WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com, welcome back to Physical Chemistry.
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Today, we are going to continue our discussion of the hydrogen atom.
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Let us go ahead and jump right on in.
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Let me go ahead and do black today.
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In the last lesson, we looked at the radial function R sub N L of R.
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In particular, we look at the one that is orbital.
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In particular, we looked at R 1, 0 of R which was 2/ A₀³/2 E ⁻R/ 2 α₀.
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This is only the radial component of the wave function ψ.
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There is also a spherical component or the angular component, the S 0, 0 of θ and φ.
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S of 0, 0 happens to be 1/ radical 4 π.
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Nice and simple, notice θ and φ going to show up.
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This is just a constant function.
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The normalization condition for the spherical component is as follows.
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The normalization condition θ runs from 0 to π, φ runs from 0 to 2 π.
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S00 conjugate × S00 and sin θ D φ D θ.
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Once again, do not forget this because you are integrating over the surface of the sphere.
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You are integrating spherical coordinates so you have to include this factor.
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When you include R, the fact it was R².
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The entire triple integral, this part becomes R² sin θ DR D φ D θ.
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Do not forget this factor.
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Again, integrating with respect to spherical coordinates.
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This is just this, we end up with the normalization condition being 1/ 4 π, the integral from 0 to π,
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the integral from 0 to 2 π sin θ D φ D θ.
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Φ comes first because we are going φ first the integral and we will go to the outer integral.
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There is nothing else going on here, that is just the standard normalization condition for the angular component.
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This is a function of two variables so we end up having a double integral.
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The normalization condition is always the integral of the conjugate of the function × the function itself.
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The complete wave orbital, so the complete 1S orbital wave function is, this is ψ 1, 0, 0.
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It is a function of R θ and φ, it is equal to the R1, 0 of R × the S0, 0 of θ and φ.
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When I put this together with this and simplify, I end up with 1/ π A₀³¹/2 E ⁻R/ α₀.
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We had the radial component S0, 0 happen to be a constant function so the entire wave function,
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now that we have put together it turns out also to be only a function of R.
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In the case of the S orbital, the 1S, 2S, 3S, 4S, it is always going to be this way.
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All of the S orbital are only going to be functions of R only because the S sub 00 term is a constant term.
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What this means is when you have three variables and it is only a function of one of the variables,
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in this case the θ and φ, they sweep out a sphere that means the 1S orbital is spherical symmetric.
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That is why we represent them as globes, bigger 3S, 4S, they are spheres.
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A spherical symmetric, that is what this means.
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This is where it comes from.
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Let me go back to black here.
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The normalization condition for ψ the entire wave function is the integral from 0 to infinity,
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the integral from 0 to π, the integral from 0 to 2 π of ψ 1, 0, 0 conjugate × ψ 1, 0, 0.
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Then, R² sin θ the inner integral is D φ D θ to the final integral is DR.
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This is the normalization condition, that is all we are doing here.
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We just wrote the following.
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We just wrote the normalization condition as the integral from 0 to infinity, the integral from 0 to π,
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the integral from 0 to 2 π ψ 1, 0,0* ψ 1, 0, 0 R² sin θ D φ D θ DR.
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If it makes more sense to you, you can actually separate variables.
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This is a triple integral, just integrating one at a time.
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Working from the inside out, the inside integral is the D φ.
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The outer integral with respect to θ and the next integral is with respect to R.
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If it is more clear, you can separate the variables and write it like this.
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You can write as the integral from 0 to infinity the R² DR, the integral from 0 to π sin θ D θ,
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the integral from 0 to 2 π ψ 1, 0,0 ψ 1,0, 0 D φ.
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In this particular case, you are just separating the variables.
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You are setting up this integrand.
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You are integrating with respect to φ first and then what you get is going to be your integrand including this
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and you can integrate with respect to θ next.
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What you get, you are going to put here integrand also including this and we are going to integrate with respect to R.
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If this makes sense to you, more than this make sense to you, you are just working from right to left.
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Here, you are working from inside out.
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If you will put integrals before, back into calculus or multi variable calculus,
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you were probably accustomed to seeing this one.
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It does not really matter which one you use.
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Since, ψ 1, 0, 0 = 1/ π A₀³ ^½ E ⁻R/ A₀ is a function only of R.
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It is spherically symmetric sphere in space.
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The electron can be found within that sphere but somewhere within that sphere.
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Let us go ahead and work on some probabilities.
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We have the integral from 0 to A, the integral from 0 to π,
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the integral from 0 to 2 π of ψ 1, 0, 0 conjugate ψ 1, 0, 0 R² sin θ D φ D θ DR.
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The ψ conjugate × the ψ 1, 0, 0, I have the function which is right there.
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I’m just going to multiply it by itself and I end up with 1/ π A₀³ E⁻² R/ Α₀.
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Let me go ahead and just call this Z so I do not have to keep writing it over and over again.
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I’m just going to call this one big fat Z.
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We are going to take this and we are just going to solve this when I put this Z into here and
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we are actually going to integrate this one integral at a time, to see what we get when we do that.
