 In today's class we are going to look at some numerical problems which basically help us understand some of the concepts that we have covered in class. So in today's assignment we are going to look at problems related to bond formation, we are going to look at problems related to density of states and other related aspects. So these will essentially cover the first 2 lectures of this course. So let us look at problem number 1. For a linear array of 6 hydrogen atoms draw the 6 possible molecular orbitals in order of increasing energy. So we have 6 hydrogen atoms and we know that when 6 atoms come together or when n atoms come together they form n orbitals which have 2n electrons which can accommodate 2n electrons. So we have 6 hydrogen atoms and we know they are going to form 6 molecular orbitals I will call them MOs and we need to draw those 6 possible configurations. So we will start with this but let me go back and start with a much simpler system that we looked at in class. So during the lecture we considered 3 hydrogen atoms. For 3 hydrogen atoms you have essentially 3 molecular orbitals. We saw that they have 0, 1 and 2 nodes and we are able to draw them in this fashion. So these dots represent the centers of the 3 hydrogen atoms. So basically the nuclei let the distance between the hydrogen atoms be A. So this is also A and just for the sake of symmetry we consider a distance A over 2 from one side and A over 2 from the other side. So the total length is 3A. So we can draw 3 molecular orbitals for this particular case. In the first case with 0 nodes we were able to draw something like this. So this has 0 nodes then the next one with increasing energy would be 1 node. So in this particular case the node goes through the center. So this has 1 node and we can draw something similar with 2 nodes. So here you have 2 nodes and these are the 3 molecular orbitals with increasing energy. So if you look at the concept behind drawing the nodes you can think of a wave that essentially encompasses all the 3 hydrogen atoms. So in this particular case you have half a wave. Here you have 1 full wave. So you have both the up and the down and here you have 1 and a half waves. So if lambda be the wavelength of this wave then in the first case with 0 node you had lambda by 2. So half the wave equal to 3A. The case of 1 node you have 2 times lambda by 2 equal to 3A. The case of 2 nodes we have 3 times lambda by 2 equal to 3A. So by knowing this value of lambda by 2 and lambda by 2 defines the position of the node you can go ahead and construct the molecular orbitals. So this is the case of 3 hydrogen atoms. We can go ahead and extend the same concept to 6 hydrogen atoms. So let me just draw that. So now you have 6 hydrogen atoms, 6 hydrogen atoms from 6 molecular orbitals, 6 MOs and the number of nodes will go from 0, 1, 2, 3, 4 and 5. So you have 6 of these going from 0 to 5. So once again let me mark the atoms. So these are my 6 hydrogen atoms. The spacing between them is A and the spacing is uniform throughout. We will also extend A over 2 on one side and A over 2 on the other side. So the dotted lines are just a guide to help me draw. So these are the 6 hydrogen atoms. The first case is 0 nodes. So that lambda over 2 is 6A. So it is the same as the 3 hydrogen atom situation where lambda over 2 was 3A. So this if you draw, so we have the 6 atomic orbitals mixing to form 1 molecular orbital which has 0 nodes. The next one is the 1 with 1 node. So that 2 times lambda over 2 is 6A. So now we have 3 up and 3 down and the function goes to 0 at essentially the middle of this. So here is where it goes to 0. So if you were to draw the wave functions, that is your 1 node. So we have 1 node, 3 of the atomic orbitals are up and 3 are down. Next you have 2 nodes. So let me write this as 0 nodes, 1 node, 2 nodes. So now it is 3 times lambda over 2 is 6A. So now you have 2 nodes. So the nodes are located here and then here and then we can draw the wave functions. So we have 2 up, 2 down and 2 up. So that overall we have 2 nodes. So let me just extend this down. So then we can keep going on. So now we are going to have 3 nodes. So that 4 times lambda over 2 is 6A. So we use this formula to find out the position of the nodes and once we know the position of the nodes, we can go back and draw the molecular orbitals. So here you have 1 node here and the next node comes in between, next node here. So this is the case with 3 nodes, 1, 2 and 3. Then we have 4 nodes. So it is 5 times lambda over 2 is 6A. So the positions, so this is slightly more complicated to draw but once you have the nodal positions, you can draw it. And then the last with 5 nodes, that is the trivial. So 6 times lambda over 2 is 6A. So you have everything going up and down. So the case of 6 hydrogen atoms, these are your 6 molecular orbitals and they go in the order of increasing energy. So the one with 0 nodes is the one that is most stable and then as we go down, we essentially increase in energy. So this answers the first part of the question. So draw the 6 possible molecular orbitals and we also see that the energy increases with the number of nodes. So when we ask the question, how does the energy depend on the number of nodes? The more the number of nodes, greater is the energy. We also want to plot a qualitative energy versus bond length curve for the system. So that we have already seen in class. We can draw these energy versus bond length. If we had only 2 hydrogen atoms, you will have 2 curves, 1 corresponding to bonding, 1 corresponding to anti-bonding. If you have 3 hydrogen atoms, you will have 3 curves and now we have 6 hydrogen atoms. We will essentially have 6 curves. So at infinity, you have the energy level corresponding