this perpendicular
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distance of the fluorine atom from the bond
axis is r b f times sine sixty ok so you get
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00:28:27,789 --> 00:28:41,490
the ri the r one r two this is two this is
three so m two r two square plus m three r
196
00:28:41,490 --> 00:28:48,140
three square we want to calculate this for
the moment of inertia i along the x axis because
197
00:28:48,140 --> 00:28:57,820
m one r one square is zero the mass of the
fluorine atom being m of f we have r two which
198
00:28:57,820 --> 00:29:11,049
is the perpendicular distances given as the
ah r b f sin sixty degree whole square and
199
00:29:11,049 --> 00:29:19,390
into two because this atom is also exactly
at the same distance from the axis therefore
200
00:29:19,390 --> 00:29:27,720
this is what you have sin sixty is root three
by two therefore you have mf three by fourth
201
00:29:27,720 --> 00:29:37,340
three by two so it is three by two mf r bf
square ok thats the first one
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00:29:37,340 --> 00:29:46,919
now what about this axis lets look at this
axis if you look at this axis all the three
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00:29:46,919 --> 00:29:54,289
atoms are away from this axis the distance
of let me now remove some of these things
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00:29:54,289 --> 00:30:04,970
make it easier
the distance is this the distance of this
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00:30:04,970 --> 00:30:11,220
atom is this they are both identical and the
its easy to see that this distance is and
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00:30:11,220 --> 00:30:27,929
also this one ok so let me redraw the boron
trifluoride this is ah the bond this is the
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00:30:27,929 --> 00:30:33,220
bond so the axis that we have is this so we
wanted to calculate this distance and we wanted
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00:30:33,220 --> 00:30:44,779
to calculate this distance they are both the
same ok and then you have this is simply rbf
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00:30:44,779 --> 00:30:54,390
whole square it contributes m f one this is
f one this is f two this is f three this is
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00:30:54,390 --> 00:31:01,690
boron ok therefore this you can see immediately
as this is again ah thirty degrees so this
211
00:31:01,690 --> 00:31:24,340
distance will turn out to be r ah bf two times
sin thirty whole square times two of them
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00:31:24,340 --> 00:31:33,340
because f two and f three so that would be
the answer so this gives us as rbf square
213
00:31:33,340 --> 00:31:40,520
m f one plus all the rbf are the same therefore
this is sin thirty is one by two so its one
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00:31:40,520 --> 00:31:55,210
by four times two it gives you rbf square
times one by two mf so this is the total moment
215
00:31:55,210 --> 00:32:07,549
of inertia i and this is also three by two
rbf square mf so this is identical to the
216
00:32:07,549 --> 00:32:11,399
moment of inertia about this axis thats also
three by two
217
00:32:11,399 --> 00:32:22,340
now what about the perpendicular axis
you talk about the perpendicular axis it is
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00:32:22,340 --> 00:32:29,340
something which is like that therefore you
see all the three atoms are away from this
219
00:32:29,340 --> 00:32:37,010
axis bu exactly their bond distances therefore
the ah perpendicular distance from this axis
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00:32:37,010 --> 00:32:47,710
is actually the bf bond distance so you have
three times mf times the rbf square so this
221
00:32:47,710 --> 00:33:00,029
is what you get this is for iz now is interesting
that ix plus iy is equal to iz ok because
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00:33:00,029 --> 00:33:04,940
you have three by two here you have three
by two for the previous case and this is three
223
00:33:04,940 --> 00:33:10,091
this is a planarity in fact this is a planarity
condition for the moment of inertia the personal
224
00:33:10,091 --> 00:33:15,520
reason for including this as an example therefore
you see you have to sit down and try to calculate
225
