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welcome back to the lecture we will
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continue with the analysis of the
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solutions that we have proposed for the
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hydrogen atom the equation being the
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Schrodinger equation H size equal to e
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sy sy or theta fee and we have proposed
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this to be a radial part and the angular
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part containing theta and fee in these
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two together written as a spherical
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harmonics of two dependents ok so this
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is the formal structure that we have for
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the solutions and the wave functions
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when we solve these differential
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equations the wave functions will depend
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on three chord in three quantum numbers
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n L M these are the standard
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representations for the quantum numbers
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the values for these quantum numbers are
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n goes from 1 to 32 in finite and the
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value of L is limited by the choice of
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any n it goes from 0 1 2 up to n minus 1
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and the value of M is also chosen by the
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values of L namely 0 plus minus 1 plus
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minus 2 up to plus minus e l therefore
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the wave functions are given by DS three
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quantum numbers and if we write this
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sigh NL em with n l and IAM or theta V
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as the radial function yin le dependent
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on both the quantum numbers and the
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spherical harmonics ylm theta fee the
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first value is sy 1 n equal to 1 and the
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only choice that we have for y ln m
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or 0 and 0 100 this is known in the
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standard representation as the 1s
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orbital the next quantum number that we
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have is n is 2 and therefore we have the
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wave function with the n quantum number
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two and yell can be 1 or 0 and etl is
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one the NEM can take the three possible
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values namely 10 and minus 1 1 2 2 and
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therefore the three wave functions will
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have this representation to 11 sigh to
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10 n sy to 1 minus 1 is 3 ok the overall
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energy is a solution in the radial part
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of the equation therefore this is the e1
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the overall energy will depend only on
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yen all of these will have the same
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energy e2 and when yen is to yell can be
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1 or 0 and therefore M will be 0 other
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wave functions I 200 this is to yes
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orbital and the L equal to 1 they are
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all known as P orbitals and this is the
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2p orbital and likewise for n equal to 3
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you will have L equal to 0 1 or 2 and L
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equal to 0 will give you Emmy quilt is e
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ro this will give you three values 0
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plus minus 1 this will have five values
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0 plus minus 1 plus males too therefore
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for any n there will be n square wave
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functions
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all of which or degenerate
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they all have the same energy according
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to the formula that en given by the
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standard formula minus HC the riddler
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constant by n square where HS appliance
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constancy is the speed of light okay
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this is something that you are familiar
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from the Bohr's model and also from the
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Schrodinger equation gives exactly this
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as the solution except to that it has n
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square degeneracy for every m and the
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wave functions are given according to
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this particular format now what we will
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do is we will see these wave functions
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in two parts the angular part first
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which brings to you the results in some
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familiar form to what you already know
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namely the the orbital forms that you
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have seen that the shapes of the various
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atomic orbitals are given by the angular
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parts okay no remember theta and fee or
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polar coordinates and in spheres eight
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on fee have the limits of theta is equal
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to 0 and pi and fee has the limits of 0
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and 2pi so with these variations we can
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create a spherical surface therefore
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these are the the what is called the
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maximum value asleep aksam um this is
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the range of the theta and fee the
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radius of course goes from yes fear of
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zero radius to infinite radius therefore
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radius goes from 0 to infinity so this
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collection of the coordinate system that
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we have reproduces the boundary
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conditions that we have namely zero
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radius to infinite radius and put each
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radius a spherical surface enclosing a
