Chapter 2 · Structure of Atom

2.3 The Quantum Mechanical Model: Quantum Numbers and Orbitals

Electrons as waves: the de Broglie relation, Heisenberg's uncertainty principle, the orbital as a cloud of probability, the four quantum numbers, the shapes of s, p and d orbitals, their nodes and their energies.

An electron microscope can photograph a single virus, and the latest instruments can map the position of every atom in a protein. They do it with a beam of electrons — not light — because electrons behave as waves whose wavelength is thousands of times shorter than that of visible light. That electrons are waves at all was a bold guess in 1924, confirmed experimentally in 1927, and it brought down Bohr's picture of an electron circling the nucleus on a neat track.

This section builds the model of the atom used in all of modern chemistry. Two ideas — the wave nature of matter and the uncertainty principle — show why a definite orbit is impossible. Schrödinger's wave mechanics replaces the orbit with the orbital, a region where the electron is likely to be found, and each orbital is labelled by a set of quantum numbers. By the end you will be able to name every orbital, picture its shape, count its nodes and say how its energy compares with others — the tools Section 2.4 uses to write electronic configurations.

The Dual Behaviour of Matter

Section 2.2 showed that radiation behaves both as a wave and as a stream of particles. In 1924 the French physicist Louis de Broglie turned the argument round: if waves can behave like particles, perhaps particles such as electrons can behave like waves. For a photon, $E=h\nu=hc/\lambda$ and its momentum is $p=E/c=h/\lambda$. De Broglie proposed that the same relation between momentum and wavelength holds for every moving particle.

📐 Law 2.8 — The de Broglie relation
A particle of mass $m$ moving with speed $v$ (momentum $p=mv$) has an associated wave of wavelength $$\lambda=\frac{h}{mv}=\frac{h}{p}.$$ For a particle with kinetic energy $K$, $p=\sqrt{2mK}$, so $\lambda=\dfrac{h}{\sqrt{2mK}}$.

De Broglie's prediction was confirmed in 1927, when C. J. Davisson and L. H. Germer in the USA and G. P. Thomson in Scotland observed that a beam of electrons is diffracted by a crystal, exactly as X-rays are — a property only waves can have. The electron microscope, which uses electron waves to form highly magnified images, is a direct application. Because $h$ is so small, the wavelength is noticeable only for particles of tiny mass: for everyday objects it is far too small to detect.

✏️ Example 1 — Electrons and cricket balls
Find the de Broglie wavelength of (a) an electron moving at $2.18\times10^{6}\ \mathrm{m\,s^{-1}}$ (its speed in the first Bohr orbit of hydrogen), and (b) a cricket ball of mass $0.16\ \mathrm{kg}$ bowled at $40\ \mathrm{m\,s^{-1}}$. Comment on each answer. ($h=6.626\times10^{-34}\ \mathrm{J\,s}$, $m_{e}=9.11\times10^{-31}\ \mathrm{kg}$; $1\ \mathrm{J}=1\ \mathrm{kg\,m^{2}\,s^{-2}}$.)
(a) $\lambda=\dfrac{h}{m_{e}v}=\dfrac{6.626\times10^{-34}\ \mathrm{kg\,m^{2}\,s^{-1}}}{(9.11\times10^{-31}\ \mathrm{kg})(2.18\times10^{6}\ \mathrm{m\,s^{-1}})}=3.34\times10^{-10}\ \mathrm{m}=334\ \mathrm{pm}$.
Comment The circumference of the first Bohr orbit is $2\pi\times52.9\ \mathrm{pm}=332\ \mathrm{pm}$: almost exactly one wavelength fits round it. This is no coincidence — see Interactive 1. An electron's wavelength is comparable to the size of an atom, so its wave nature dominates its behaviour inside atoms.
(b) $\lambda=\dfrac{6.626\times10^{-34}}{0.16\times40}=1.0\times10^{-34}\ \mathrm{m}$.
Comment This is $10^{19}$ times smaller than a nucleus — impossible to detect. The ball's wave nature can be ignored completely; it behaves as a classical particle.

De Broglie's idea also explains Bohr's mysterious third postulate. If the electron is a wave travelling round a circular orbit, the wave must join smoothly onto itself after one lap; otherwise successive laps interfere destructively and the wave dies away. So a whole number of wavelengths must fit round the circumference: $2\pi r=n\lambda=n\dfrac{h}{m_{e}v}$, which rearranges to $m_{e}vr=n\dfrac{h}{2\pi}$ — exactly Bohr's quantisation condition.