00:33:15,520 --> 00:33:20,790
the moments of inertia from simple geometries
and you must know he actual structure the
226
00:33:20,790 --> 00:33:25,210
equilibrium structure of the molecule the
bond is the bond angles and then go back and
227
00:33:25,210 --> 00:33:30,520
calculate the rotational constants using that
and see in experiments whether thats the rotational
228
00:33:30,520 --> 00:33:35,240
constants you get and so on that is a reason
for this particular problem
229
00:33:35,240 --> 00:33:42,130
next the next problem talks about rotational
spacing between successive lines of hc l as
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00:33:42,130 --> 00:33:46,550
twenty point eight centimeter inverse and
its asking you to calculate the bond distance
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00:33:46,550 --> 00:33:53,960
ah its also asking you to determine the number
of molecules in the level j is equal to four
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00:33:53,960 --> 00:34:00,140
relative to j is equal to three and then one
more thing namely determine the level in which
233
00:34:00,140 --> 00:34:06,240
there is maximum population that is t is equal
to three hundred k may be assumed ok its important
234
00:34:06,240 --> 00:34:17,710
to remember that in the case of rotations
the line intensities actually increase as
235
00:34:17,710 --> 00:34:26,229
you go from j is equal to zero one two three
etcetera and if you talk about the intensity
236
00:34:26,229 --> 00:34:33,429
here the rotational line intensity is increases
something like that and there is a maximum
237
00:34:33,429 --> 00:34:42,950
and this we remember is due to the fact that
nj divided by nj lets write it as prime for
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00:34:42,950 --> 00:34:48,490
the upper level double prime for the lower
level this is given as the nu j prime the
239
00:34:48,490 --> 00:34:55,039
degeneracy of j prime and the degeneracy of
nu j double prime therefore the ratio times
240
00:34:55,039 --> 00:35:08,849
exponential minus delta e from j prime to
J double prime divided by kb t so the difference
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00:35:08,849 --> 00:35:15,140
between vibrational population and the rotational
population in this case is that here is a
242
00:35:15,140 --> 00:35:21,869
degeneracy factor associated with nu ah nu
j prime and j double prime and in the absence
243
00:35:21,869 --> 00:35:28,059
of any external field the j prime level has
a degeneracy j prime level has a degeneracy
244
00:35:28,059 --> 00:35:33,559
of two j prime plus one and like wise the
j double prime level has a degeneracy of two
245
00:35:33,559 --> 00:35:48,099
j double prime plus one therefore this problem
ask you to calculate the energy level n j
246
00:35:48,099 --> 00:35:55,779
is equal to four divided by n j is equal to
three this is a second subdivision the first
247
00:35:55,779 --> 00:36:04,430
division is the bond distance itself for r
h cl this is easy please remember the successive
248
00:36:04,430 --> 00:36:14,059
rotational lines when you write the intensities
the successive rotational lines are all differ
249
00:36:14,059 --> 00:36:24,130
by two b ok this is intensity this is the
wave number they all differ by two b and therefore
250
00:36:24,130 --> 00:36:31,069
two b is equal to twenty point twenty point
eight centimeter inverse and you know b is
251
00:36:31,069 --> 00:36:39,210
given by the formula h by eight pi square
i c and therefore h by eight pi square ic
252
00:36:39,210 --> 00:36:47,380
is ten point four centimeter inverse and since
i is equal to mu r square and in the case
253
00:36:47,380 --> 00:36:56,849
of hcl mu of hcl is one point zero zero eight
into thirty five point four five six divided
254
00:36:56,849 --> 00:37:06,670
by thirty six point four six four times one
point six six six one into ten raise to minus
255
00:37:06,670 --> 00:37:14,160
twenty seven kilograms ok
therefore mu is known r square is what you
256
00:37:14,160 --> 00:37:25,279
are asked to calculate so you write h by eight
pi square mu r square c is equal to ten point
257
00:37:25,279 --> 00:37:32,769
four into ten raise to two meter inverse make
sure that you use the same si units because
258
00:37:32,769 --> 00:37:41,089
we are using kilogram and r is in meters therefore
you see immediately r is given as square root
259
00:37:41,089 --> 00:37:54,670
of h by eight pi square mu c times ten point
four into ten raise to two ok this is and
