Problem with solution: Excited Hydrogen Atom: Quantum Numbers and Orbitals

The electron of the hydrogenium atom (hydrogen atom) cannot have an arbitrary binding energy, but can only take on discrete energies $ W_n $, which are characterized by the principal quantum number $ n $. The individual energy states of the electron are designated by letters K, L, M, N etc., especially in chemistry.

What is the energy of the electron in the K, L, and M shells?
How many orbitals $ N $ does the Hydrogen atom have if the energy of the electron is $ W_n \lt -0.5 \,\mathrm{eV} $?
Enumerate all possible orbitals by specifying their $ (n, l, m_{\text l}) $ state. How many orbitals of these have the minor quantum number $ l=2 $?
What energy must the electron have so that the Hydrogen atom has at least 200 orbitals but at most 300 orbitals?

Solution tips

Use the Rydberg formula to determine the individual energy states for the respective shells: \[ W_n = -W_{\text R} \, \frac{1}{n^2} \] hierbei ist $ W_{\text R} = 13.6 \, \text{eV} $ der Energiebetrag des Grundzustands $ n = 1 $, also der Betrag der tiefsten Energie, die das Elektron im H-Atom annehmen kann.

Das Schalenmodell ist eine Vereinfachung des Orbitalmodells. Die Schale K steht für $ n = 1 $. Die Schale L steht für $ n = 2 $. Die Schale M steht für $ n = 3 $.

Zu (b): Anzahl der Orbitale $ N $ pro Hauptquantenzahl $ n $ ist gegeben durch $ n^2 $.

Zu (c): Die Nebenquantenzahl $ l $ hat die Einschränkung: $ l \lt n $ und startet mit $l=0$. Die magnetische Quantenzahl $ m_{\text l} $ kann die Werte $ -l \leq m_{\text l} \leq l $ annehmen.

Solution for (a)

The energy of the electron is given by the Rydberg formula:

$$ \begin{align} W_n = -W_{\text R} \, \frac{1}{n^2} \end{align} $$

Here, $ W_{\text R} = 13.6 \, \text{eV} $ is the Rydberg energy, which gives the energy of an electron in the ground state ($ n = 1 $) of the Hydrogen atom. The minus sign in Eq. 1 says that the electron is bound in the H atom. This means that the energy must be absorbed to ionize the H atom (to remove the electron from the H atom).

Using 1, you can determine the energy of the electron in the K shell ($ n = 1 $):

$$ \begin{align} W_n = -W_{\text R} \cdot \frac{1}{1^2} = -13.6 \, \text{eV} \end{align} $$

Energy of the electron in the L-shell ($ n = 2 $):

$$ \begin{align} W_n = -W_{\text R} \cdot \frac{1}{2^2} = -3.4 \, \text{eV} \end{align} $$

Energy of the electron in the M shell ($ n = 3 $):

$$ \begin{align} W_n = -W_{\text R} \cdot \frac{1}{3^2} = -1.51 \, \text{eV} \end{align} $$

Solution for (b)

The number $ N_{n} $ of orbitals (without spin) for just one energy level $ n $ is given by the principal quantum number squared and is indirectly in the Rydberg formula (a) 1:

$$ \begin{align} N_{n} ~=~ n^2 \end{align} $$

The principal quantum number $ n $, if you look at the Rydberg formula (a) 1, is larger the larger the energy of the electron is. Thus according to eq. (b) 1 also the number of orbitals is larger. According to the problem the energy is given by $W_n$ and should be smaller than $ W_n \lt -0.5 \, \text{eV} $. "Less than" means that you have to use the less sign instead of the equal sign. Insert in this inequality for $W_n$ the Rydberg formula (a) 1:

$$ \begin{align} - W_{\text R} \, \frac{1}{n^2} ~\lt~ -0.5 \, \text{eV} \end{align} $$

Rearrange the equation for $n$:

$$ \begin{align} n \lt \sqrt{ \frac{13.6 \, \mathrm{eV} }{ 0.5 \, \mathrm{eV} } } \end{align} $$