Interactive 1 — Matter Waves on a Bohr Orbit

The slider sets the size of the orbit, measured by how many de Broglie wavelengths fit round it. For a whole number the wave closes on itself (green) — an allowed orbit with $m_{e}vr=nh/2\pi$. For any other size the end of the wave does not meet its start (red mark), and the wave would cancel itself on the next lap. The readouts use Bohr's radius $r=52.9\,n^{2}\ \mathrm{pm}$ with the speed from $m_{e}vr=nh/2\pi$ (about $2.19\times10^{6}/n\ \mathrm{m\,s^{-1}}$), where $n$ is the slider value.

2.50
orbit radius $r$ (pm)—
speed $v$ (m/s)—
$\lambda=h/m_{e}v$ (pm)—
$2\pi r/\lambda$—
allowed?—

Heisenberg's Uncertainty Principle

To locate an electron we must "see" it — bounce light off it. But light short enough in wavelength to locate something as small as an electron has photons so energetic that the collision changes the electron's momentum unpredictably. Light of longer wavelength disturbs it less but locates it less precisely. In 1927 Werner Heisenberg showed that this is not a fault of our instruments but a basic law of nature, a consequence of the wave nature of matter.

📐 Principle 2.9 — Heisenberg's uncertainty principle
It is impossible to determine simultaneously, with unlimited precision, both the exact position and the exact momentum (or velocity) of an electron or any other microscopic particle. If $\Delta x$ is the uncertainty in position and $\Delta p_{x}$ the uncertainty in momentum along the same direction, then $$\Delta x\,\Delta p_{x}\ge\frac{h}{4\pi},\qquad\text{equivalently}\qquad\Delta x\,\Delta v_{x}\ge\frac{h}{4\pi m}.$$ The more precisely one quantity is known, the less precisely the other can be.
✏️ Example 2 — How uncertain?
(a) An electron's position is known to within $\Delta x=1.0\times10^{-10}\ \mathrm{m}$, about the size of an atom. Find the minimum uncertainty in its velocity, and compare it with the electron's speed in the first Bohr orbit, $2.18\times10^{6}\ \mathrm{m\,s^{-1}}$. (b) Repeat for a speck of dust of mass $1.0\times10^{-9}\ \mathrm{kg}$ located to within $1.0\ \mathrm{\mu m}$. ($h=6.626\times10^{-34}\ \mathrm{J\,s}$, $m_{e}=9.11\times10^{-31}\ \mathrm{kg}$.)
(a) $\Delta v\ge\dfrac{h}{4\pi m_{e}\Delta x}=\dfrac{6.626\times10^{-34}}{4\pi(9.11\times10^{-31})(1.0\times10^{-10})}=5.8\times10^{5}\ \mathrm{m\,s^{-1}}$.
Compare That is about a quarter of the speed itself ($5.8\times10^{5}/2.18\times10^{6}\approx0.27$). If we pin the electron down to atomic size, its velocity — and so its future path — is hugely uncertain. A precisely defined orbit is meaningless.
(b) $\Delta v\ge\dfrac{6.626\times10^{-34}}{4\pi(1.0\times10^{-9})(1.0\times10^{-6})}=5.3\times10^{-20}\ \mathrm{m\,s^{-1}}$ — utterly negligible. For everyday objects the uncertainty principle has no practical effect.
⚠️ Why the uncertainty principle rules out Bohr orbits

An orbit is a precise path: at every instant the electron would have an exact position and an exact velocity. The uncertainty principle forbids exactly that for an object as light as an electron. So we must give up the idea of trajectories inside atoms and speak instead of the probability of finding the electron in a region of space. Bohr's model, which ignored both the wave nature of the electron and the uncertainty principle, had to be replaced — even though its energies for hydrogen are correct.

Students sometimes say the principle is "only about measurement error". It is not: an electron does not have a sharply defined position and momentum at the same time, whatever instrument is used.

The Quantum Mechanical Model: Orbitals and Quantum Numbers

In 1926 Erwin Schrödinger wrote down the equation that governs the waves of electrons, as Newton's laws govern the motion of cricket balls. For an electron in an atom it has the form $\hat{H}\psi=E\psi$, where $\hat{H}$ is a mathematical operator (the Hamiltonian) describing the electron's kinetic and potential energy, $\psi$ (psi) is the wave function and $E$ the energy. You will not solve it in this course; what matters are its results.

📘 Definition 2.9 — Atomic orbital

An atomic orbital is a one-electron wave function $\psi$ for an electron in an atom. In practice we picture it as the region of space around the nucleus where the probability of finding that electron is high — a boundary-surface diagram encloses, say, $90\%$ of the probability. An orbital can hold at most two electrons.

