2
$\begingroup$

I have been having trouble with understanding and solving the following problem.

For two-dimensional conjugated systems, we may use the particle in a two-dimensional box model. In this case the total energy can be written as follows:

$$E_{n_1,n_2} = \frac{h^2}{8m} \left( \frac{n^2_1}{L^2_1}+\frac{n^2_2}{L^2_2} \right)$$

where $L_1$ and $L_2$ are the lengths and $n_1$ and $n_2$ are the quantum numbers of the first and second dimensions, respectively, $m$ is the mass of the particle and $h$ is the Planck constant.

Graphene is a sheet of carbon atoms in the form of a two-dimensional hexagonal lattice in which one atom forms each vertex.

enter image description here

The distance between two adjacent carbons in the hexagonal 6-carbon unit is approximately $1.4$ Å. For a square shaped graphene sheet with $L_1$ = $L_2$ = $11$ Å:

1) Calculate the number of electrons in a ($11$ Å $×$ $11$ Å) sheet of graphene. For this problem you may ignore edge electrons.

2) Calculate the energy of the HOMO.

3) Calculate the energy of the LUMO.

For the first question, I simply divided the area of the graphene sheet by the area of a hexagon shaped graphene unit. This resulted in approximately 24 units, which means there is a total of 48 carbon atoms, since each unit contributes 2 carbon atoms. Therefore there would be a total of 288 electrons in the sheet (from multiplying 48 by 6, since each carbon atom has 6 electrons).

In case this is right, I don't know how to go about calculating the HOMO and LUMO energies. I'm having a bit trouble understanding the quantum numbers in the formula given above: I don't know what the values of the quantum numbers should be when calculating the HOMO or LUMO energy. I do understand how the quantum numbers combinations and energy are related, but I don't know, which specific combinations are used when calculating the HOMO and LUMO energies.

$\endgroup$
  • 2
    $\begingroup$ You can use the number of occupied MOs to get to the HOMO. Then it'll straightforward to get to the LUMO. Part of the exercise is to relate the quantum numbers to the MOs. $\endgroup$ – rbw May 20 at 22:58
1
$\begingroup$

The key to answering this is to remember the following principles:

1) electrons are fermions and therefore two electrons cannot share all quantum numbers (degeneracy of all QM numbers is forbidden)

2) electrons have spin quantum number, which means an energy state corresponding to a given set of $n_1$ and $n_2$ quantum numbers can be doubly occupied (doubly degenerate)

3) the Aufbau principle, which means lowest energy states are occupied first

Once you know the total number of electrons, you have to assign each a pair of quantum numbers $n_1$ and $n_2$ (ignoring $n_s$, the spin quantum #). The first two principles listed above mean that to each $(n_1,n_2)$ state you can assign two and only two electrons (with opposite spin quantum #). The third principle means that, since the energy increases with $n_i$, and the sides of the box are identical (making states that permute the $n_i$ degenerate), you should assign the $n_i$ in increasing order of energy as follows: the lowest E state has $n_1=1,n_2=1$, or (1,1); followed by (2,1) or (1,2) - these are degenerate in energy; then (2,2); then (3,1) or (1,3) -again degenerate; then (2,3) and so on.

In tabular form the energies increase as follows:

$$\begin{array}{c|c|c|c}\hline i & n_1 & n_2 & n_s \\ \hline 1 & 1 & 1 & \alpha \\ & 1 & 1 & \beta \\ 2 & 2 & 1 & \alpha \\ & 2 & 1 & \beta \\ & 1 & 2 & \alpha \\ & 1 & 2 & \beta \\ 3 & 2 & 2 & \alpha \\ & 2 & 2 & \beta\\ 4 & 1 & 3 & \alpha \\ ...&...&...& \\ \hline\end{array} $$

If you counted 288 electrons you need 144 states that are doubly degenerate (permutations of $n_i$, differing only in $n_s$) to populate the box. The HOMO is the occupied state with highest energy.