to the hydrogen atom. There is a particular position at which the energy is minimum. This is the equilibrium distance. So we can plot energy versus bond length. So the lowest energy is the one with 0 nodes. So this will be the one with 0 nodes. The highest energy will be the one with 6 nodes. So this one will be the one with 6 nodes, sorry, with 5 nodes. So the one with the last one and everything else will come in between. So 1 node and then 2 nodes, 3, 4 and 5. So we essentially have 6 curves and they go again in order of increasing energy. So we then need to fill this system with the appropriate level and position of the electrons. So if you have 6 hydrogen atoms, each hydrogen atom can provide 1 electron. So we have 6 electrons and each molecular orbital can take 2 electrons of opposite spin. So this means you will fill 3 molecular orbitals. So each MO has 2 electrons of opposite spin. So these electrons we can put. So the first 2 electrons go here, then the next 2 electrons go here and then the final 2. So the lowest 3 molecular orbitals are full and the highest 3 are empty. So the last part of the question, what would you expect to be the equilibrium configuration of 6 hydrogen atoms? So if you have 6 hydrogen atoms, in reality 2 hydrogen atoms come together to form a hydrogen molecule. So if you have 6 hydrogen atoms, the equilibrium configuration is essentially 3 hydrogen molecules. So rather than form this linear chain of 6 hydrogen atoms, the actual configuration would just be 3 hydrogen molecules. So let us now go to the next question. So question 2, we want to draw energy versus bond length curves for various elements and compounds. So we have argon, sodium, sodium chloride and magnesium. So the important thing to note here is that sodium and magnesium are metals and we know in the case of metals, the valence band and the conduction band should overlap and argon and sodium chloride are insulators. So that we should show a band gap there. So let us draw the energy versus bond length curves. So let us first start with argon. So we can draw, we can write the electronic configuration for argon. So the previous inert gas is neon. So we have helium, neon, argon and so on. Neon followed by a 3s2, 3p6 and then the next higher level is empty. So 4s0. So 3p would form your valence band and 4s would form your conduction band and there will be a gap between these two. Another important thing to note is that argon is an inert gas which has something called a van der Waals bonding. We do not talk too much about van der Waals bonding in class but the important thing to note here is that the overlap is very small which means the energy bonds or the energy bands are all very small. So let us draw the energy versus bond length. So I have energy on my y-axis and then I have bond length on the x-axis. So this is my 3p6 which is the full level and then I have 4s0. There is a certain equilibrium distance which we will keep. So I am going to call this equilibrium. So we again have to show that there is some overlap but the overlap is very small. So this is a full level. So I will just mark it full but you can see that the overlap here is really small and then you have a 4s level which is empty. So this is full, this is empty and there is a band gap making this an insulator. So the next material we will look at is sodium chloride. Sodium chloride is also an insulator but it is a case of an ionic bond. So you have an ionic bond between Na plus and Cl minus. So instead of writing the electronic configuration of the atoms, we need the electronic configuration of the ions. So Na plus has the electronic configuration of neon 3s0, Cl minus has the electronic configuration of neon 3s2 and 3p6. So this is your full level which is 3p6. This is the empty level and there is a gap between those two which forms your band gap. So we can draw again energy versus bond length. This is an equilibrium distance. So I will just say equi to mean equilibrium distance. So I have the 3p6 of chlorine which is full and I have the 3s0 of sodium that is empty. So once again these come together to form a band. So this is a band that is full because the chlorine level is full. The case of sodium, let me just redraw this a bit. The case of sodium, so this is my 3s0 of sodium. You have a band that is empty and once again you have a band gap. So both sodium chloride and argon are essentially insulators and the way we depict them in your energy versus bond length diagram is to show a gap between a filled state and an empty state. So the next two are essentially metals. So you have sodium and magnesium and we have to show an overlap between the valence and the conduction band. So in C we will look at sodium. So now we have sodium that is a metal. So it is electronic configuration is neon 3s1. So we have s1. So we have a band that is half full. So the easiest way to depict this is to have your energy E versus bond length. You have a 3s1 that has only one electron. There is your equilibrium distance and when the atoms come together they form a band. So let me just slightly modify this. The atoms form a band and the band is only half full. So this makes sodium a metal because we have a band that is half full. The last one we are going to look at is magnesium. Magnesium is also a metal and it has the configuration neon 3s2. So s2 is essentially a full band. So to make sure that it is a metal to show the overlap between the valence and the conduction band, we consider the next level 3p0 and we say that there is a mixing between the s and the p. So this again we can show energy versus bond length. So this is the 3s level. So that is full. Now you have a 3p0 that is empty and this overlaps with the 3s. So now you have a full