Orbit (Bohr)Orbital (quantum mechanics)
a definite circular path round the nucleusa three-dimensional region where the electron is likely to be found
electron's position and velocity both exact — contradicts the uncertainty principlegives only the probability of finding the electron — consistent with it
all orbits are circular and flatorbitals have different shapes (spherical s, dumbbell p, cloverleaf d) and orientations
an orbit can hold up to $2n^{2}$ electronsan orbital holds at most $2$ electrons
📘 Definition 2.10 — The four quantum numbers
Principal quantum number $n$$n=1,2,3,\ldots$ Fixes the shell (K, L, M, N… for $n=1,2,3,4$), the size of the orbital and, in hydrogen, its energy. Larger $n$: larger, higher-energy orbital.
Azimuthal quantum number $l$(orbital angular momentum quantum number) $l=0,1,2,\ldots,n-1$. Fixes the subshell and the shape of the orbital: $l=0,1,2,3$ are called s, p, d, f.
Magnetic quantum number $m_{l}$$m_{l}=-l,\ldots,-1,0,+1,\ldots,+l$ — that is, $2l+1$ values. Fixes the orientation of the orbital in space.
Spin quantum number $m_{s}$$m_{s}=+\tfrac12$ or $-\tfrac12$ (shown $\uparrow$ and $\downarrow$). Describes the electron's spin, its intrinsic angular momentum, proposed by Uhlenbeck and Goudsmit in 1925. It belongs to the electron, not to the orbital.

An orbital is named by $n$ followed by the letter for $l$: $n=3$, $l=2$ is a $3d$ orbital. The three quantum numbers $n,l,m_{l}$ identify one orbital; adding $m_{s}$ identifies one electron.

📐 Rule 2.10 — Counting orbitals and electrons
number of orbitalsmaximum electrons
subshell with quantum number $l$$2l+1$ (s: 1, p: 3, d: 5, f: 7)$2(2l+1)$ (s: 2, p: 6, d: 10, f: 14)
shell with quantum number $n$$n^{2}$ (it has $n$ subshells)$2n^{2}$
Orbitals of the same subshell (for example the three $2p$ orbitals) have the same energy; orbitals with the same energy are called degenerate.
✏️ Example 3 — Which sets of quantum numbers are allowed?
Say whether each set $(n,l,m_{l},m_{s})$ is allowed. If it is, name the orbital; if not, say which rule it breaks. (a) $(2,2,0,+\tfrac12)$ (b) $(3,1,-1,+\tfrac12)$ (c) $(1,0,+1,-\tfrac12)$ (d) $(4,3,-3,-\tfrac12)$ (e) $(3,2,0,0)$
(a) Not allowed: $l$ can be at most $n-1=1$.
(b) Allowed: $l=1\le2$, $m_{l}=-1$ lies in $-1\ldots+1$. A $3p$ electron.
(c) Not allowed: with $l=0$ the only value of $m_{l}$ is $0$.
(d) Allowed: $l=3\le3$, $m_{l}=-3$ lies in $-3\ldots+3$. A $4f$ electron.
(e) Not allowed: $m_{s}$ must be $+\tfrac12$ or $-\tfrac12$, never $0$. (The first three numbers describe a valid $3d$ orbital.)
✏️ Example 4 — Counting with quantum numbers
(a) How many orbitals are there in the $n=3$ shell, and how many electrons can it hold? List its subshells. (b) Name the subshell with $n=4$, $l=2$; how many orbitals and electrons does it hold? (c) How many electrons in an atom can have $n=3$ and $m_{l}=-1$? (d) How many electrons can have $n=4$ and $m_{s}=-\tfrac12$?
(a) $l=0,1,2$: subshells $3s$, $3p$, $3d$ with $1+3+5=9=n^{2}$ orbitals and $2n^{2}=18$ electrons.
(b) $4d$: $2l+1=5$ orbitals, $10$ electrons.
(c) $m_{l}=-1$ needs $l\ge1$, so $l=1$ ($3p$) or $l=2$ ($3d$): two orbitals, each holding $2$ electrons: 4 electrons.
(d) The $n=4$ shell has $n^{2}=16$ orbitals; one electron in each can have $m_{s}=-\tfrac12$: 16 electrons.

Shapes of Orbitals, Nodes and Orbital Energies

The probability of finding an electron can be shown in two ways. A charge-cloud picture shades each point according to $\psi^{2}$ — dense where the electron is likely, faint where it is not. A boundary-surface diagram draws a surface of constant $\psi^{2}$ that encloses about $90\%$ of the probability; it gives the familiar "shape" of the orbital.

A node is a region where the probability density $\psi^{2}$ is zero. A radial node is a sphere at a fixed distance from the nucleus (the $2s$ orbital has one: an inner sphere of probability, a zero, then an outer shell); an angular node is a plane or a cone through the nucleus (the nodal plane of a p orbital, the two cones of $d_{z^{2}}$).

