Question

Suppose two electrons in an atom have quantum numbers n = 2 and l = 1 . (a) How many states are possible for those two electrons? (Keep in mind that the electrons are indistinguishable.) (b) If the Pauli Exclusion Principle did not apply to the electrons, how many states would be possible?

Short answer

1. There are 15 possible states for those two electrons.
2. If the Pauli Exclusion Principle not applied to the electrons, the number of possible states would be 21 .

Step-by-step solution

The given data:

There are two electrons in an atom with quantum numbersn = 2 and l = 1.

Understanding the concept of quantum number of an electron:

Pauli's exclusive principle states that no two electrons in the same atom can have the same values in all four of their quantum numbers.

The set of numbers used to describe the position and power of an electron atom is called quantum numbers. There are four quantum numbers, namely, prime numbers, azimuthal, magnetic numbers, and spin quantum. The stored values of the quantum system are given by quantum numbers.

For every value of the principal quantum number, there are n values of ranging from 0 to $\left(\mathrm{n}-1\right)$. For every value of the orbital quantum number, there exist $\left(2\mathrm{l}+1\right)$values of magnetic quantum number ranging from $-\mathrm{l}\mathrm{to}+\mathrm{l}$. Thus, the maximum value of this magnetic quantum number is equal to the value of the orbital quantum number. Now, every electron has two spin orientations, thus the value of the magnetic quantum number is determined by these spins.

(a) Calculation of the number of possible states of the two electrons:

Using the given quantum state, n = 2 and l = 1

The possible cases of the magnetic quantum number are $\left(-1,0+1\right)$.

Now, for every electron there are 2 spins orientations; thus the possible cases of spin quantum number are

${\mathrm{m}}_{\mathrm{s}}=\pm \frac{1}{2}$

Thus, the possible states if Pauli’s Exclusion Principle is applied are:

role="math" localid="1661499684099" $$\begin{gathered} \left(m_{l1},m_{s1},m_{l2},m_{s2}\right)=\left(1,+1/2,1,-1/2\right),\left(1,+1/2,0+1/2\right),\left(1,+1/2,0,-1/2\right),\left(1,+1/2-1,+1/2\right), \\ \left(1,+1/2,-1,-1/2\right),\left(1,-1/2,0,+1/2\right),\left(1,-1/2,0-1/2\right),\left(1,-1/2,-1,+1/2\right), \\ \left(1,-1/2,-1,-1/2\right),\left(0,+1/2,0,-1/2\right),\left(0,+1/2,-1,+1/2\right),\left(0,+1/2,-1,-1/2\right), \\ \left(0,-1/2,-1,+1/2\right),\left(0,-1/2,-1,-1/2\right),\left(-1,+1/2,-1,-1/2\right) \end{gathered}$$

Hence, there are total 15 possible states.

(b) Calculation of the possible states if Pauli Exclusion Principle is not applied

In the above possible states, there are 6 not possible states if Pauli’s Exclusion is applied.

Those possible states if Pauli’s Exclusion Principle is not applied are:

$$\begin{gathered} \left(m_{l1},m_{s1},m_{l2},m_{s2}\right)=\left(1,+1/2,1,+1/2\right),\left(1,-1/2,1,-1/2\right),\left(0,+1/2,0,+1/2\right) \\ \left(0,-1/2,0,-1/2\right),\left(-1,+1/2,-1/2\right),\left(-1,-1/2,-1,-1/2\right) \end{gathered}$$

Hence, there are total 6 possible states.

Thus, the total number of possible states including the above possible states is given by:

15+6 = 21 electron states.

Hence, there are21 possible electron states.