260
00:37:54,670 --> 00:38:00,039
h of course is known mu is known and c is
known therefore you can calculate r therefore
261
00:38:00,039 --> 00:38:03,519
this is a very straight forward substitution
formula ok
262
00:38:03,519 --> 00:38:11,369
now the second is of course n j is equal to
four divided by nj is equal to three please
263
00:38:11,369 --> 00:38:19,809
remember e j is equal to four if you recall
for the ah microwave spectrum of rigid diatomic
264
00:38:19,809 --> 00:38:28,920
molecules it is hc b j into j plus one ok
therefore for j is equal to four it is twenty
265
00:38:28,920 --> 00:38:39,010
hc b in terms of joules ok thats why the h
and c are here for e j is equal to three it
266
00:38:39,010 --> 00:38:48,309
will be four three into four so it will be
twelve hcb therefore the delta e between j
267
00:38:48,309 --> 00:39:01,119
is equal to four to j is equal to three is
eight hc b and you are given two b as twenty
268
00:39:01,119 --> 00:39:09,339
point eight centimeter inverse therefore eight
the delta e is given as eight into hc into
269
00:39:09,339 --> 00:39:18,789
this is two b therefore i would say four times
four hc times twenty point eight into ten
270
00:39:18,789 --> 00:39:28,959
raise to two meter inverse so delta e is known
and nj is equal to three to four to three
271
00:39:28,959 --> 00:39:37,119
yes nj is equal to four divided by n j is
equal to three is nine to j plus one j is
272
00:39:37,119 --> 00:39:44,609
equal to four two j plus one seven and then
you have e to the minus delta ej is equal
273
00:39:44,609 --> 00:39:54,329
to four to j is equal to three divided by
kb t where kb is one point three eight into
274
00:39:54,329 --> 00:40:02,979
ten raise to minus twenty three minus twenty
three joule per kelvin and t is three hundred
275
00:40:02,979 --> 00:40:08,749
kelvin therefore you can calculate this number
by doing the numerical work ok
276
00:40:08,749 --> 00:40:18,440
now the last section of this problem is to
ask you what is the maximum what is the value
277
00:40:18,440 --> 00:40:33,199
of j for which the intensity is maximum the
intensity is maximum thats what is you are
278
00:40:33,199 --> 00:40:41,709
asked to do and 9i think you must remember
this formula that the nu ah the j max this
279
00:40:41,709 --> 00:40:50,359
was discussed in the in the lectures j max
is approximately the nearest integer to kb
280
00:40:50,359 --> 00:41:03,900
t by two hc b minus one half you have already
been given b you know t you know hc therefore
281
00:41:03,900 --> 00:41:08,180
whatever is the integer nearest to this number
this will not be an integer because the square
282
00:41:08,180 --> 00:41:14,640
root factor will not give you ah round integer
value or half integer value for you to substitute
283
00:41:14,640 --> 00:41:30,390
therefore take the nearest integer to this
number as the j for which the population is
284
00:41:30,390 --> 00:41:47,029
for which the intensity is maximum ok thats
that so we will come to the last problem this
285
00:41:47,029 --> 00:41:54,289
is purely a classification of the molecules
as spherical symmetric or asymmetric tops
286
00:41:54,289 --> 00:42:02,059
which is important for ah understanding the
molecular spectroscopy and therefore ah we
287
00:42:02,059 --> 00:42:08,400
need to make sure that the geometry of the
molecules are visualized and that you see
288
00:42:08,400 --> 00:42:14,249
that the moments of inertia about the three
different axis whether they are the same or
289
00:42:14,249 --> 00:42:17,940
whether two of them are the same and different
from the third or if all the three of them
290
00:42:17,940 --> 00:42:22,699
are different you have to have a mental picture
and its very clear from some of these
291
00:42:22,699 --> 00:42:31,710
for example the tetrahedron now cbr four you
should be immediately obvious that this is
292
00:42:31,710 --> 00:42:37,660
the spherical top the reason is tetrahedron
is the half the symmetry of the cube so you
293
00:42:37,660 --> 00:42:45,089
know hat if you place this molecule in the
cube ok and then you start worrying about
294
00:42:45,089 --> 00:42:49,400
what those moments of inertia are you will
immediately see that the moments of inertia
295
00:42:49,400 --> 00:42:54,859
are identical about any three axis because
this is one br this is another br and the
296
00:42:54,859 --> 00:42:59,339
third br is going to be here and the fourth
br is going to be here and the carbon is right
297
00:42:59,339 --> 00:43:06,160
