Quiz

Question 1 — Valid or Invalid Quantum Numbers?

For each set of quantum numbers below, state whether it is valid or invalid, and explain why.

Recall the allowed ranges:

Quantum number	Symbol	Allowed values
Principal	$n$	$1,2,3,\dots$ (positive integers only)
Azimuthal	$\ell$	$0,1,\dots ,n-1$
Magnetic	$m_{\ell }$	$-\ell ,\dots ,0,\dots ,+\ell$
Spin	$m_s$	$+\frac{1}{2}$ or $-\frac{1}{2}$ only

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Set A: $n=1,\ell =1$

Answer

Invalid. $\ell$ must satisfy $0\le \ell \le n-1$. For $n=1$, the only allowed value is $\ell =0$. Setting $\ell =1$ violates this constraint — there is no $p$ subshell in the first shell.

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Set B: $n=2,\ell =1,m_{\ell }=-1,m_s=+\frac{3}{2}$

Answer

Invalid. The spin quantum number $m_s$ can only take the values $+\frac{1}{2}$ or $-\frac{1}{2}$. The value $+\frac{3}{2}$ is not allowed — it would correspond to a higher spin than an electron possesses. The other three numbers ($n=2$, $\ell =1$, $m_{\ell }=-1$) are all valid on their own.

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Set C: $n=3,\ell =1,m_{\ell }=2$

Answer

Invalid. $m_{\ell }$ must satisfy $-\ell \le m_{\ell }\le +\ell$. For $\ell =1$, the allowed values are $m_{\ell }\in \{-1,0,+1\}$. The value $m_{\ell }=2$ lies outside this range.

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Set D: $n=0$

Answer

Invalid. $n$ must be a positive integer: $n\ge 1$. Zero is not a valid principal quantum number — there is no zeroth shell.

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Question 2 — Ground-State Configuration of Calcium

What is the ground-state electron configuration of Calcium (Ca, $Z=20$)?

Answer

$$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^2$$

Why $4s$ before $3d$? The Madelung (diagonal) rule fills subshells in order of increasing $n+\ell$:

• $4s$: $n+\ell =4+0=4$
• $3d$: $n+\ell =3+2=5$

Since $4<5$, the $4s$ subshell is lower in energy for light atoms and fills first. After $3p^6$ (at $Z=18$, argon), the next two electrons go into $4s$, completing the configuration above.

The shorthand notation using the previous noble gas is:

$$[\text{Ar}]4s^2$$

Calcium is the last element before the $3d$ transition metals begin filling. At $Z=21$ (Scandium), the first $3d$ electron appears.

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Question 3 — Meaning of the Wave Function

The wave function $\mathrm{\Psi }$ is complex-valued, yet every measurement of a quantum system returns a real number. What does $\mathrm{\Psi }$ represent, and how is a real, observable prediction extracted from it?

Answer

The wave function is not itself an observable quantity — it is a probability amplitude. The observable is its absolute square,

$$|\mathrm{\Psi }(\mathbf{r},t){|}^2={\mathrm{\Psi }}^{\ast }\mathrm{\Psi },$$

which is real and non-negative. $|\mathrm{\Psi }(\mathbf{r},t){|}^{2}dV$ gives the probability of finding the particle in the small volume $dV$ around $\mathbf{r}$ at time $t$.

Because the particle must be found somewhere, $\mathrm{\Psi }$ is normalized so that the total probability is 1:

$$\int |\mathrm{\Psi }{|}^{2}dV=1.$$

The complex phase of $\mathrm{\Psi }$ never shows up in a single measurement, but it is essential: it governs how quantum states interfere, which is what makes quantum mechanics differ from a classical probability theory.

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Question 4 — Particle in a Box: Energies

A particle is confined to a one-dimensional box of length $L$.

(a) What is the ratio $E_3/E_1$ of the third energy level to the ground state? (b) Why can the particle never have exactly zero energy?

Answer

(a) The allowed energies are

$$E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2},$$

so $E_n\propto n^2$. Therefore

$$\frac{E_3}{E_1}=\frac{3^2}{1^2}=9.$$

(b) The lowest allowed state is $n=1$, not $n=0$. Setting $n=0$ gives $\psi (x)=A\sin (0)=0$ everywhere — a particle that exists nowhere. The smallest genuine standing wave fits half a wavelength across the box and still carries

$$E_1=\frac{{\pi }^2{\hslash }^2}{2m{L}^2}>0.$$

This irreducible zero-point energy is a direct consequence of confinement: a perfectly localized, perfectly motionless particle is forbidden by the uncertainty principle.

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Question 5 — Why Spherical Coordinates?

Why is the Schrödinger equation for the hydrogen atom solved in spherical coordinates rather than Cartesian, and what two equations does separation of variables produce?

Answer

The Coulomb potential depends only on the distance from the nucleus:

$$V(r)=-\frac{e^2}{4\pi {ϵ}_{0}r}.$$

In Cartesian coordinates that distance is $r=\sqrt{x^2+y^2+z^2}$, which tangles all three variables together. Spherical coordinates match the natural spherical symmetry of the atom, so the potential involves $r$ alone.