band and that is only half full. So we have overlap between s and p. So this way magnesium becomes a metal. So let us now go to the next question. So question 3, there are 10 electrons in a slab. So we have 10 electrons and the dimensions of the slab are given. So we have length of 0.5 nanometers and then width is 1 nanometer and height is 2 nanometers. So for simplicity, let us just call these a, b and c. So for an electron in a slab, there are essentially 3 quantum numbers n1, n2 and n3. So we have a set of quantum numbers n1, n2 and n3 and they are used to define the energy of the electron. So e in this case is nothing but h square over 8 Me, n1 square over a square plus n2 square over b square, n3 square over c square. So these represent the energy of the electron as a function of n1, n2 and n3 and each of these we can think or each set of these we can think of as an orbital which can accommodate 2 electrons. So if we have 10 electrons in a slab, we essentially have 5 orbitals which can each take 2 electrons so that all the 10 electrons are filled and these orbitals go in order of increasing energy. So we have 5 orbitals with 2 electrons each in order of increasing e. So we need to find 5 sets of n1, n2 and n3 which are all go in the order of increasing energy. An important thing is n1, n2 and n3 are all integers and they all should be greater than 0. So we can simplify this expression which we have here when we realize that b is nothing but 2 times a and c is nothing but 4 times a. Then energy e can be written as h square by 8 Me a square n1 square plus n2 square over 4 plus n3 square over 16. This is just by substituting b and c in terms of a and taking a out by writing by taking a common factor out. This can be written as h square over 128 Me square 16 n1 square plus 4 n2 square plus n3 square. So the first term here is essentially a constant let us call it k. So this is a constant times a variable that depends upon n1, n2 and n3. So we can plug in different values and then basically see the lowest 5 energy levels and then we can put the electrons in them. Let me write that in the form of a table. So we have the energy e is nothing but a constant times 16 n1 square plus n2 square plus n3 square. k we can evaluate and k comes out to be 0.094 electron volts. So let me write a table the first state next is energy in terms of k and then finally energy in electron volts. So we said that n1, n2 and n3 should all be integers and they should all be greater than 0. So 0 is not an acceptable number. So the lowest energy state is typically 1, 1, 1. So n1, n2 and n3 are all ones. In that particular case this expression this should be 4 n2 square sorry so 16 plus 4 plus n3 square. So if it is all 1, 1, 1 this expression is essentially 21 and the energy is 1.974. So this is your lowest energy level. The next energy level so one of these numbers must become 2 because they can only be integer numbers. So if you put 2 in the first case for n1 you find that the energy is actually multiplied by 16. So the next energy state would be 1, 1 and 2. That gives you a number 24 and the energy is 2.256. Then we have 113. Again it is just a question of plugging the numbers and checking the math 726. Then we have 121 that is 33, 3.102 and then actually we find that we have 2 states 1, 2, 2 and 114 both have the same energy. So the number is 36 and the energy is 3.384. So these states are called degenerate states because you have 2 sets of quantum numbers which have the same energy. So now we have our 5 orbitals. So if you fill in this gets 2 electrons, this gets 2, this gets 2 and this gets 2. So we have 8 electrons and both of them have the same energy. So each gets 1 electron. So if you look at the first part of the question we have to assign the quantum numbers to the electrons. Quantum numbers are your numbers n1, n2 and n3. So they have all be assigned. We want to deduce EF from this energy distribution. So the next thing we want to find is EF. EF is the Fermi energy and if you go back to the definition of EF the Fermi energy is the highest energy state. So in this particular example the highest energy state is this. So EF is just 3.384 electron volts. Part C we want to determine the density of states at EF. So density of states once again we will go back to the definition. So density of states is the total number of states per unit volume. So this is the number of states per unit volume. So the question says we want the density of states including spin. So if you look at 3.384 we have essentially 2 states. Each state can take 2 electrons. So we have a total of 4 available states. So the number of states is 4. The volume of the slab is nothing but ABC. So the product of all 3. So the density of states is nothing but 4 over ABC. So we can express this either in electron volts per Newton meter cube or joules per meter cube. Part D we want the total kinetic energy of all the electrons. So the total kinetic energy is nothing but the sum of all these energy values. So you just add all these numbers. The average kinetic energy is nothing but the total divided by 10. So the idea is even if you have a discrete system with a specific number of electrons instead of a solid where we have 10 to the 23 electrons we can still define terms like a Fermi energy or a density of states. Let us now go to problem 4. So we want to derive an expression for the density of states for a 2 dimensional and a 1 dimensional solid. We want to compare this with the derivation for a 3D solid. So we have looked at how to derive for a 3 dimensional solid in class. So let me refresh that briefly and then we will come back to how we do this for a 2D and a 1D solid. So in the case of a 3 dimensional solid we once again defined 3 