📐 Rule 2.11 — Counting nodes
For an orbital with quantum numbers $n$ and $l$: $$\text{radial nodes}=n-l-1,\qquad\text{angular nodes}=l,\qquad\text{total nodes}=n-1.$$ So $1s$, $2p$, $3d$ and $4f$ (the first orbital of each type) have no radial node; $2s$ has one radial node, $3s$ two, $3p$ one; every p orbital has one angular node and every d orbital two.
Interactive 2 — Orbital Explorer: Shapes, Signs and Nodes

Left: a slice through a hydrogen orbital, shaded by the probability density $\psi^{2}$ (red where $\psi\gt0$, blue where $\psi\lt0$); dashed circles are radial nodes and dashed lines angular nodes. Right: the radial probability $r^{2}R^{2}$ — the chance of finding the electron at distance $r$ from the nucleus — with its radial nodes marked. Compare $1s$, $2s$ and $3s$, then $2p$ and $3p$, then the d orbitals.

$n,\ l$—
radial nodes $n-l-1$—
angular nodes $l$—
most probable $r$ (pm)—
slice shown—
✏️ Example 5 — Nodes
(a) Find the number of radial, angular and total nodes in $3s$, $3p$, $4d$ and $5f$ orbitals. (b) An orbital has two radial nodes and one angular node. Identify it. (c) Which orbital of the $n=4$ shell has no radial node?
(a) $3s$: radial $3-0-1=2$, angular $0$, total $2$. $3p$: $1$, $1$, total $2$. $4d$: $4-2-1=1$, $2$, total $3$. $5f$: $5-3-1=1$, $3$, total $4$.
(b) Angular nodes $=l=1$ (a p orbital); radial nodes $n-1-1=2\Rightarrow n=4$: a $4p$ orbital.
(c) $n-l-1=0\Rightarrow l=3$: the $4f$ orbitals.

Energies of orbitals

📐 Rule 2.12 — Energies of orbitals

Hydrogen (one electron). The energy depends only on $n$: $1s\lt2s=2p\lt3s=3p=3d\lt4s=4p=4d=4f\lt\ldots$ All orbitals of one shell are degenerate.

Atoms with more than one electron. The energy depends on both $n$ and $l$. Within a shell, $ns\lt np\lt nd\lt nf$. The subshells of neighbouring shells can overlap: for example $4s$ lies below $3d$ in potassium and calcium. For a given orbital, the energy also falls as the atomic number (nuclear charge) increases.

Why does $l$ matter once there are several electrons? Each electron is attracted by the nucleus but repelled by the other electrons. The inner electrons partly shield (screen) an outer electron from the nucleus, so it feels a smaller effective nuclear charge, $Z_{\text{eff}}$, than the full $+Ze$. An s electron has a significant probability of being found very close to the nucleus — it penetrates the inner shells — so it is shielded less, feels a larger $Z_{\text{eff}}$, and is held more tightly (lower energy) than a p electron of the same shell; a p electron penetrates more than a d electron, and so on. Section 2.4 turns this ordering into the $(n+l)$ rule for filling orbitals.

✏️ Example 6 — Comparing orbital energies
Arrange in order of increasing energy: (a) $3d$, $4s$, $3p$ in a hydrogen atom; (b) the same three in a calcium atom; (c) say whether $2s$ and $2p$ are degenerate in $\mathrm{H}$, in $\mathrm{He^{+}}$ and in $\mathrm{Li}$, and why.
(a) In hydrogen only $n$ matters: $3p=3d\lt4s$.
(b) In a many-electron atom $l$ matters too, and $4s$ penetrates strongly: $3p\lt4s\lt3d$.
(c) $\mathrm{H}$ and $\mathrm{He^{+}}$ have one electron each, so there is no shielding and $2s$, $2p$ are degenerate. In $\mathrm{Li}$ ($1s^{2}$ plus one more electron) the $1s$ electrons shield the outer electron; $2s$ penetrates the $1s$ cloud more than $2p$, so $2s\lt2p$.