Notice that the less sign has flipped because both sides have been multiplied by "-". Calculate the square root:

$$ \begin{align} n ~\lt~ 5.2 \end{align} $$

Thus, the largest principal quantum number according to Eq. 9 is: $ n = 5 $. Consider that such principal quantum numbers as $ n=5.2$ do NOT exist. From this you determine the orbital number with the help of eq. (b) 1. You sum up the number from 1 to 5:

$$ \begin{align} N ~=~ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55 \end{align} $$

You can calculate the sum of squares $ n^2 $ as in Eq. 5, which tells you the number of orbitals, generally as follows:

$$ \begin{align} N ~=~ \frac{n(n+1)(2n+1)}{6} \end{align} $$

Solution for (c)

After determining the number of orbitals in (b), you can also ask yourself, what are the exact states (i.e., quantum numbers) of these 55 orbitals? An orbital is determined by the three quantum numbers:

Principal quantum number $ n $, where it can only take integers $n = 1,2,3... $.
Angular moment quantum number $ l $ is always smaller than the principal quantum number. And the maximum principal quantum number is: $ l = n-1 $. Like any other quantum number, it can also take only integers. It can have the value $ l = 0 $ in contrast to $ n $.
Magnetic quantum number $ m_{\text l} $ is between minimum angular momentum quantum number $-l $ and maximum angular momentum quantum number $l$ $ -l \leq m_{\text l} \leq l $.

To determine the concrete orbitals, you have to specify the number triplet $ (n,l,m_{\text l})$ and in this case 55 number triplets, because of 55 possible orbitals. In (b) you have determined the maximum major quantum number $ n = 5 $ for the energy $ E \lt -0.5 \, \text{eV} $. Thus, the maximum angular momentum quantum number is $ l = n-1 = 5-1 = 4 $.

The magnetic quantum number can only take the following values for $ l = 4 $:

$$ \begin{align} m_{\text l} = \{-4,-3,-2,-1,0,1,2,3,4\} \end{align} $$

The magnetic quantum number for $ l = 3 $, on the other hand:

$$ \begin{align} m_{\text l} = \{-3,-2,-1,0,1,2,3 \} \end{align} $$

The magnetic quantum number for $ l = 2 $:

$$ \begin{align} m_{\text l} = \{-2,-1,0,1,2 \} \end{align} $$

The magnetic quantum number for $ l = 1 $:

$$ \begin{align} m_{\text l} = \{-1,0,1 \} \end{align} $$

The magnetic quantum number for $ l = 0 $ is $ m_{\text l} = \{ 0 \} $.

If you go through all possibilities, considering the above conditions for the quantum numbers, then you get exactly 55 orbitals. So start at $ n = 1 $ and go through the conditions. Then go to $ n = 2 $ and go through the conditions and so on. The following table summarizes all possible triplets up to $n=5$:

Table : All orbitals up to the 5th energy level
Principal quantum number $ n$	Angular momentum quantum number $l$	Orbital $ (n,l,m_{\text l})$
1	0	(1,0,0)
2	0	(2,0,0)
2	1	(2,1,-1)
2	1	(2,1,0)
2	1	(2,1,1)
3	0	(3,0,0)
3	1	(3,1,-1)
3	1	(3,1,0)
3	1	(3,2,1)
3	2	(3,2,-2)
3	2	(3,2,-1)
3	2	(3,2,0)
3	2	(3,2,1)
3	2	(3,2,2)
4	0	(4,0,0)
4	1	(4,1,-1)
4	1	(4,1,0)
4	1	(4,1,1)
4	2	(4,2,-2)
4	2	(4,2,-1)
4	2	(4,2,0)
4	2	(4,2,1)
4	2	(4,2,2)
4	3	(4,3,-3)
4	3	(4,3,-2)
4	3	(4,3,-1)
4	3	(4,3,0)
4	3	(4,3,1)
4	3	(4,3,2)
4	3	(4,3,3)
5	0	(5,0,0)
5	1	(5,1,-1)
5	1	(5,1,0)
5	1	(5,1,1)
5	2	(5,2,-2)
5	2	(5,2,-1)
5	2	(5,2,0)
5	2	(5,2,1)
5	2	(5,2,2)
5	3	(5,3,-3)
5	3	(5,3,-2)
5	3	(5,3,-1)
5	3	(5,3,0)
5	3	(5,3,1)
5	3	(5,3,2)
5	3	(5,3,3)
5	4	(5,4,-4)
5	4	(5,4,-3)
5	4	(5,4,-2)
5	4	(5,4,-1)
5	4	(5,4,0)
5	4	(5,4,1)
5	4	(5,4,2)
5	4	(5,4,3)
5	4	(5,4,4)