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Let us go ahead and we are going to integrate with respect to φ first.
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This integral right here is going to end up being, or if you want this one.
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We would do that one first.
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We are actually doing the integration with respect to φ.
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It is going to be 0 to 2 π of Z, whatever this thing is.
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D φ, that is going to end up being 2 π × Z when we do that integral.
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We are going to integrate with respect to θ next.
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We are going to put what we get, the 2 π Z into here and we are going to solve this integral with respect to θ, the 2 π Z.
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Z does not contain any θ so I could just pull it out.
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0 to π sin θ D θ is going to give me 2 π Z × - cos θ from 0 to π.
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When I end up solving this part, I'm going to get 4 π Z.
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I got 4 π Z and that is going to equal, I have Z right here which is that thing.
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Therefore, I will go ahead and write these again.
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I will stick with red.
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4 π Z is actually equal to 4 π and we said the Z is equal to 1/ π A₀³ E ⁻2R/ α₀.
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The π cancel leaving me with 4/ A₀³ E ⁻2R/ Α₀.
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The final integral gives us the probability of finding the electron within A of the radius.
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The final integral is the integral from 0 to A of 4/ π A₀³ E ⁻2R/ α₀ R² DR.
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This right here, remember this represents the probability that the electron is between R and DR.
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When I integrate that probability from 0 to A, it gives me the probability that the electron is within A of the nucleus.
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This expression, we have already seen it.
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We saw it in the last lesson, it is the same expression for just the radial function.
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Let me go back to black here.
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We saw this already in the last lesson when dealing with only the R1, 0 of R.
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It turned out to be the same probability for the entire wave function ψ 1, 0, 0
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precisely because θ and φ are not involved in the 1S orbital.
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Because θ and φ are not part of the function, they are not part of ψ 1, 0, 0.
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In other words, the S orbital are spherically symmetric.
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Let us talk a little bit about average values.
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Let me go ahead and put the line of demarcation here.
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Let us move here, what about average values?
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Recall that the average value for A is the integral of ψ conjugate × A × ψ × DV.
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The average value R, the average value of where we are going to find the radius
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is going to equal the integral from 0 to infinity, the integral from 0 to π.
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This integral right here is just the radicals symbol for integral.
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Since, we are dealing with function of three variables, we would have a triple integral.
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0 to π, 0 to 2 π of ψ 1, 0, 0 conjugate × R × ψ 1, 0, 0 × R² sin θ D φ D θ DR.
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When we actually end up taking the ψ × R × ψ conjugate is just multiplication.
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It is ψ conjugate × ψ which we already know we will just multiply it by R.
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We end up with the integral 0 to infinity when we take care of all of this, we get 4/ π A₀³ E ⁻2R/ Α₀ R,
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which is this R, R² DR, which is going to end up equaling 4 / π A₀³.
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The integral from 0 to infinity of E ⁻2R/ Α₀ R³ DR.
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When we actually solve this integral, we get 3/2 A₀.
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For this particular orbital, on average if I take a bunch of measurements of where the electron is going to be,
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I'm going to find it is actually ½ bohr radius away from the nucleus.
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Remember, we had a distribution that looks like this for the 1S orbital, something like that.
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It is going to be between here and here, on average.
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We are going to find it in a bunch of different places.
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There is a place of highest probability, that on average for 100 measurement, 1000 measurement,
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a million measurements, it is going to end up being ½ bohr radius is away from the nucleus.
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That is all these means.
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It is just an integral, it is just the function, the conjugate of the function × whatever it is that you are looking for.
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In this case, we are looking for R radius how far away × the function and then just solve the integral
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or let use mass software solve the integral.
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Let us go ahead and look at an example here.
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For the 1S orbital, calculate the region of the highest probability for finding the electron.
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Highest probability means we want to maximize the probability.
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We are looking for the peak, we are looking for the X value where the peak occurs.
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The probability, the 1S orbital was founded already, it is 1/ π A₀³.
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I will go ahead and put the R² E ⁻2R/ α₀ DR.
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Remember, this part is the probability density.
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The probability is the probability density × the differential element.
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What we need to do is we need to maximize the probability density.
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Remember the graph that we were looking at, the 1S orbital.
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Probability density was on the Y axis, R was on this axis.
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The probability density was on this axis.
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We got something like that, we want to find that R value.
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This is a probability density vs. R, that was the graph was.
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In case we need to maximize that function.
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We maximize something by doing exactly what you did in calculus.
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We take the derivative, we set it equal to 0.
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DDR of this function 4/ π A₀³ R² E ⁻2R/ α₀.
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When you take the derivative of this function, you get 4/ π A₀³.
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This is the product rule so it is going to be this × the derivative of that + that × the derivative of this.
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We are going to get R² × E ⁻2R/ α₀ × -2/ α₀ + 2R × E ⁻2R/ Α₀.
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This is going to equal 4/ π × A 0³ × E ⁻2R/ α₀ × - 2R²/ α₀ + 2R.
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That is the derivative, we are going to set the derivative equal to 0.
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This constant goes away, this E ⁻2R/ α₀ is never 0.