Exercises

I
Matter waves and the uncertainty principle. Use $h=6.626\times10^{-34}\ \mathrm{J\,s}$, $m_{e}=9.11\times10^{-31}\ \mathrm{kg}$, $m_{p}=1.673\times10^{-27}\ \mathrm{kg}$, $1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$.
  • 1 Find the de Broglie wavelength of an electron moving at $1.0\times10^{6}\ \mathrm{m\,s^{-1}}$.
  • 2 Find the de Broglie wavelength of a proton moving at the same speed.
  • 3 Find the de Broglie wavelength of a $60\ \mathrm{kg}$ sprinter running at $10\ \mathrm{m\,s^{-1}}$.
  • 4 At what speed must an electron travel to have a wavelength of $0.10\ \mathrm{nm}$?
  • 5 An electron has a kinetic energy of $4.5\times10^{-25}\ \mathrm{J}$. Find its de Broglie wavelength.
  • 6 Electrons are accelerated from rest through a potential difference of $100\ \mathrm{V}$, gaining $100\ \mathrm{eV}$ of kinetic energy. Find their wavelength and explain why such electrons are diffracted by crystals.
  • 7 The position of an electron is known to within $1.0\ \mathrm{pm}$. Find the minimum uncertainty in its velocity.
  • 8 The speed of a $25\ \mathrm{g}$ ball is known to within $1.0\times10^{-5}\ \mathrm{m\,s^{-1}}$. Find the minimum uncertainty in its position.
  • 9 Explain why the uncertainty principle is essential for describing electrons in atoms but irrelevant for describing the motion of a car.
Check: de Broglie wavelength calculator
momentum (kg m/s)—
wavelength (m)—
compared with an atom (100 pm)—
1 $\lambda=\dfrac{6.626\times10^{-34}}{9.11\times10^{-31}\times1.0\times10^{6}}=7.3\times10^{-10}\ \mathrm{m}=0.73\ \mathrm{nm}$.
2 $\lambda=\dfrac{6.626\times10^{-34}}{1.673\times10^{-27}\times1.0\times10^{6}}=4.0\times10^{-13}\ \mathrm{m}$ — $1836$ times shorter than the electron's.
3 $\lambda=\dfrac{6.626\times10^{-34}}{60\times10}=1.1\times10^{-36}\ \mathrm{m}$ — undetectable.
4 $v=\dfrac{h}{m_{e}\lambda}=\dfrac{6.626\times10^{-34}}{9.11\times10^{-31}\times1.0\times10^{-10}}=7.3\times10^{6}\ \mathrm{m\,s^{-1}}$.
5 $\lambda=\dfrac{h}{\sqrt{2m_{e}K}}=\dfrac{6.626\times10^{-34}}{\sqrt{2\times9.11\times10^{-31}\times4.5\times10^{-25}}}=\dfrac{6.626\times10^{-34}}{9.06\times10^{-28}}=7.3\times10^{-7}\ \mathrm{m}$ ($732\ \mathrm{nm}$ unrounded) — a very slow electron.
6 $K=100\times1.602\times10^{-19}=1.602\times10^{-17}\ \mathrm{J}$; $\lambda=\dfrac{6.626\times10^{-34}}{\sqrt{2\times9.11\times10^{-31}\times1.602\times10^{-17}}}=1.23\times10^{-10}\ \mathrm{m}=123\ \mathrm{pm}$. This is similar to the spacing of atoms in a crystal ($100$–$300\ \mathrm{pm}$), and a wave is diffracted strongly by a grating whose spacing is comparable to its wavelength.
7 $\Delta v\ge\dfrac{6.626\times10^{-34}}{4\pi\times9.11\times10^{-31}\times1.0\times10^{-12}}=5.8\times10^{7}\ \mathrm{m\,s^{-1}}$ — about a fifth of the speed of light.
8 $\Delta x\ge\dfrac{6.626\times10^{-34}}{4\pi\times0.025\times1.0\times10^{-5}}=2.1\times10^{-28}\ \mathrm{m}$.
9 The minimum uncertainties are inversely proportional to mass. For an electron ($10^{-30}\ \mathrm{kg}$) confined to an atom, $\Delta v$ is comparable to its speed, so its path cannot be defined; for a car ($10^{3}\ \mathrm{kg}$) the uncertainties are some $10^{-37}$ of everyday values and far below anything measurable.
II
Orbitals and quantum numbers.
  • 1 Give two differences between an orbit and an orbital.
  • 2 List the possible values of $l$ for $n=3$, and of $m_{l}$ for each.
  • 3 Name the subshell with (a) $n=2,l=1$; (b) $n=4,l=0$; (c) $n=5,l=3$; (d) $n=3,l=2$.
  • 4 Which of these designations are impossible: $1p$, $2s$, $2d$, $3f$, $4f$? Explain.
  • 5 Give $n$, $l$ and all the possible values of $m_{l}$ for a $4f$ orbital.
  • 6 What is the maximum number of electrons that can have (a) $n=3$, $l=1$; (b) $n=3$; (c) $n=2$, $l=1$, $m_{l}=0$?
  • 7 Which of these sets $(n,l,m_{l},m_{s})$ are allowed? (a) $(3,0,0,+\tfrac12)$ (b) $(2,1,2,-\tfrac12)$ (c) $(4,2,-2,+\tfrac12)$ (d) $(0,0,0,+\tfrac12)$. Name the orbital for each allowed set.
  • 8 How many orbitals are there in the $n=5$ shell? Name its subshells.
  • 9 Which quantum number fixes (a) the size, (b) the shape, (c) the orientation of an orbital?
  • 10 How many electrons in an atom can have $n+l=5$?
1 An orbit is a definite circular path; an orbital is a three-dimensional region of high probability. An orbit implies exact position and velocity (forbidden by the uncertainty principle); an orbital gives only probabilities. (Also: an orbital holds at most two electrons and has a characteristic shape.)