The second question was how many orbitals have the angular momentum quantum number $ l =2 $. You can easily count this using the table. Of course the energies with $ n=1$ and $n=2$ cannot have a state with a angular momentum quantum number $ l=2$, because otherwise they would not fulfill the condition $ l \lt n $. So, only principal quantum numbers greater than 2 can be considered. In our case $ n = 3,4,5$ (in total 3 principal quantum numbers).

Each energy state with $ n \gt 2 $ contains exactly five $l=2$ states, for example for $ n = 3 $ they are (3,2,-2), (3,2,-1), (3,2,0), (3,2,1) and (3,2,2). Three principal quantum numbers thus have $3 \cdot 5=15$ orbitals with $l=2$. Alternatively, you can calculate the number $N(l)$ of orbitals with any $l$ as follows:

$$ \begin{align} N(l) ~=~ (n_{\text{max}}-l) \cdot (2l + 1) \end{align} $$

Of course, the condition $ n \gt l $ must always be fulfilled.

Solution for (d)

We are looking for the energy $ W_n $ where the minimum orbital number is $ N_{\text{min}} = 200 $ and the maximum orbital number is $ N_{\text{max}} = 300 $. Thus, the energy $W_n$ must be in a certain energy range for these two conditions to be satisfied. For this you need the formula from exercise (b), which generally gives you the number of orbitals in the H atom. Here is the formula:

$$ \begin{align} N ~=~ \frac{n(n+1)(2n+1)}{6} \end{align} $$

We are looking for the principal quantum number $ n $, so that you can determine the minimum energy with the Rydberg formula. Rearrange the equation 1 in such a way that the equation has the following form:

$$ \begin{align} 0 ~=~ 2n^3 + 3n^2 + n - 6N \end{align} $$

As you can see from Eq. 2, you cannot simply solve for $ n $ because it is a 3rd degree polynomial. Use your calculator or the internet, with which you can find the zeros (i.e. the solutions). You get the following real solution, if you use $N=N_{\text{min}}=200$: n=7.944. And, if you use $N=N_{\text{max}}=300$, you get the following solution: $n=9.164$. This means: The Hydrogen atom has between 200 and 300 orbitals exactly when the principal quantum number is between 7,944 and 9,164. But since it must be an integer, it is then either 8 or 9.

To calculate the minimum and maximum energy, simply substitute these two quantum numbers into the Rydberg formula:

$$ \begin{align} W_{\text{min}} &= -13.6 \, \text{eV} \cdot \frac{1}{8^2} = -0.2125 \, \text{eV} \\\\
W_{\text{max}} &= -13.6 \, \text{eV} \cdot \frac{1}{9^2} = -0.1679 \, \text{eV} \end{align} $$

That means: For at least 200 orbitals to be occupied, the energy $ W_n $ of the electron must be greater than $ - 0.2125 \, \text{eV} $. However, there must also be no more than 300 orbitals, so the energy must be less than $ - 0.1679 \, \text{eV} $. Overall, the energy lies between these two values: $ - 0.2125 \, \text{eV} ~\leq~ W_n ~\leq~ - 0.1679 \, \text{eV} $.