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spherical volume there for the entire
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three-dimensional volume is reproduced
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this is seen by a very simple animation
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that one came view here so let me show
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you that these are the very
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for theta and fee the polar coordinate
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has zero to PI so it Wayne just that way
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and then the fight coordinate taking the
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semicircle throughout it generates the
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whole circle surface therefore please
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remember the angles are limited by this
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unsymmetrical or asymmetrical choice one
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is from 0 to PI the other is from 0 to 2
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pi if you put both of them 0 to 2pi you
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will generate the spherical surface
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twice you will generate the infinite
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volume twice therefore you do it you get
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the value of two times that therefore
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it's not correct the silikal coordinate
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coordinates have this be as the limits
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now let's look at the series of
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functions that we wanted to see
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pictorially okay so let me write some of
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these spherical harmonics as solutions
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when L is 0 and m is 0 the spherical
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harmonics y is 0 0 is not dependent on
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any angle and it has a value 1 by root
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for pie no angular dependence
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when L is zero when L is one when L is 1
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and m is zero the spherical harmonic is
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why LM is 1 0 theta fee and it has the
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value that root 3 by 4 pi cos theta
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independent of Phi when L is 1 and m is
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equal to plus or minus 1 the spherical
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harmonics is y 1 plus or minus 1 theta P
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and it has the form minus R plus root 3
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by 8 5 pi sine theta e to the plus or
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minus V this is PI this is V comply
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complex functions and in general for any
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M the Phi dependence is given by this
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function e to the i M Phi and you can
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see that here is plus or minus 15 it is
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plus amba is 0 5 which is of course one
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therefore these are what are called the
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spherical harmonics for the p orbital
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and what are the values for the l equal
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to tool which are known as the D
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orbitals
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these are all three of these are P
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orbitals and these are the 5 D orbitals
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you have L equal to 2 if M is 0 you have
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the spherical harmonics way to 0 theta
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fee and the value is given by root 5 by
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16 pi times 3 cos square theta minus 1
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and when M is plus or minus 1 y 2 plus
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or minus 1 theta fee has a minus plus n
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square root of 15 by a pi sine theta cos
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theta e to the plus or minus I Phi P and
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when M is plus minus 2 the spherical
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harmonic is y2 plus minus 2 theta fee
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and that's given by root 15 x 32 pie
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sine square theta e to the plus minus
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two I fee so you can see that the P
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orbitals are all functions of cos theta
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R sine theta raise to power one that is
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the it's a monomial if l is equal to two
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you can see that it's 3 cos square theta
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but one is nothing but sine square theta
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plus cos square theta therefore it is 2
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cos square theta minus sine squared
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theta so it is a function of cos theta
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sin theta but degree two polynomial of
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degree 2 and likewise for sine theta cos
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theta sine square theta so all the yells
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the spherical harmonics for each and
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every yellow you will have the theta
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dependent part as a Yelp degree
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polynomial homogeneous okay it will
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involve sine theta and cos theta but the
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total power of sine and cos theta will
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be n the five part is why LM the five
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part is a to the im5 that's it for the
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structure of the spherical harmonics and
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the patterns are clear how do we get
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these constants in front of it and how
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do we get the plus minus signs etc
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that's if I mean more mathematics
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business through the normalization of
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the spherical harmonics to unity over
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these this fear and therefore these
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constants would be shown in the next
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part as the actual numbers that come out
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when you normalize the spherical
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harmonics like the way you normalize the
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wave functions by taking psystar psy.d
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tell the integral as one here you would
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take the spherical harmonics ylm star
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ylm and taking through the spherical