2 $l=0$: $m_{l}=0$. $l=1$: $m_{l}=-1,0,+1$. $l=2$: $m_{l}=-2,-1,0,+1,+2$.
3 (a) $2p$ (b) $4s$ (c) $5f$ (d) $3d$.
4 $1p$ (needs $l=1\gt n-1=0$), $2d$ ($l=2\gt1$) and $3f$ ($l=3\gt2$) are impossible; $2s$ and $4f$ exist.
5 $n=4$, $l=3$, $m_{l}=-3,-2,-1,0,+1,+2,+3$ (seven $4f$ orbitals).
6 (a) $3p$: $6$. (b) $2n^{2}=18$. (c) one orbital ($2p_{z}$): $2$.
7 (a) allowed, $3s$. (b) not allowed ($|m_{l}|\le l=1$). (c) allowed, $4d$. (d) not allowed ($n\ge1$).
8 $n^{2}=25$ orbitals; subshells $5s,5p,5d,5f,5g$ ($l=0$ to $4$).
9 (a) $n$ (b) $l$ (c) $m_{l}$.
10 $n+l=5$: $5s$ $(5,0)$, $4p$ $(4,1)$, $3d$ $(3,2)$ — not $2f$, which does not exist. Electrons: $2+6+10=18$.
III
Shapes, nodes and energies.
  • 1 Describe the boundary-surface diagrams of $2s$, $2p_{z}$ and $3d_{xy}$ orbitals.
  • 2 Find the numbers of radial and angular nodes in $2s$, $4p$, $3d$ and $6s$ orbitals.
  • 3 Identify the orbital that has one radial node and two angular nodes.
  • 4 Which of $3p$, $3d$, $4s$, $4f$ have no radial nodes?
  • 5 Name the nodal plane of a $p_{z}$ orbital and the two nodal planes of a $d_{xy}$ orbital.
  • 6 Arrange $4s$, $3d$ and $4p$ in order of increasing energy (a) in a hydrogen atom, (b) in a potassium atom.
  • 7 Explain why $2s$ has lower energy than $2p$ in lithium, though the two have the same energy in hydrogen.
  • 8 Which of $3p_{x}$, $3p_{y}$, $3d_{xy}$, $3p_{z}$ are degenerate in a many-electron atom?
  • 9 Use Interactive 2 to find the most probable distance of a $1s$ electron from the nucleus in hydrogen. Which quantity from Bohr's model does it match?
1 $2s$: a sphere (larger than $1s$), with an inner spherical node. $2p_{z}$: two lobes along the $z$ axis, opposite in sign, separated by the $xy$ nodal plane. $3d_{xy}$: four lobes in the $xy$-plane pointing between the $x$ and $y$ axes, alternating in sign.
2 $2s$: $1$ radial, $0$ angular. $4p$: $4-1-1=2$ radial, $1$ angular. $3d$: $0$ radial, $2$ angular. $6s$: $5$ radial, $0$ angular.
3 $l=2$ and $n-2-1=1\Rightarrow n=4$: a $4d$ orbital.
4 $3d$ ($3-2-1=0$) and $4f$ ($4-3-1=0$).
5 $p_{z}$: the $xy$-plane. $d_{xy}$: the $xz$- and $yz$-planes.
6 (a) Hydrogen: $3d\lt4s=4p$ (energy depends only on $n$). (b) Potassium: $4s\lt3d\lt4p$.
7 In lithium the two $1s$ electrons shield the outer electron. A $2s$ electron penetrates closer to the nucleus than a $2p$ electron, so it is shielded less, feels a larger effective nuclear charge and has lower energy. Hydrogen has no other electrons, so there is no shielding and the energy depends only on $n$.
8 $3p_{x}$, $3p_{y}$ and $3p_{z}$ (same $n$ and $l$); $3d_{xy}$ has higher energy.
9 The radial probability $r^{2}R^{2}$ for $1s$ peaks at $r=a_{0}=52.9\ \mathrm{pm}$ — the radius of Bohr's first orbit. In quantum mechanics this is the most probable distance, not a fixed one.
IV
Harder problems.
  • 1 Using $r_{n}=52.9n^{2}\ \mathrm{pm}$ and $v_{n}=2.18\times10^{6}/n\ \mathrm{m\,s^{-1}}$, find the de Broglie wavelength of the electron in the third Bohr orbit of hydrogen, and show that about three wavelengths fit round the orbit.
  • 2 An α-particle and a proton move at the same speed. Find the ratio of their de Broglie wavelengths, $\lambda_{\alpha}/\lambda_{p}$ (take $m_{\alpha}\approx4m_{p}$).
  • 3 An electron and a proton have the same de Broglie wavelength. Which has the greater kinetic energy, and by what factor?
  • 4 The speed of an electron is measured as $450\ \mathrm{m\,s^{-1}}$ with an uncertainty of $0.010\%$. Find the minimum uncertainty in its position.
1 $v_{3}=7.27\times10^{5}\ \mathrm{m\,s^{-1}}$; $\lambda=\dfrac{6.626\times10^{-34}}{9.11\times10^{-31}\times7.27\times10^{5}}=1.00\times10^{-9}\ \mathrm{m}=1000\ \mathrm{pm}$. Circumference $=2\pi\times52.9\times9=2991\ \mathrm{pm}\approx3\lambda$ ✓ (the small difference comes from rounding the constants).
2 Same $v$: $\lambda\propto\dfrac1m$, so $\dfrac{\lambda_{\alpha}}{\lambda_{p}}=\dfrac{m_{p}}{m_{\alpha}}\approx\dfrac14$.
3 Same $\lambda$ means the same momentum $p$. $K=\dfrac{p^{2}}{2m}$, so the lighter electron has the greater kinetic energy, by a factor $\dfrac{m_{p}}{m_{e}}=1836$.
4 $\Delta v=450\times\dfrac{0.010}{100}=0.045\ \mathrm{m\,s^{-1}}$; $\Delta x\ge\dfrac{6.626\times10^{-34}}{4\pi\times9.11\times10^{-31}\times0.045}=1.3\times10^{-3}\ \mathrm{m}$ — about a millimetre, enormous on the atomic scale.
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E1. NEET Which set of quantum numbers $(n,l,m_{l},m_{s})$ is not possible? (A) $(3,2,-2,+\tfrac12)$ (B) $(2,1,0,-\tfrac12)$ (C) $(4,4,0,+\tfrac12)$ (D) $(5,3,+3,-\tfrac12)$

