WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to again continue our discussion of the hydrogen atom.
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Let us jump right on in.
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We have been dealing with this hydrogen atom wave function, this ψ which is a function of R θ and φ,
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which we separated into a radial component R of R and a spherical component S of θ and φ.
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In the last three lessons, we have been dealing with S of θ and φ.
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We are going to go ahead and deal with R.
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We find this R of R.
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The equation that we need to solve, you will see this, know this, or understand it completely but it is nice to see it.
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The equation to solve here to find the radial component which is a function of R single variable is -H ̅² / 2 ×
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the mass of the electron R² DDR R² D of R DR + H ̅² L × L + 1/ 2 mass of the electron R² - E² / 4 π E₀ × R - E R = 0.
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When we solve this equation, we find that the energy for the state N,
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there is a yet another quantum number M E sub N = - M sub E,
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this is a subscript E so the mass of the electron E × E⁴ / 32 π² ε² H ̅² N².
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Or N take some of values 0, 1, 2, 3, and so fourth.
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If we introduce a bore radius which is symbolized A₀ which is equal to 4 π ε H ̅² /
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the mass of the electron E², we get for the energy.
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This is a perfectly, it is not a problem or you can use this one, E sub N = - E²/ 8 π E₀ A₀ N².
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Again, N take some of values 0, 1, 2, and so forth.
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This gives us the energy of that particle for that with respect to the radial function.
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We also found the following.
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Notice, we have a L in that equation, we also discover and we have introduced this new quantum number N.
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There is a relationship between N and L.
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We also find that the relation between N and L is N is greater than or equal to L + 1.
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But we write it more familiarly this way, this definitely remember.
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It is written as L which is going to be greater than or equal to 0, less than or equal to N – 1.
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There you go.
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This is the important relation between L and N -1.
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The solutions for the radial equation.
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We have discussed the energies for the radial equation, now the solutions for the radial equation.
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We go ahead and do this in blue.
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The solutions for the radial component, the radial function they depend on two quantum numbers.
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Let me write this our.
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On two quantum numbers, N and L.
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The symbolism is going to be the following.
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We have R and L which is a function of R is equal to -.
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Is there a minus in there?
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I have to double check that.
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We get N – L – 1!/ 2 N × N + L!³ × 2 / N A₀ ⁺L + 3/ 2.
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Do not worry about it, this is just the general solution, what it looks like.
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We want you to see what it looks like.
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The actual functions themselves are a lot easier to deal with, when you put in the L, the N,
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and all of the number sort of condensing become something.
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Do not get me wrong, I’m not saying this is not complicated, but it is not as bad as it looks.
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R ⁺L E ⁻R/ N A₀,
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I would make a little bit more room here so let me make these a little bit smaller.
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This is going to be × R ⁺L × E ⁻R/ A N₀ × L sub N + L super 2 O + 1 the function of the 2R/ N A₀.
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I have these things right here, do it in red.
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These are called, these L, N + L, super 20 + 1, they are functions of 2R/ N A₀,
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these are called the associated Laguerre polynomials.
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We had the associated Legendre polynomials.
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We have the Legendre polynomials.
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We have the associated Legendre functions
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These are the associated Laguerre polynomial or associated Laguerre functions.
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The first few are as follows.
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The first few associated Laguerre polynomials.
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Again, notice it depends on L and it depends on N, L and N.
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For N = 1, I’m sorry I think in the previous slide it is N starts at 0.
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It does not start at 0, N starts at 1.
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For N = 1, L = 0.
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We get L of 1, 1 of X is equal to 1.
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For N = 2, we have the possibility L = 0 and we have a possibility L = 1, because L runs from 0 all the way to N – 1.
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If N is 2, starts at 0 and move all the way to N -1, 2 -1 is 1.
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For N2 L0, we get L2₂ super 1 of X is equal to – 2! × 2 – X.
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Here we have L, again N + L 2 + 1 this is 3.
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The superscript is 2L + 1, 2 × 1 is 2 + 1 is 3, so L3 3 of X is going to equal -3!.
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Let us do N = 3, for N = 3 we have L equal 0, we have L = 1, and we have L = 2.
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For the N3 L0, we have L3 1 of X is equal to -3! × 3 -3 X + ½ X².