Using a script to crack the problem (rather than searching for a mathematical formula that encodes the pattern) produced the following list of states (each can be populated by two electrons, one with $n_s=\alpha$, the other $n_s=\beta$). The HOMO is triply degenerate (ignoring spin) and the LUMO doubly degenerate.

$$\begin{array}{c|c|c|c}\hline i & n_1^2 + n_2^2 & n_1 & n_2 \\ \hline 1 & 2 & 1 & 1 \\ 2 & 5 & 1 & 2 \\ 3 & 5 & 2 & 1 \\ 4 & 8 & 2 & 2 \\ 5 & 10 & 1 & 3 \\ 6 & 10 & 3 & 1 \\ 7 & 13 & 2 & 3 \\ 8 & 13 & 3 & 2 \\ 9 & 17 & 1 & 4 \\ 10 & 17 & 4 & 1 \\ 11 & 18 & 3 & 3 \\ 12 & 20 & 2 & 4 \\ 13 & 20 & 4 & 2 \\ 14 & 25 & 3 & 4 \\ 15 & 25 & 4 & 3 \\ 16 & 26 & 1 & 5 \\ 17 & 26 & 5 & 1 \\ 18 & 29 & 2 & 5 \\ 19 & 29 & 5 & 2 \\ 20 & 32 & 4 & 4 \\ 21 & 34 & 3 & 5 \\ 22 & 34 & 5 & 3 \\ 23 & 37 & 1 & 6 \\ 24 & 37 & 6 & 1 \\ 25 & 40 & 2 & 6 \\ 26 & 40 & 6 & 2 \\ 27 & 41 & 4 & 5 \\ 28 & 41 & 5 & 4 \\ 29 & 45 & 3 & 6 \\ 30 & 45 & 6 & 3 \\ 31 & 50 & 1 & 7 \\ 32 & 50 & 5 & 5 \\ 33 & 50 & 7 & 1 \\ 34 & 52 & 4 & 6 \\ 35 & 52 & 6 & 4 \\ 36 & 53 & 2 & 7 \\ 37 & 53 & 7 & 2 \\ 38 & 58 & 3 & 7 \\ 39 & 58 & 7 & 3 \\ 40 & 61 & 5 & 6 \\ 41 & 61 & 6 & 5 \\ 42 & 65 & 1 & 8 \\ 43 & 65 & 4 & 7 \\ 44 & 65 & 7 & 4 \\ 45 & 65 & 8 & 1 \\ 46 & 68 & 2 & 8 \\ 47 & 68 & 8 & 2 \\ 48 & 72 & 6 & 6 \\ 49 & 73 & 3 & 8 \\ 50 & 73 & 8 & 3 \\ 51 & 74 & 5 & 7 \\ 52 & 74 & 7 & 5 \\ 53 & 80 & 4 & 8 \\ 54 & 80 & 8 & 4 \\ 55 & 82 & 1 & 9 \\ 56 & 82 & 9 & 1 \\ 57 & 85 & 2 & 9 \\ 58 & 85 & 6 & 7 \\ 59 & 85 & 7 & 6 \\ 60 & 85 & 9 & 2 \\ 61 & 89 & 5 & 8 \\ 62 & 89 & 8 & 5 \\ 63 & 90 & 3 & 9 \\ 64 & 90 & 9 & 3 \\ 65 & 97 & 4 & 9 \\ 66 & 97 & 9 & 4 \\ 67 & 98 & 7 & 7 \\ 68 & 100 & 6 & 8 \\ 69 & 100 & 8 & 6 \\ 70 & 101 & 1 & 10 \\ 71 & 101 & 10 & 1 \\ 72 & 104 & 2 & 10 \\ 73 & 104 & 10 & 2 \\ 74 & 106 & 5 & 9 \\ 75 & 106 & 9 & 5 \\ 76 & 109 & 3 & 10 \\ 77 & 109 & 10 & 3 \\ 78 & 113 & 7 & 8 \\ 79 & 113 & 8 & 7 \\ 80 & 116 & 4 & 10 \\ 81 & 116 & 10 & 4 \\ 82 & 117 & 6 & 9 \\ 83 & 117 & 9 & 6 \\ 84 & 122 & 1 & 11 \\ 85 & 122 & 11 & 1 \\ 86 & 125 & 2 & 11 \\ 87 & 125 & 5 & 10 \\ 88 & 125 & 10 & 5 \\ 89 & 125 & 11 & 2 \\ 90 & 128 & 8 & 8 \\ 91 & 130 & 3 & 11 \\ 92 & 130 & 7 & 9 \\ 93 & 130 & 9 & 7 \\ 94 & 130 & 11 & 3 \\ 95 & 136 & 6 & 10 \\ 96 & 136 & 10 & 6 \\ 97 & 137 & 4 & 11 \\ 98 & 137 & 11 & 4 \\ 99 & 145 & 1 & 12 \\ 100 & 145 & 8 & 9 \\ 101 & 145 & 9 & 8 \\ 102 & 145 & 12 & 1 \\ 103 & 146 & 5 & 11 \\ 104 & 146 & 11 & 5 \\ 105 & 148 & 2 & 12 \\ 106 & 148 & 12 & 2 \\ 107 & 149 & 7 & 10 \\ 108 & 149 & 10 & 7 \\ 109 & 153 & 3 & 12 \\ 110 & 153 & 12 & 3 \\ 111 & 157 & 6 & 11 \\ 112 & 157 & 11 & 6 \\ 113 & 160 & 4 & 12 \\ 114 & 160 & 12 & 4 \\ 115 & 162 & 9 & 9 \\ 116 & 164 & 8 & 10 \\ 117 & 164 & 10 & 8 \\ 118 & 169 & 5 & 12 \\ 119 & 169 & 12 & 5 \\ 120 & 170 & 1 & 13 \\ 121 & 170 & 7 & 11 \\ 122 & 170 & 11 & 7 \\ 123 & 170 & 13 & 1 \\ 124 & 173 & 2 & 13 \\ 125 & 173 & 13 & 2 \\ 126 & 178 & 3 & 13 \\ 127 & 178 & 13 & 3 \\ 128 & 180 & 6 & 12 \\ 129 & 180 & 12 & 6 \\ 130 & 181 & 9 & 10 \\ 131 & 181 & 10 & 9 \\ 132 & 185 & 4 & 13 \\ 133 & 185 & 8 & 11 \\ 134 & 185 & 11 & 8 \\ 135 & 185 & 13 & 4 \\ 136 & 193 & 7 & 12 \\ 137 & 193 & 12 & 7 \\ 138 & 194 & 5 & 13 \\ 139 & 194 & 13 & 5 \\ 140 & 197 & 1 & 14 \\ 141 & 197 & 14 & 1 \\ \text{HOMO}\,142 & 200 & 2 & 14 \\ \text{HOMO}\,143 & 200 & 10 & 10 \\ \text{HOMO}\, 144 & 200 & 14 & 2 \\ \text{LUMO}\, 145 & 202 & 9 & 11 \\ \text{LUMO}\, 146 & 202 & 11 & 9 \\ 147 & 205 & 3 & 14 \\ 148 & 205 & 6 & 13 \\ 149 & 205 & 13 & 6 \\ 150 & 205 & 14 & 3 \\ 151 & 208 & 8 & 12 \\ 152 & 208 & 12 & 8 \\ 153 & 212 & 4 & 14 \\ 154 & 212 & 14 & 4 \\ 155 & 218 & 7 & 13 \\ 156 & 218 & 13 & 7 \\ 157 & 221 & 5 & 14 \\ 158 & 221 & 10 & 11 \\ 159 & 221 & 11 & 10 \\ 160 & 221 & 14 & 5 \\ \hline\end{array} $$