Rules $0\le l\le n-1$; $-l\le m_{l}\le+l$; $m_{s}=\pm\tfrac12$.
Answer (C): $l=4$ is not allowed when $n=4$ (maximum $l=3$). (A) is a $3d$, (B) a $2p$ and (D) a $5f$ electron.

E2. JEE Main Numerical answer. What is the maximum number of electrons in an atom that can have $n+l=4$?

List $(n,l)$ with $n+l=4$ and $l\le n-1$: $(4,0)=4s$, $(3,1)=3p$. $(2,2)$ is not allowed.
Answer $2+6=$ 8.

E3. Board · 3 marks Give three differences between an orbit and an orbital.

1 mark An orbit is a well-defined circular path of the electron round the nucleus; an orbital is a three-dimensional region in which the probability of finding the electron is maximum.
1 mark An orbit implies that the position and velocity of the electron are known exactly, contradicting the uncertainty principle; an orbital describes only probabilities and is consistent with it.
1 mark Orbits are all circular and planar; orbitals have different shapes (s spherical, p dumbbell, d cloverleaf) and orientations. (Also accepted: an orbit can hold $2n^{2}$ electrons, an orbital at most $2$.)

E4. JEE Main NEET An electron and a proton have the same kinetic energy. The ratio of their de Broglie wavelengths, $\lambda_{e}/\lambda_{p}$, is about ($m_{p}/m_{e}=1836$) (A) $1$ (B) $1/1836$ (C) $43$ (D) $1836$

Relation $\lambda=\dfrac{h}{\sqrt{2mK}}$; with equal $K$, $\lambda\propto\dfrac{1}{\sqrt{m}}$.
Ratio $\dfrac{\lambda_{e}}{\lambda_{p}}=\sqrt{\dfrac{m_{p}}{m_{e}}}=\sqrt{1836}=42.8$.
Answer (C). Option (D) would be right for equal speeds, not equal kinetic energies.

E5. JEE Advanced One or more options correct. (A) The angular nodes of a $d_{z^{2}}$ orbital are two cones. (B) A $4f$ orbital has one radial node. (C) In the hydrogen atom the $2s$ and $2p$ orbitals have the same energy. (D) The probability density of a $2p_{x}$ electron is zero everywhere in the $yz$-plane.

