 Okay, so now that we know how to write down the wave functions for the hydrogen atom, I've written down a few of them here. Let's take a closer look and see if we can understand what those wave functions look like graphically or spatially. So what I've written here is the first few wave functions with L equals 0 and M equals 0. So with quantum number N equals 1, N equals 2, N equals 3. The reason I've chosen these three to write down, of course there's many more that we could write down. And I've chosen these three to write down is these are the ones that don't have any theta or phi dependence because the L equals 0 and M equals 0. There's no terms that involve sine or cosine theta or e to the i phi. So there's only our dependence, no theta phi dependence. What that means is these wave functions are spherically symmetric. The size of the wave function depends on how far the electron is away from the nucleus. But it doesn't depend on what direction the electron is relative to the nucleus. So no matter what direction you look outwards from the nucleus, you have the same probability of finding the electron in any direction. That gives these wave functions spherical symmetry. And for that reason we call them S wave functions. So the 1, 0, 0 wave function, we sometimes call the N equals 1 spherical wave function, the 1S wave function. Likewise the 2, 0, 0 wave function is a 2S wave function and 3, 0, 0 corresponds to the psi 3S wave function. Those are the same 1S, 2S, 3S orbitals or 1S, 2S, 3S wave functions that you're familiar with from thinking about hydrogen atoms at earlier levels of chemistry courses. So to understand the shapes of these 1S, 2S, 3S orbitals, we need to look in more detail with these wave functions and see what these wave functions look like if we graph them. So there's only one variable, the theta and phi are not present. So we can just graph a function of an exponential. We know what exponential functions look like if I graph the 1S wave function as a function of distance. That's just an exponential function. It's going to decay exponentially with distance from the origin or distance from the nucleus. So that seems to tell us that the farther away from the nucleus we get the less likely we are to find the wave function because remember the, let's write this over here, the probability of finding a wave function, an electron at some position is proportional to the wave function squared. But to really ask the probability, if I want to draw a picture, here's a three-dimensional graph x, y, and z with a nucleus at the middle and I want to ask where am I likely to find that electron relative to the nucleus. I'm asking what's the probability of finding it in some little region of space, some little dx, dy, dz component of three-dimensional space. But since we've written the wave functions in terms of r's and theta's and phi's, I don't really want to write dx, dy, dz. Instead I'll write that as dx, dy, dz when I convert it to polar coordinates that's r squared sine theta dr, d theta, d phi. So that's an important difference. The probability of finding an electron in some three-dimensional volume of space, some small little three-dimensional volume of space is not just the wave function squared, it's the wave function squared times r squared times sine theta dr, d theta, d phi. And if I really only care about how likely I am to find the wave function in some r-coordinate close to the nucleus, far from the nucleus, all I'm really asking about is the probability is that some particular r without worrying about theta and phi. So what I need to do to get that, that's the... I can go ahead and integrate over theta and phi, so the integral looks like r squared sine theta dr, d theta, d phi. If I go ahead and do these integrals, theta from 0 to pi, phi from 0 to 2 pi, so I'm going to perform the theta and phi integrals, integrate away theta and phi, not perform the r integral. So we can do that integral for these s orbitals because there's no theta or phi dependence in the wave function. The wave function squared doesn't have any thetas and phi's in it, r squared doesn't have any thetas or phi's in it, the integral of sine theta, d theta, d phi is just 4 pi. So I've done the theta and phi integrals, integral of sine theta, d theta, d phi is 4 pi, so I've still got an r squared and I've still got a wave function squared. So the graph I'm really interested in is not what the wave function looks like as a function of distance, but what the probability of finding the 1s electron looks like as a function of distance. And that's not just an exponential, that is, in this case, it's some constants times r squared times the wave function squared. So this graph is going to look like r squared times my wave function squared, e to the minus constants, I just, when I square it, I get e to the minus twice as much r in the numerator over a naught. So the graph without worrying about the constants too much, the graph is not just an exponential anymore, now it's r squared times an exponential. So it rises at the beginning as r squared's getting bigger, it's 0 when r equals 0 because the r squared is 0 there, but then it decays at long distances because of the exponential. So this is more like what we think of the wave function as being, I don't have the highest probability of being found at the origin, the probability of being at the origin is actually very small because there's very few points that are directly at the origin. At some distance away from the origin there's more points with a particular radius away from the origin, so if I move a little bit away from the origin I'm more likely to find the electron, if I move even further away from the electron I've got more probability of finding the electron there, but then as I move farther and farther away from the electron I become less and less likely. So it's a spherical cloud of electron density surrounding the nucleus that decays the further away we get from the nucleus and that's what the picture you probably already have in your head of a 1s orbital matches this picture and mathematically can be described by this particular function. Notice that there's a maximum in this curve, if I ask you at what distance you're most likely to find the electron, it's not r equals 0, it's not r equals infinity, it's somewhere in between, so there's some particular distance where we're most likely to find the electron around a 1s hydrogen orbital. So to ask ourselves where that position is, let's say specifically for an H atom, so I'm going to let z equals 1, I'm going to look for the place where this curve goes through a maximum, so where the slope of that curve is equal to 0, so since I know how to write down the probability, r squared e to the minus some constants times r, I