00:01
In this problem, we are looking at the wave function for the 2s orbital of the hydrogen atom.
00:07
And we are looking at whether it satisfies the stronger equation.
00:14
That wording is a little ambiguous.
00:16
It could mean just showing that the wave function is mathematically well -behaved enough that it could be a solution to the equation.
00:23
Or it could mean confirming that this specific wave function actually is the correct wave function for the 2s state.
00:35
In that, it gives the correct energy for n equals 2, which we amount to showing that it produces the same energy value as given by the bore model, which is what we had to do in problem 16 for the 1s case.
00:55
So this is kinetic energy.
01:04
This is potential energy.
01:07
The e squared comes from the charges of the electron and the proton nucleus.
01:16
This is the total energy.
01:25
So by inspection, this doesn't change under differentiation.
01:37
So if we want to find the total energy later, this term is going to divide out both the right -hand side and the left -hand side.
01:45
This is a constant.
01:47
That's going to divide out and cancel out on the left -hand right -hand side.
01:52
The derivatives in the exponential are going to be pulling out a minus 1 over 2 a 0.
01:58
So those factors will remain.
02:00
But this middle term 2 minus r divided by a knot is a little bit more complicated.
02:07
So we can't do this entirely by inspection.
02:14
So to show that this is mathematically well behaved, we need to satisfy some conditions continuity the way function can't have any abrupt kinks or discontinuities in it otherwise it's not physical it has to be at least twice differentiable so the first and second derivatives of si have to exist but also size squared we need size squared to be single valued so there can't be size square gives you the probability of finding particles so there can only be one probability at any given differential width in the radius so basically size squared has to pass the vertical line test and it probably should the side terms and it should also pass the vertical line test.
03:54
And it also has to satisfy the born rule.
04:06
So the integral of zero and infinity of the size square, it has to be finite.
04:13
Because remember the solutions to the stronger equation are representing probabilities and all of them together have to add up to one for 100 % probability.
04:25
This is a particular state.
04:28
Only one possible say it could be in, so it's going to have to be less than one.
04:39
So this aspect of the problem is just a lot of math.
04:45
So first derivative, you just do these radial derivative of the side 2s term we have up there.
05:08
I'm going to switch back to using blue.
05:13
So if you actually do that derivative, it's going to work out to be 8 square root 2.
05:23
For a nought seven halves r minus four a knot e minus r two a knot so i said this exponential term is not going to change under differentiation but does pull constant factors out so one a second derivative so then that becomes the ddr of what of this term.
06:11
You have to take the r -druth of this.
06:14
So when you do that, you get over 16.
06:46
So these functions exist.
06:52
So r goes to zero.
06:58
These side terms are finite.
07:08
They are finite.
07:18
For it goes infinity, the exponential term dominates over this r term.
07:26
And i'm running side here, but i mean these derivative terms, though it's also true for a side 2s itself.
07:36
So that looks good.
07:42
And then just to be thorough, we're going to want to do that to a squared.
07:52
So that's just taking this equation up here and squaring it.
08:08
Of force of habit, i write these as partial derivatives, but you can use ordinary derivatives because there's only one dimension that we're dealing with in this problem.
08:28
So what i'm writing here in the parentheses is side 2s squared.
09:03
So that becomes, it pulls out minus 1 over a factors.
09:13
This whole thing is negative.
09:38
So then the second derivative of si2s squared is the radial derivative of what we just calculated there.
10:20
That works out to positive 1 over 32 pi 8 to the 7th.
10:36
So this is getting kind of ugly looking and complicated.
10:47
But the important thing is that these are, this exponential term dominates, and these are basically just decaying exponential terms.
10:58
So the graphs are very simple, and they pass vertical line tests.
11:02
And again, and r goes to zero.
11:08
These are finite.
11:21
Argos infinity.
11:23
Si goes to z...
11:26
By si again, i also mean the derivatives in here, not just si squared.
11:37
So those parts look good.
11:55
So let's see.
12:14
So this looks...
12:36
That's what psi 2s looks like.
12:41
The derivative of side 2s looks something like that and the second derivative side 2s square looks something like that so again a vertical line test there's no problem these are single value so that is well behaved sorry i made a mistake here that's i meant the second derivative of side 2s so now we'll do the same thing for side 2 s squared.
14:41
That's what side 2s squared looks like.
14:51
This is what the dr of side 2s squared looks like.
14:58
And the second derivative looks something like that.
15:30
So you can do vertical line tests, it passes that.
15:33
So this is, so it's well behaved enough to be stronger equation.
15:43
Though we need to check one more thing, we need to see that obeys the born rule.
15:56
So that is zero to infinity of this term.
16:42
So that becomes, i'm trying to write the not terms on all the a's, but i might be dropping them here and there.
17:22
So that the limits of integration, infinity is zero.
17:29
So let's see, when it goes to infinity, this exponential term dominates and the whole thing goes to zero.
17:37
And that's minus, let's see...