quantum numbers nx, ny and nz and just like a particle in a box which we just saw in the previous problem. We were able to write an expression for the energy which is nothing but h square over 8 m l square nx square plus ny square plus nz square. This is nothing but h square n square over 8 m l square. So if you look at it the total number of states for a given value of n is essentially a sphere but we only consider the first quadrant of the sphere because all the values of n should be positive that is there should be greater than 0. So if you want to write the total number of states so s orbital of n. So the total number of states with energy less than n is nothing but the volume of the sphere and only considering the first quadrant so 1 over 8. If you also include spin each state can essentially take two electrons of opposite spin. So s of n including spin is nothing but 1 over 3 pi n cube. So we can write n in terms of energy. So if we use this equation n is nothing but 8 m l square over h square times e whole to the square root. So s of n we can write in terms of energy by just substituting the value of n here. So s of n 3 pi 8 m l square over h square energy whole to the 3 over 2. So this represents the total number of states in terms of the energy. If you want to know the total number of states per unit volume you divide this by l cube which will essentially cancel this expression l square. So you are left with this if you want to find the density of states then the density of states g of e is nothing but d s over d e. So we differentiate this with respect to energy and the final expression which we write is 8 pi square root of 2 m e over h square whole power 3 over 2 square root of e. So this is the derivation for a 3d solid. So we can now modify this to apply to a 2d and a 1d solid. For a 2d solid we can once again write energy e as h square n square over 8 m l square. But now instead of 3 quantum numbers you only have 2 nx square plus ny square. So instead of a sphere you will essentially have only a circle nx and ny and we will once again consider only the first quadrant of the circle. So the total number of states whose energy is less than that of n is again the area of the circle. So in 3d it was the volume of the sphere. Now it is the area of the circle pi n square and only the first quadrant so 1 4th. So it is 1 4th pi n square. If we take a round spin so this is 2 times s orbital of n. So this is pi n square over 2. So n square we can once again substitute in terms of energy. So s of e we can write the expression is 4 pi m l square over h square times e. If you want to do this per unit volume the l square term goes off and to find the density of states we differentiate s respect to e so that this is just a constant. So 4 pi m over h square. So for a 3d solid we found that the density of states increases with energy. It goes as a square root of e. For a 2d solid the density of states is essentially just a constant. For a 1d solid we can once again write a similar expression but there is only 1 n. So it is only n x. In that particular case s orbital is nothing but n. If you take spin into account s of n is 2n. We can substitute the value of n using this expression we will go through the same math. The final answer is the density of states is square root of 8m over h square 1 over root e. So for a 1d solid the density of states actually goes down with increase in energy. So let us now look at the last problem. So we want to plot the Fermi function for temperatures of 0, 500 and 2000 Kelvin. It is a semi qualitative plot and we want to do this on the same plot. So the Fermi function if you remember f of e is nothing but 1 over 1 plus exponential e minus ef over kT. So when the energy is less than ef the probability is 1. When the energy is more than ef at 0 Kelvin probability is 0 and at e equal to ef probability is half. So f of e was this temperature sorry f of e was this energy. So we want to plot the Fermi function as function of energy. So f of e was this energy. Let me mark ef and 0 and Fermi function goes from 0 to 1 and I will mark 1 half. So at 0 Kelvin it has a value of 1 and then it has a value of 0. So it is a delta function. So this is 0 Kelvin. We now increase the temperature slightly. So we go to 500 Kelvin. So now you have a slight broadening. So you have some states that can get excited. So here you have a certain occupation probability above the Fermi energy and you have some electrons that are lost below the Fermi energy. So this is 500 Kelvin. We can do the same if we increase the temperature further. This deviation is even more but all cases it passes through half. So whatever be the temperature the occupation probability is always half at the Fermi energy. So this one is 2000. So these spreads are essentially qualitative they are just to show that with increase in temperature more and more higher energy states are being occupied. So the next part we want to compare the Fermi function and the Boltzmann function. So f of E we have written E f over kT. The Boltzmann function is just exponential minus E over E f over kT. So when E minus E f is 3 kT we can substitute and the numbers are 3 minus 3 kT. The Fermi function is 0.047. The Boltzmann function is 0.049. So they are close but there is a small deviation. For 15 kT the Fermi function is 3.06 times 10 to the minus 7 and the Boltzmann function is also 3.06 times 10 to the minus 7. So further the deviation from the Fermi energy so if you want to look at it 15 kT at room temperature is approximately 0.37 electron volts which is smaller than the band gap of silicon or half the band gap of silicon. So in the case of solids the Boltzmann function is a good enough approximation of the Fermi function.