Writing the wave function as a product $\mathrm{\Psi }(r,\theta ,\varphi )=R(r)Y(\theta ,\varphi )$ and separating splits one hard 3-D problem into two simpler ones:

• the radial equation (in $r$) — determines the energy levels and the principal quantum number $n$;
• the angular equation (in $\theta ,\varphi$) — determines the orbital shapes and the quantum numbers $\ell$ and $m_{\ell }$.

The two equations are linked by the separation constant $\ell (\ell +1)$.

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Question 6 — Orbitals in a Subshell

For the $d$ subshell:

(a) What are the allowed values of $m_{\ell }$? (b) How many orbitals does the subshell contain? (c) How many electrons can it hold in total?

Answer

A $d$ subshell has $\ell =2$.

(a) $m_{\ell }$ takes every integer from $-\ell$ to $+\ell$:

$$m_{\ell }\in \{-2,-1,0,+1,+2\}.$$

(b) Each value of $m_{\ell }$ labels one orbital, so there are $2\ell +1=5$ orbitals.

(c) By the Pauli exclusion principle each orbital holds 2 electrons (opposite spins), giving

$$2(2\ell +1)=10\text{ electrons}$$

— exactly the width of the $d$-block.

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Question 7 — Nodes of a 3p Orbital

How many radial nodes and how many angular nodes does a $3p$ orbital have? What is the total number of nodes?

Answer

A $3p$ orbital has $n=3$ and $\ell =1$.

• Angular nodes $=\ell =1$ — one nodal plane through the nucleus, which is what splits a $p$ orbital into two lobes.
• Radial nodes $=n-\ell -1=3-1-1=1$ — one spherical shell of zero probability.
• Total nodes $=n-1=2$.

The rule total nodes $=n-1$ holds for every orbital. The value of $\ell$ decides how many of those nodes are angular; the remainder, $n-\ell -1$, are radial.

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Question 8 — Hydrogen Energy and Subshell Splitting

(a) Using $E_n=-13.6\text{eV}/n^2$, find the energy of the $n=2$ level of hydrogen. (b) In hydrogen the $2s$ and $2p$ subshells have the same energy, but in a multi-electron atom $2s$ lies below $2p$. Why?

Answer

(a)

$$E_2=-\frac{13.6\text{eV}}{2^2}=-\frac{13.6\text{eV}}{4}=-3.40\text{eV}.$$

(b) In hydrogen — a single electron — the energy depends only on $n$, so $2s$ and $2p$ are degenerate. In a multi-electron atom this degeneracy is broken by penetration and shielding. The $2s$ radial distribution has a small inner peak that lets the electron penetrate the $1s$ core and feel a larger effective nuclear charge $Z_{\text{eff}}$. The $2p$ orbital has a node at $r=0$ and is held further out, so it is shielded more effectively. Penetrating deeper, the $2s$ electron is more tightly bound — lower in energy — hence $E_{2s}<E_{2p}$.

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Question 9 — Hund's Rule and Nitrogen

State the Pauli exclusion principle and Hund's rule. Then apply them to nitrogen ($Z=7$): give its ground-state $2p$ occupation, the number of unpaired electrons, and whether it is paramagnetic or diamagnetic.

Answer

• Pauli exclusion principle: no two electrons in an atom may share all four quantum numbers $(n,\ell ,m_{\ell },m_s)$. Hence an orbital holds at most 2 electrons, with opposite spins.
• Hund's rule: within a subshell, electrons occupy separate orbitals with parallel spins before any orbital is doubly filled — this minimizes electron–electron repulsion.

Nitrogen's configuration is $1s^{2}2s^{2}2p^3$. The three $2p$ electrons each take a different $2p$ orbital with parallel spins:

$$2p:\uparrow \uparrow \uparrow$$

That leaves 3 unpaired electrons, so nitrogen is paramagnetic. With its half-filled $2p$ subshell, it is in fact the most paramagnetic atom of period 2.

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Question 10 — Why Periods 6 and 7 Hold 32 Elements

Periods 6 and 7 of the periodic table each contain 32 elements. Derive this number from the quantum-mechanical subshell capacities.

Answer

A period corresponds to filling one complete set of subshells in Aufbau order. Each subshell holds $2(2\ell +1)$ electrons:

$$s=2,p=6,d=10,f=14.$$

Periods 6 and 7 are the first to draw on all four block types — they fill an $s$, an $f$, a $d$ and a $p$ subshell:

$$2+14+10+6=32.$$

For period 6 the sequence is $6s\to 4f\to 5d\to 6p$; for period 7 it is $7s\to 5f\to 6d\to 7p$. The 14 contributed by the $f$ subshell is exactly the length of the lanthanide and actinide rows.

(Periods 4–5 omit the $f$ block and hold $2+10+6=18$; periods 2–3 omit both $f$ and $d$ and hold $2+6=8$.)