on the middle c therefore you see that this
is a tetrahedral molecule inside the cube
298
00:43:06,160 --> 00:43:11,059
if you want to do that it has half the symmetry
of the cubes therefore you see that with respect
299
00:43:11,059 --> 00:43:16,430
to the structure whether the axis as this
or whether the axis is this or whether the
300
00:43:16,430 --> 00:43:21,749
axis is perpendicular to this phase the moments
of inertia is same because the molecules are
301
00:43:21,749 --> 00:43:28,390
ideally displaced from these axis by the same
distance therefore something should be immediately
302
00:43:28,390 --> 00:43:34,950
seen as based on the geometrical conservations
that it is a spherical top in this case now
303
00:43:34,950 --> 00:43:43,059
ch cl three again you should see that this
is actually a symmetric top the reason is
304
00:43:43,059 --> 00:43:52,509
the three chlorines actually form the base
of an equilateral triangle and then the carbon
305
00:43:52,509 --> 00:43:57,920
is the hydrogen bond is above tetrahedral
molecule if you put the three molecules as
306
00:43:57,920 --> 00:44:03,869
the base of a ah of an equilateral triangle
then it does not matter whether the axis is
307
00:44:03,869 --> 00:44:09,749
this whether the axis is that all the three
atoms are away from hose two axis by he same
308
00:44:09,749 --> 00:44:14,200
distance and so is the hydrogen because its
in the plane therefore two of the moment of
309
00:44:14,200 --> 00:44:19,190
inertia right away equal and the third moment
of inertia which is the moment of inertia
310
00:44:19,190 --> 00:44:26,579
along the ch bond the the symmetry axis of
the molecule is different ok so this is the
311
00:44:26,579 --> 00:44:33,729
symmetric top
and s of six octahedron when i talked about
312
00:44:33,729 --> 00:44:39,250
cube and symmetry of the tetrahedron is half
of this symmetric of the cube octahedron has
313
00:44:39,250 --> 00:44:43,799
the same geometry of as that of the cube because
those are the six mid points of the faces
314
00:44:43,799 --> 00:44:49,119
therefore it does not matter whether you draw
the axis about any two of these atoms the
315
00:44:49,119 --> 00:44:55,329
moments of inertia for all these three axis
will be identical so s of six will be a spherical
316
00:44:55,329 --> 00:45:04,650
top ok ch two cl two we notice that it is
an asymmetric top because it has only the
317
00:45:04,650 --> 00:45:10,160
symmetry of a two fold axis molecules which
have only two fold symmetry as a rule you
318
00:45:10,160 --> 00:45:16,880
can also try and remember will have to be
asymmetric top molecules need a minimum of
319
00:45:16,880 --> 00:45:22,440
a three fold axis in order for them to be
asymmetric top o zone has only a two fold
320
00:45:22,440 --> 00:45:28,309
axis symmetry the bond ax the bisecting the
bond of the oxygen oxygen oxygen and therefore
321
00:45:28,309 --> 00:45:39,910
that you can see right away that o zone is
these two are asymmetric tops hcn no problem
322
00:45:39,910 --> 00:45:45,729
its a linear molecule therefore two of its
moment of inertia are the same about the axis
323
00:45:45,729 --> 00:45:54,079
both axis are perpendicular to the hcn bond
its a linear molecule therefore there is he
324
00:45:54,079 --> 00:46:00,249
carbon hydrogen so i would say that the ah
center mass is here and the other axis is
325
00:46:00,249 --> 00:46:07,670
that but about this axis the moment of inertia
is zero therefore its a linear molecule and
326
00:46:07,670 --> 00:46:15,789
last c two h four also has only two fold axis
symmetry so despite the fact that it looks
327
00:46:15,789 --> 00:46:29,769
to be very symmetric this is an asymmetric
top because the axis here and the axis here
328
00:46:29,769 --> 00:46:35,390
the atoms are at different distances compared
to those and the third axis is of course perpendicular
329
00:46:35,390 --> 00:46:38,799
to the plane of this molecule therefore its
an asymmetric top
330
00:46:38,799 --> 00:46:44,729
so the purpose of this video tutorial is to
tell you a little bit about how to look at
331
00:46:44,729 --> 00:46:49,930
problems from the point of view of not just
solving them but in terms of relevance of
332
00:46:49,930 --> 00:46:56,180
these problems in understanding the concepts
and so on and therefore i suggest that you
333
00:46:56,180 --> 00:47:01,309
solve many such problems by yourself and if
you have questions please feel free to write
334
00:47:01,309 --> 00:47:04,770
that in the forum i wish you all the best
thank you