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volume elements namely theta is equal to
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zero to PI and Phi is equal to 0 to 2 pi
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and the spherical differential element
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sine theta D theta D Phi when you do
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that you will get all these constants
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clearly now let us see the pic
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oriol representations for the real part
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of y1 plus or minus one and the
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imaginary part of y1 plus or minus one
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and the function y 10 which is real
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anyway it is a function of cos theta and
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here this will contain sine theta the
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real part will contain sine theta cos
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fee and the imaginary part will contain
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sine theta sine fee so we shall see this
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in the spherical coordinate pictures
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representations let's look at the Y 11
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so first let me see why 10 here i am
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plotting y 10 on one value of Phi but
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this function you know is cos theta y 10
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therefore it's the same for all values
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of Phi so if you know the shape of this
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function for theta then we can reproduce
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that shape for all the values of Phi and
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what is done here is cos theta is
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plotted on the theta coordinate remember
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the theta coordinate for the polar axis
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system starts with some Z direction
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where theta is zero and then theta is
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some value from value then it's 90 and
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then it's 180 so the value of cos theta
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is plotted on that value of theta radius
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and then you connect them okay so this
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is theta is equal to 0 cos theta is 1 15
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cos theta is fine line you mark it on
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the radius that the entire length okay
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and then you connect all these points to
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get a representation for cos theta on a
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polar system sorry kill polar system
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this is Pollard once you do it for all
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values of Y you will get it for silicon
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polar I have given a different color
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because cos theta is negative for theta
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greater than 90 degrees but the values
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are symmetrical on either side of the
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x-axis ok so that's the shape of cos
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theta in a polar coordinate system and
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now in a complete sphere how does this
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look like it's the same graph for all
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values of Phi therefore if you plot this
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ok so here you will see why 10 plotted
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for all values of the azimuthal angle
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Phi and this is what you have seen for a
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given value of theta from 0 to PI
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therefore if you plot it for all values
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of Phi you will get the same graph with
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of course the plus and minus signs
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marked on the either side of the x axis
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because you know cos theta is negative
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for tita greater than 90 and that's
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below the x-axis and for above the
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x-axis cos theta is positive therefore
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this is the standard representation of
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the p orbital that you see plus minus
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lobe which is nothing but the PZ orbital
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sine theta cos p + sine theta sine p
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which are along the other two directions
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so let's first of all see sine theta
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plot okay cause fee I have kept it as 1
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by choosing p is equal to 0 so this plot
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is along the x-axis and then this plot
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needs to be rotated as you go for
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various values of Phi while rotating it
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00:17:34,740 --> 00:17:38,460
for various values of Phi you must also
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multiply the sine theta plot by kospi
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therefore it will go to 0 it will go to
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negative and become 0 and you come to
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positive and so on so you can see first
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of all the sine theta on the polar graph
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here which does not have any negative
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values because sine theta is positive
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00:17:54,870 --> 00:18:01,370
in the range 0 to 180 it starts from
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zero
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00:18:11,600 --> 00:18:17,520
again I remind you the value of sine
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theta is plotted along that theta
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direction by marking the point and the
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00:18:22,860 --> 00:18:33,630
table gives you what the value is for a
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00:18:25,380 --> 00:18:36,000
few of the Thetis so that's sine theta
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00:18:33,630 --> 00:18:41,130
in one direction with costs fees equal
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00:18:36,000 --> 00:18:44,760
to one that is along the x-axis if we
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00:18:41,130 --> 00:18:47,580
plot this for all values of fee you will
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00:18:44,760 --> 00:18:49,950
actually see that this graph is x cos
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fee therefore it shrinks to zero when it
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00:18:49,950 --> 00:18:55,770
comes to ye because along the y-axis