(A) True: $d_{z^{2}}$ has $l=2$ angular nodes, which are two cones (at about $54.7^{\circ}$ to the $z$ axis), not planes.
(B) False: radial nodes $=4-3-1=0$.
(C) True: in a one-electron atom the energy depends only on $n$.
(D) True: the $yz$-plane ($x=0$) is the nodal plane of $p_{x}$.
Answer (A), (C), (D).

E6. Board · 2 marks State Heisenberg's uncertainty principle. Calculate the minimum uncertainty in the position of an electron whose velocity is uncertain by $5.7\times10^{5}\ \mathrm{m\,s^{-1}}$ ($h=6.626\times10^{-34}\ \mathrm{J\,s}$, $m_{e}=9.11\times10^{-31}\ \mathrm{kg}$).

1 mark It is impossible to determine simultaneously the exact position and exact momentum of a microscopic particle: $\Delta x\,\Delta p\ge\dfrac{h}{4\pi}$.
1 mark $\Delta x\ge\dfrac{h}{4\pi m_{e}\Delta v}=\dfrac{6.626\times10^{-34}}{4\times3.1416\times9.11\times10^{-31}\times5.7\times10^{5}}=1.0\times10^{-10}\ \mathrm{m}$ — about the size of an atom.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How can microscopes that use electron waves show individual atoms in a protein, when the best light microscopes cannot see anything smaller than about $200\ \mathrm{nm}$?

Why it matters. In 2020 laboratories in the USA, the UK, China and elsewhere used cryo-electron microscopy (cryo-EM, Nobel Prize in Chemistry 2017) to map the "spike" protein of the virus that causes COVID-19 within weeks of its discovery, guiding vaccine design. The same year, cryo-EM reached a resolution of about $0.12\ \mathrm{nm}$, enough to see individual atoms. The secret is the de Broglie wavelength.

What to do.

  1. As a rough rule, a microscope cannot resolve details much smaller than about half the wavelength it uses. Estimate the limit for green light, $\lambda=500\ \mathrm{nm}$.
  2. Electrons accelerated through a voltage $V$ gain kinetic energy $eV$. Using $\lambda=\dfrac{h}{\sqrt{2m_{e}eV}}$, find their wavelength for $V=100\ \mathrm{V}$ and $V=10\ \mathrm{kV}$.
  3. Microscopes for proteins use $V=200$–$300\ \mathrm{kV}$. Find the speed predicted by $\tfrac12m_{e}v^{2}=eV$ for $200\ \mathrm{kV}$ and compare it with the speed of light. Why must the simple formula be corrected (by Einstein's relativity) at these energies? The corrected wavelength is $2.51\ \mathrm{pm}$; compare it with your uncorrected value.
  4. If the electron wavelength is a few picometres, why is the best resolution "only" about $0.1\ \mathrm{nm}$? Find out about two practical limits (lens imperfections, radiation damage to the sample).

Tools: Calculator; the de Broglie check-lab above; a reliable source such as the Nobel Prize website's popular-science summary for 2017.

Hand in: A table of wavelengths and resolution limits, and a paragraph explaining to a younger student why "electrons are waves" lets us see atoms.

Going further. Neutrons are also used for diffraction (for example to locate hydrogen atoms in crystals, which X-rays barely see). What speed must a neutron have for a wavelength of $0.15\ \mathrm{nm}$?

Instructor note: light limit $\approx250\ \mathrm{nm}$. $\lambda(100\ \mathrm{V})=1.23\times10^{-10}\ \mathrm{m}$; $\lambda(10\ \mathrm{kV})=1.23\times10^{-11}\ \mathrm{m}$; uncorrected $\lambda(200\ \mathrm{kV})=2.74\ \mathrm{pm}$ with $v=2.65\times10^{8}\ \mathrm{m\,s^{-1}}$ (about $0.88c$, so relativity matters). Neutron: $v=\dfrac{h}{m_{n}\lambda}=\dfrac{6.626\times10^{-34}}{1.675\times10^{-27}\times1.5\times10^{-10}}=2.6\times10^{3}\ \mathrm{m\,s^{-1}}$ ("thermal" neutrons).

🧪 Self-check quiz

0/5 correct

Q1. An electron, a proton, a neutron and an α-particle move at the same speed. Which has the longest de Broglie wavelength?

Q2. According to Heisenberg's uncertainty principle, $\Delta x\,\Delta p_{x}$ is

Q3. How many orbitals are there in the shell with $n=4$?

Q4. How many angular nodes does a $3d$ orbital have?

Q5. In a many-electron atom, the energy of an orbital depends on

← 2.2 Light, Quantisation and Bohr's Model2.4 Electronic Configuration of Atoms →