quantum numbers and Pauli principle

Quantized energy levels and line spectra show that an electron in an atom can occupy only discrete states. Each state is labeled by a set of quantum numbers: the principal number $n=1,2,3,\dots$ sets the energy and size of the region, the orbital number $l=0,1,\dots,n-1$ gives the magnitude of orbital angular momentum, the magnetic number $m_l=-l,\dots,+l$ gives its projection on a chosen axis, and the spin number $m_s=\pm\tfrac12$ describes the electron's intrinsic angular-momentum projection. The Pauli principle forbids two electrons in the same atom from sharing the same complete set $(n,l,m_l,m_s)$. This exclusion rule fixes subshell and shell capacities $2(2l+1)$ and $2n^2$, and it underlies electron filling rules seen in the periodic table and in the fine structure of spectral lines.

For an electron with principal quantum number $n=3$, list all allowed values of $l$, $m_l$, and $m_s$. How many distinct one-electron states exist for $n=3$?

Check whether the set $(n=2,\ l=2,\ m_l=0,\ m_s=+\tfrac12)$ is allowed. If it is not, state which rule is violated. Then determine the maximum number of electrons in the $3p$ subshell.

A hypothetical atomic configuration is written as $1s^3$. Using quantum numbers and the Pauli principle, explain why this configuration is impossible. Write the correct ground-state configuration of lithium ($Z=3$).

Use an interactive quantum-number simulation (for example, PhET “Quantum Numbers” or an electron-configuration builder).

• Build a table for $n=1,2,3$: list the allowed $l$, $m_l$, and $m_s$, and count the number of states.
• Compare the counted states with $2n^2$ and explain the match.
• Try to place electrons into subshells while respecting the Pauli principle, and note which attempts the simulation rejects.
  Do not use real radioactive or high-voltage sources; a simulation is sufficient to observe the quantum constraints.

Show how quantum numbers form discrete states and how the Pauli principle restricts the filling of these states by electrons.

Diagram: a vertical energy scale with levels $n=1,2,3$; beside it, subshell boxes labeled $(n,l)$; arrows $\uparrow$ and $\downarrow$ denote $m_s=+\tfrac12$ and $m_s=-\tfrac12$; one box with two identical arrows is crossed out as forbidden by the Pauli principle.

A night street is lit by yellow lamps. In the tube of a $150\ \text{W}$ street lamp a discharge burns at a current of about $1\ \text{A}$: sodium atoms emit almost a single wavelength — $589.0\ \text{nm}$ and $589.6\ \text{nm}$, that is photons of energy $E=hc/\lambda=2.1\ \text{eV}$.

The state of the outer electron of sodium is described by four quantum numbers: the ground state has $n=3$, $l=0$, $m_l=0$; the excited $3p$ state has $n=3$, $l=1$, $m_l=-1,0,+1$. The Pauli principle allows at most two electrons per orbital, with $m_s=+1/2$ and $m_s=-1/2$, so the three $p$ orbitals provide $3\times 2=6$ seats rather than three. This set of allowed states decides which transitions are possible at all: the $3p\to 3s$ transition is allowed and gives the bright yellow line.

A balloon of volume $0.03\ \text{m}^3$ is filled with helium. The density of helium is $0.17\ \text{kg/m}^3$, of air $1.2\ \text{kg/m}^3$, so the lift is $(1.2-0.17)\cdot 0.03\cdot 9.8\approx 0.30\ \text{N}$ and the balloon holds up a garland of about $30\ \text{g}$. To obtain liquid helium the gas must be cooled to $4.2\ \text{K}$.

Helium has two electrons and both occupy the $1s$ orbital ($n=1$, $l=0$, $m_l=0$) with opposite spins $m_s=+1/2$ and $m_s=-1/2$. The Pauli principle forbids a third electron from occupying $n=1$, so the $1s^2$ shell is closed and helium forms no chemical bonds — at $293\ \text{K}$ it stays a monatomic gas. The attraction between closed shells is so weak that at atmospheric pressure the atoms hold together only below $4.2\ \text{K}$.

In a kitchen a magnet with a pull-off force of $5\ \text{N}$ sticks to the steel door of the fridge and immediately slides off an aluminium drink can. Iron has $26$ electrons and the outer configuration $3d^6 4s^2$; aluminium has $13$ electrons and the configuration $3s^2 3p^1$. Both are solid metals, yet they behave differently with a magnet.

For $l=2$ (the $d$ subshell) the magnetic quantum number takes five values $m_l=-2,-1,0,+1,+2$, and the Pauli principle allows two electrons ($m_s=\pm 1/2$) in each — ten seats in $3d$ in total. In iron the first five of the six $3d$ electrons occupy all five orbitals singly with parallel spins, and four stay unpaired: their magnetic moments add up and the metal becomes magnetised. In aluminium the single $3p$ electron joins the common electron gas of the metal and its spin is paired, so the net magnetic moment is almost zero and the magnet does not stick.