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fees zero 90 degrees therefore cost 90
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is zero along the minus x axis fee is
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180 therefore costs fee is minus 1
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00:19:01,470 --> 00:19:09,840
therefore from why all the way to minus
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y when the fee values are between 90 and
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00:19:09,840 --> 00:19:16,230
270 cos B is negative therefore whatever
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00:19:13,920 --> 00:19:18,450
you see here will have the negative sign
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00:19:16,230 --> 00:19:21,440
and whatever you see on this side will
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have the positive sign
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so sine theta cos V Y 11 y 10 these are
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the two plots the third function which
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00:19:45,470 --> 00:19:52,669
is the imaginary part of the UI 11 is
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sine theta sine V the difference between
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sign fee and kospi is 90 degrees
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therefore all you would see when you
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plot that function is that this is
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rotated by 90 degrees so we start with
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the y-axis because sine theta sine T if
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00:20:08,000 --> 00:20:12,019
you want to plot you plot it on a sine
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00:20:10,129 --> 00:20:15,909
Phi maximum which is along the y-axis
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00:20:12,019 --> 00:20:18,320
and then you will see that sign fee is
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positive between fee is equal to 0 and
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180 0 and 180 as you see it is between
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the X and minus x axis and 180 to 360
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when sine fee is negative it's along the
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other side namely minus X to the plus x
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axis along the minus y and therefore you
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see the natural function that plot that
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00:20:40,610 --> 00:20:48,769
you see here and sine theta goes to sign
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00:20:45,830 --> 00:20:52,460
fee goes to zero or fees equal to 180
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and p is equal to 360 now for this is
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plotted along the x-axis I mean they
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pros the point the trigonometric
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00:20:56,419 --> 00:21:00,350
functions look different but the
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00:20:58,820 --> 00:21:04,490
functions have the same representation
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graphical representations on a sphere
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00:21:04,490 --> 00:21:09,200
because the sphere doesn't care for what
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00:21:07,070 --> 00:21:11,240
is x-axis or y-axis our third axis all
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00:21:09,200 --> 00:21:13,009
three are the same therefore the
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00:21:11,240 --> 00:21:15,350
orbitals are symmetric about the three
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00:21:13,009 --> 00:21:17,570
mutually perpendicular directions and
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00:21:15,350 --> 00:21:20,299
it's your convention to choose a
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00:21:17,570 --> 00:21:23,360
right-handed coordinate system and a
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00:21:20,299 --> 00:21:26,269
standing up axis because most of us see
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00:21:23,360 --> 00:21:29,090
standing up I mean it can be lying down
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00:21:26,269 --> 00:21:33,409
or it can be standing down the z-axis is
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00:21:29,090 --> 00:21:36,769
as arbitrary as the the spheres
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00:21:33,409 --> 00:21:38,860
Direction is therefore if you go to the
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00:21:36,769 --> 00:21:41,500
nuclear magnetic resonance research lab
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00:21:38,860 --> 00:21:44,330
you will see the z axis is horizontal
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00:21:41,500 --> 00:21:46,039
because the magnets are like this and
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therefore this is the said and you are
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00:21:46,039 --> 00:21:50,000
XY plane is this way
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it's it's your choice on a spherical
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00:21:50,000 --> 00:21:55,730
coordinate system when you block the
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00:21:52,160 --> 00:21:58,400
silikal harmonics ylm you get exactly
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00:21:55,730 --> 00:22:00,800
the distributions and the shapes that
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00:21:58,400 --> 00:22:02,750
you get you have seen in your textbooks
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00:22:00,800 --> 00:22:04,490
and that's a beauty of it and these
349
00:22:02,750 --> 00:22:06,320
functions are exact solutions for the
350
00:22:04,490 --> 00:22:08,030
hydrogen atom in the next part of this
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00:22:06,320 --> 00:22:11,030
lecture we will continue with the D
352
00:22:08,030 --> 00:22:13,970
orbitals and I will also show one F
353
00:22:11,030 --> 00:22:16,550
orbital so the next lecture is purely an
354
00:22:13,970 --> 00:22:19,190
extension of this lecture you don't need
355
00:22:16,550 --> 00:22:21,740
to see that part if you wish go to the
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00:22:19,190 --> 00:22:24,710
website and see all the 15 clots that we
357
00:22:21,740 --> 00:22:29,090
have that I have put up the 15 plots are
358
00:22:24,710 --> 00:22:32,090
for the 3p orbitals the 5 D orbitals and
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00:22:29,090 --> 00:22:34,490
the 7 F orbitals chemistry and chemical
360
00:22:32,090 --> 00:22:36,980
systems do not require see orbitals
361
00:22:34,490 --> 00:22:39,200
right now because the atomic number that
362
00:22:36,980 --> 00:22:42,680
we know maximum atomic number that we
363
00:22:39,200 --> 00:22:44,630
know 120 still does not warrant a stable
364
00:22:42,680 --> 00:22:46,460
atom with a G orbital so we don't worry
365
00:22:44,630 --> 00:22:48,170
about it but spherical harmonics is
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00:22:46,460 --> 00:22:50,390
fundamentally important in all of
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00:22:48,170 --> 00:22:51,890
physics and all of engineering what you
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00:22:50,390 --> 00:22:54,770
see here is nothing but the
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00:22:51,890 --> 00:22:57,290
representation of a spherical harmonic
370
00:22:54,770 --> 00:22:59,450
the real and imaginary part of it on a
371
00:22:57,290 --> 00:23:01,190
spherical coordinate system therefore
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00:22:59,450 --> 00:23:03,410
these pictures may be useful to anybody
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00:23:01,190 --> 00:23:05,240
who wants to look at them ok not just
374
00:23:03,410 --> 00:23:07,010
the chemistry part of it will continue
375
00:23:05,240 --> 00:23:10,870
with the D orbitals in the next lecture
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00:23:07,010 --> 00:23:10,870
until then thank you very much
377
00:23:26,600 --> 00:23:28,660
you