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For N = 3 L = 1, we would end up with L4 3.
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The 4 comes from N + L 3 + 1 is 4.
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The 3 comes from 2L + 1, 2L + 1 is 3.
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This is going to equal -4! × 4 – X.
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Here, N + L the subscript is 5, the superscript is 2L + 1.
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2 × 2 + 1 this is going to be 5.
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This as a function of X is going to equal -5!.
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Let us go back to blue.
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The complete wave function ψ which is a function of R θ and φ for the atomic orbitals of the hydrogen atom R,
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ψ the subscript depends on 3 quantum numbers N, L, and M = R sub NL
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which is a function of R × S sub L super M which is a function of θ and φ.
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N runs from 1, 2, 3, and so on.
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L goes from 0, 1, 2, all the way to N -1 and M takes on the value 0, + or -1, + or – 2, all the way to + or –L.
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N is the quantum number.
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L depends on N, M depends on L, this is the symbolism for it.
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The energy of the particle = - E/ 8 π ε R radius N².
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Notice that the energy only depends on the N quantum number.
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The normalization condition is the integral from 0 to infinity, the integral from 0 to π.
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The integral from 0 to 2 π of ψ conjugate N LM × ψ N LM R² sin θ D φ D θ DR D φ, because this inside when it is φ.
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D θ, the next one is θ, the final one DR it is that.
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Make sure you place in order.
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These functions the ψ N LM, they form an orthonormal set.
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The integral of 1 × the other = 0.
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They form an orthonormal set.
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This is what we have come to, this is what the last 3 lessons a bit about.
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We have been working up to this, this ψ N LM which is a function of R θ φ.
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This is why we have had all of this machinery.
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It is truly extraordinary when you sort of step back from this and you realize that
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the human mind has elucidated all of this and done so successfully.
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It is absolutely staggering.
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I'm still amazed by every single day.
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For the orthonormal set, what else can we say about this?
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I would like to notice something here.
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Let me go back to blue.
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Notice that the R² sin θ D φ D θ DR is the differential volume element.
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In other words, the equivalent of DX DY DZ but you notice it is not just D φ D θ DR, that is DX DY DZ.
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There is this extra factor that we have to multiply by, in order for it to actually work,
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when we take integrals in spherical coordinates.
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We have three variables to triple integral.
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For a single variable function, it is a single integral.
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Two variable function, double integral.
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Three variable function, triple integral.
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We integrate one at the time.
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Now in Cartesian coordinates, we have this DX, DY, and DZ.
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Under a change of variable for spherical coordinates, the relationship is as follows.
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The relationship between the X, Y, and Z, and R θ of φ goes like this.
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X = R sin θ cos φ, Y is equal to R sin θ sin φ, and Z is equal to R × the cos of θ.
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Given this change of variable, the DX DY DZ is actually equal to R² sin θ D φ D θ DR.
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There has to be this extra thing that we have to multiply by when we do a change of variable.
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When we are doing these integrals in polar coordinates, you cannot just write D φ D θ DR,
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you have to write R² θ D φ D θ DR at the end.
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That is the differential volume element.
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I want to actually show you where does that this actually comes from.
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I’m going to show you where this comes from.
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It comes from the following.
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This comes from the following.
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Is it really all that necessary?
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No, it is not necessary.
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If you want, you are more than welcome to get in touch with me and I'm happy to send you a brief little half page,
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one page showing you where this actually comes from.
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But it is actually not that important for our purposes.
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What is important is that, that actually has to be there.
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The moral of the story is when working in polar coordinates,
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in other words when doing integrals in polar coordinates,
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please make sure to use the R² sin θ D φ D θ DR as your differential volume element.
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It is very important, it must be there.
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Let us see, the hydrogen wave functions ψ N LM depend on three quantum numbers.
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The first one N, it is called the principle quantum number.
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It takes all on the values 1, 2, 3, 0, on and on.
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The energies depend only on this principle quantum number.
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E sub N depends only on an M.
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The next one we have is L, this is called the angular momentum quantum number.
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We have come to the point where this is the important stuff right here.
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This is what we want to know.
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This is what we are going to use to actually solve our problems.
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We have the wave function and we want to be able to understand what the relationship is
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among these quantum numbers and how it relates to a particular energy, orbital, and things like that.