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Therefore, this is the equation that we have to solve.
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We get -2R²/ Α₀ + 2R.
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We set it equal to 0, we end up with -2R × R/ Α₀ -1 is equal to 0.
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We end up with R equal to 0 or we end up with R equal to α₀.
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0 of course does not matter, that is out of the nucleus.
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For the 1S orbital, calculate the region of highest probability for finding the electron.
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The place we are most likely going to find electronic is 1 bohr radius away from the nucleus.
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That all that is going on here.
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We maximize the probability density because we were looking for the maximum region of highest probability.
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Let us look at the 2S orbital.
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We are looking at the 1S orbital, and the one that is complicated is the 2S orbital.
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In other words, we are going to be looking for ψ 2, 0, 0.
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Ψ 2, 0, 0 is equal to R 2, 0 × S 0, 0.
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This S 0, 0 term shows up again and it is going to show up to all of the S orbital because the 3S orbital is going to be ψ 3, 0, 0.
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4S orbital is going to be ψ 4, 0, 0, the S, 00 that term always shows up in the S orbital.
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The S 0,0 term is part of the wave function.
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In fact, for every S orbital the spherical harmonic will be S00 precisely because L = 0 and M = 0.
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That is the S orbital.
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The only thing that is going to change is the primary quantum number, 1, 2, 3, 4, 5, that is all that changes.
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For every S orbital, the only variable for the wave function ψ is going to be R.
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The total wave function, total hydrogen wave function for all S orbitals are only going to the functions of R.
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They are only going to depend on the radius.
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They are spherically symmetric.
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Every S orbital is spherically symmetric.
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For ψ 2, 0, 0 we end up with v1/ 32 π × 1/α₀³/2 × 2 - R/α₀ × E ⁻R/ 2 Α₀.
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The average value of R for 2, 0, 0 is equal to exactly what you think.
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It is the integral of the ψ conjugate of 2, 0, 0 × R × the ψ 2, 0, 0.
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We solve that integral.
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Ψ conjugate × ψ itself is just this thing multiplied by itself because this is a real function.
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The conjugate is just ψ itself.
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This becomes ψ² × R.
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This thing × R.
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We have ψ 2, 0, 0 conjugate × ψ 2, 0, 0 is equal to 1/ 32 π × 2 - R/ A₀² × R × E ⁻R/ α₀.
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The 2’s cancel.
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We are solving a triple integral.
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We still have to solve a triple integral.
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Even though it is a function of R, we still have to do the entire integral.
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We already know that the integral from 0 to π, the integral from 0 to 2 π of the sin θ D φ D θ is equal to 4 π.
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The inner integral, the angular part is going to equal 4 π.
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We will just throw that in.
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What we get is the average value of R for 2, 0, 0 is going to equal 4 π × this thing / 32 π A₀³.
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Our final integral is the integration with respect to R.
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I have done the triple integral, I just know that in this particular case because θ and φ are not involved in this function,
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I will just go ahead and use this value in place of the inside two integrals.
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It is going to be that 2 – R/ α₀² E ⁻R/ α₀.
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Again, we have our R² DR, do not forget that term.
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When I actually put this in a mathematical software, I'm going to get 6 Α₀.
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For the 2S orbital, on average you will find the electron 6 bohr radius away from the nucleus.
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Remember, for the 1S orbital it was ½ × the bohr radius so it was going to be ½.
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For the 2S orbital, it is going to be 6 bohr radius away.
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On average, it is going to be farther from the nucleus.
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3S on average is going to be farther from the nucleus which is why
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the pictures in your book represent the 1S as a small sphere, the 2S is a bigger sphere, the 3S is a bigger sphere.
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On average, the electron is going to be further away.
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That is why it looks the way that it does.
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There is a general formula.
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The average value of R for whatever S orbital 1S, 2S, 3S, 4S, 5S, is equal to 3/2 A₀ N².
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Remember, the picture that we saw in the last lesson with the S orbitals, the 2S orbital has 1 node.
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That means that if I take a sphere and I split it in half, the 1S orbital did not have a node,
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which means that the electron can be anywhere in that sphere.
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The 2S orbital had 1 node that means the electrons you are going to be in this region.
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It is not going to be here or in this region.
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I sliced it in half, you are looking at half a sphere that way.
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The 3S orbital has 2 nodes.
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There is going to be a place where the electron is not going to be.
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There is going to be another place where the electron is not going to be.
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The electron is either going to be this region or in this region or in this region.
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That is what this means.
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Do not forget that there are nodes.
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When we talk about spherical symmetry, it does not mean that they can be anywhere in the sphere.
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That only possible for the 1S orbital because it has no nodes.
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But within the sphere itself, the 2S orbital has 1 node.
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That means there is 1 sphere within that bigger sphere where you will never find the electron.
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The 3S orbital has 2 nodes but if this is the main sphere, there are going to be 2 smaller spheres where you will not find the electron.
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Again, pictorial representations of these electrons orbital are not very good because they are not giving you all the information.
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Thank you for joining us here at www.educator.com.
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We will see you next time, bye.