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I'm still a bit confused how you ended up with the combinations (47,48) and (48,47). Do I need to list 144 different combinations in increasing order? $\endgroup$ – Pöytä Laatikko May 21 at 7:21
  • 1
    $\begingroup$ @PöytäLaatikko Well, I saw that that would involve a lot of effort, so I tried finding a way of computing the highest QM number. Each ni QM # occurs a max of 6 times, and from your question 6x48=288, so the HOMO has to have QM n=48 in one dimension. It looks like I made a mistake at this point, however. It seems the HOMO should be a combination of (48,49) and (49,48), and the LUMO is (49,49). I changed my answer to refect this and again, while the principles I state should guide you, ... $\endgroup$ – Buck Thorn May 21 at 8:17
  • $\begingroup$ ... enumerating states explicitly is not a good way to solve the problem, you should instead find some other way to count them. Look for a pattern in the way the states are filled. $\endgroup$ – Buck Thorn May 21 at 8:19
  • $\begingroup$ Have you figured the right combinations out yet? I'm still very unsure about them. $\endgroup$ – Pöytä Laatikko May 24 at 14:30
  • 1
    $\begingroup$ Thanks for the effort! After some inspection of my own, I think I ended up with the same result! $\endgroup$ – Pöytä Laatikko May 24 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.