need to find the place where this derivative, the derivative of r squared e to the minus 2r over a naught is equal to 0. That will tell me what this special distance is where I'm most likely to find the electron. So let's go ahead and take that derivative, notice that I've let z equals 1 so I'm specifically talking about hydrogen here. If I take this derivative using product rule, derivative of r squared is 2r and that's still got a e to the minus 2r over a naught. Or I can leave the r squared alone, take the derivative and I've forgotten the minus sign here. If I take the derivative of e to the minus 2r over a naught, I get minus 2 over a naught times the exponential. So that's the derivative, that's the thing I want to equal 0. e to the minus 2r over a naught, e to the minus 2r over a naught I can factor that out. Divide that out because it's 0 on the other side, I can get rid of 1 multiple of r from each of these terms, get rid of an r and knock that one down to r from r squared. So what I'm left with is 2 and this term looks like a minus 2 over a naught r needs to equal 0 so 2 needs to equal 2 over a naught r, the 2's I could have cancelled at any time up until this point and then rearranging that one more time I find out that r is equal to a naught or a naught remember is the Bohr radius. So that's our answer to this question, where is the probability largest of finding the electron at which r is, are we most likely to find the electron? The answer is at the Bohr radius. So the peak in this function is at a naught. If I didn't use z equals 1, if I use z equals 2 I'd get a different answer, I'd have some 2's entering the equation that weren't here before so we can work that problem for helium plus ions or lithium 2 plus ions. But the point is there's a maximum in this wave function that we can determine where that maximum is and that maximum either is the Bohr radius or is closely related to the Bohr radius. That covers the 1s orbital. Let's spend just a minute talking about these higher level spherical orbitals, the 2s and the 3s orbitals. The only difference between these wave functions and the 1s wave function is there's this polynomial in front of the wave function. So if we were to back up now and imagine repeating what we've just done for the 1s orbital, now for the 2s, the wave function, if I were to plot the wave function, this polynomial in front, what that does to the graph of the wave function, if I graph the 2s wave function, now not only do I have an exponential, but the polynomial can occasionally be 0. There's a particular value of r for which this term cancels the 2 and can make this polynomial just linear function reach 0. So what that means is I have an exponential, but it's multiplied by something that might be large or might be small, might be 0. So the function goes through 0 at what we call a node and then still exponentially decays towards 0 as the r, the value of the radius gets larger. We know it's coming to 0 from the negative direction because when r is very large, this negative term is dominating. So the wave function is not what we're particularly interested in. We're more interested in the probability of finding an electron in a 2s orbital at certain radii. So again, we need to think of the probability not just as the wave function squared. Squaring it will make the positive terms and the negative terms both positive, but we need to remember that we need to multiply by this r squared to calculate the probability of finding the electron. Not only is the wave function 0 at this node, but it's also 0 at the origin when r equals 0. So it's going to start at 0, it's going to go up and then come back down. That's what happens when I square this positive part after multiplying by r squared. The negative part, I'll multiply by r squared and square it and it's going to look something like this. So again, we have a node, a particular radius at which we never find the electron, and the probability of finding the electron there is 0. We could work through the algebra, we won't do any more algebra in this video, but we could work through the algebra and determine where exactly is the node, how many multiples of the Bohr radius is that particular node. We could ask, where are we most likely to find the electron? Turns out this second peak is higher than the first peak. There's a local maximum here and then a larger global maximum out here. If I ask where are the locations where the probability function is flat, there's going to be two of them, one of them is going to be higher than the other. So what that means is when I draw a picture of the 2s wave function containing an electron like this picture of the 1s wave function, if I draw what that looks like, so again I've got some coordinates. This first peak I can have the electron appearing at small distances from the nucleus and then there's this node, there's a particular radius at which we never find the electron. So at that particular value of the radius, at every point on this sphere, the radius is a constant and then I'm even more likely to find the electron outside that sphere than inside it. So I again have a decaying as I go further away from the nucleus less and less likely to find the electron, quite likely to find it out here in this second shell, not at all likely to find it on the nodal surface on this radial node and then somewhat likely to find it inside that node. So again we have a spherical symmetry but the difference between the 1s orbital and the 2s orbital is the 2s orbital has this radial node. And now without even writing down any math we can extrapolate a little further and ask ourselves what are things going to look like for this 3s orbital. So this was all for the 2s, if I draw one more set of pictures for the 3s I'll just draw what the wave function looks, I'm sorry not the wave function, I'll draw what the probability looks like for the 3s orbital. The fact that this linear polynomial had 1,0 led to one node in the wave function, one node in the probability density over here. This 3s orbital has a quadratic polynomial, that's going to have two roots, two zeros. So that means the 3s wave function is going to have two nodes. It's going to go through a maximum, come back down to zero, a larger maximum, come back down to zero and a third maximum. So I'm going to have two separate radial nodes in the 3s wave function. So again if we cared about the details we could solve for the locations of those nodes, we could solve for the positions of these maximum, the relative heights of these maximum. But qualitatively speaking as the value of the quantum number n increases we get successively more peaks and successively more nodes in the wave function we get successively more shells in the graphical picture of the orbital.