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This is called the angular momentum quantum number and values that take on are 0, 1, 2, all the way to M -1.
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L depends on M.
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Now, some things to know about this one.
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The angular momentum L of the electron is completely determined by L quantum number.
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It is completely determined by L, according to the following.
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The magnitude of the angular momentum is equal to H ̅ × √ L × L + 1.
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That is the relation, when you have L you have the magnitude of the angular momentum of the electron.
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The direction of the angular momentum but you had the magnitude of the angular momentum.
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I will start on the next page here.
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Note that the function itself, the radial function NL which is the function of R, the radial component of ψ depends on both N and L.
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We assign letter values to various values of L.
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We have seen this already.
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We have L = 0, that is the S orbital.
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L = 1 that is the P orbital.
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L = 2 that is the D orbital.
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L = 3 that is the F orbital.
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L = 4 that is the G orbital.
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L = 5 that is the H orbital, and just so on.
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Our final quantum number, we have M.
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We have M, that is our final quantum number, that is called the magnetic quantum number.
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M takes on the values 0, + or -1, + or – 2, all the way to + or – L.
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M depends on L.
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M has 2L + 1 values.
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M completely determines the Z component of the angular momentum.
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The way that L completely determines the magnitude of the angular momentum,
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M determines the magnitude of the Z component of the angular momentum.
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Z is equal to M × H ̅ so with some constant, some integral multiple of H ̅.
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M is called the magnetic quantum number because in the presence of a magnetic field,
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the energy E sub N ends up depending not only on N but it ends up depending on the value of M also.
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The energy E sub N depends on both N and M.
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In the absence of the magnetic field, the E sub N has a 2 O + 1 degeneracy.
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If L is equal to 1, 2 × 1 is 2 + 3, all of those 3 orbitals, they have the same energy E sub N.
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That is what that means, that is what degeneracy means.
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If let us say, L = 2, then 2 × 2 is 4, 4 + 1 is 5.
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That means that those 5 D orbitals for N = 3, they all have the same energy.
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They are all degenerate.
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When you put that in a magnetic field, all of those 5 levels, those 5 orbitals, they no longer have the same energy.
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The magnetic field actually ends up splitting up the energies, it separates them.
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When you have taken out the magnetic field, all those orbitals collapse so they become the same energy.
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We will show you what it looks like in just a second.
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In the absence of the magnetic field, the energy has a 2 0 + 1 degeneracy.
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The splitting of the energy of the E sub N into various levels, it is not the E sub N that we are splitting into various levels.
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It is the splitting of the energies into various levels instead of just the one level that was based on N.
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The various level is called the Zeeman effect.
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Let us go ahead and take a look at with this looks like.
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Do I have one more page here?
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Yes, I do.
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I think I will go ahead and do it on the next page.
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For N = 2 and L = 1, M is going to equal 0, -1 + 1.
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When we have no field over here, when we turn on the magnetic field over here,
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when there is no field this is the N = 2, this is the L = 1, N = 0 -1 + 1.
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These 3 orbitals, they all have the same energy level.
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It just depends on N = 2 because there is no field.
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The I places the magnetic field, this field actually ends up splitting the energies.
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Now all of these orbitals, they end up having different energies.
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Here is what it looks like.
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I have got N = +1, N = 0, N = -1.
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Now, the energies no longer just depend on N, they also depend on what N is.
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The absence of the field, you get a 2 0 + 1 degeneracy.
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These are all orbitals, that is an orbital, that is an orbital.
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They all have the same energy to degenerate.
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You put in a magnetic field then all of the sudden they have different energies.
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PX PY PZ that is what is going on here.
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There are 2 × 1 + 1 = 3 levels with the same energy, that is what degeneracy means.
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Same energy that just depends on N.
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Each has a different energy.
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The energy depends on N and it depends on M.
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In this case, we said N = 2, L = 1.
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L = 1 is the P orbital so these 3 are going to end up being our PX PY PZ.
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In the absence of the magnetic field, the P orbitals all have the same energy.
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If they are there, the PX PY PZ based on X =0, 1, and -1.
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Put a magnetic field, the energy split.
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The PX PY PZ has different energies.
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That is all that is happening here.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time.