Atomic Structure

Atomic structure is the foundation of chemistry, covering the composition of atoms, electron Configuration, atomic spectra, and periodic trends.

Subatomic Particles (OL/HL)

Particle	Symbol	Relative mass	Relative charge	Location
Proton	$p^+$	1	$+1$	Nucleus
Neutron	$n^0$	1	$0$	Nucleus
Electron	$e^-$	$\approx 0$ ( $1/1836$ )	$-1$	Shells

Atomic Number and Mass Number

Atomic number ( $Z$ ): number of protons.
Mass number ( $A$ ): number of protons + neutrons.

Notation: $^A_Z X$

Isotopes (OL/HL)

Atoms of the same element with different numbers of neutrons (same $Z$ Different $A$ ).

Example (OL): Carbon has two common isotopes: $^{12}_6\mathrm{C$ (98.9%) and $^{13}_6\mathrm{C$ (1.1%).

Relative Atomic Mass (OL/HL)

A_r = \frac{\sum(\mathrm{isotope abundance \times \mathrm{isotope mass)}{100}

Example (OL): Chlorine has two isotopes: $^{35}\mathrm{Cl$ (75%) and $^{37}\mathrm{Cl$ (25%).

A_r = \frac{75 \times 35 + 25 \times 37}{100} = \frac{2625 + 925}{100} = 35.5

Worked Example 1 (HL): The relative atomic mass of boron is 10.81. If boron has two isotopes, $^{10}\mathrm{B$ and $^{11}\mathrm{B$ Calculate their percentage abundances.

Let $x$ = percentage of $^{10}\mathrm{B$ and $(100 - x)$ = percentage of $^{11}\mathrm{B$ .

$10x + 11(100 - x) = 10.81 \times 100$

$10x + 1100 - 11x = 1081$

$-x = -19 \implies x = 19\%$

$^{10}\mathrm{B = 19\%$ , $^{11}\mathrm{B = 81\%$ .

Worked Example 2 (OL): Calculate the relative atomic mass of neon from its isotopes: $^{20}\mathrm{Ne$ (90.5%) and $^{22}\mathrm{Ne$ (9.5%).

$A_r = \frac{90.5 \times 20 + 9.5 \times 22}{100} = \frac{1810 + 209}{100} = \frac{2019}{100} = 20.19$

The Bohr Model and Electron Shells (OL/HL)

Electron Shell Structure

Electrons occupy shells (energy levels) around the nucleus:

Shell	$n$	Maximum electrons
K	1	2
L	2	8
M	3	18
N	4	32

Maximum electrons in shell $n$ : $2n^2$ .

Derivation: Each shell can hold a maximum of $2n^2$ electrons because there are $n^2$ orbitals In shell $n$ and each orbital holds 2 electrons.

Electron Configuration (OL/HL)

Write the number of electrons in each shell from the nucleus outward.

Example (OL): Sodium ( $Z = 11$ ): 2, 8, 1.

Example (OL): Calcium ( $Z = 20$ ): 2, 8, 8, 2.

Worked Example 3 (OL): Write the electron configuration for potassium ( $Z = 19$ ).

Potassium has 19 electrons. Filling shells: K (2), L (8), M (8), N (1). Configuration: 2, 8, 8, 1.

Subshells and Orbitals (HL)

Shells are divided into subshells: $s$ , $p$ , $d$ , $f$ .

Subshell	Orbitals	Max electrons
$s$	1	2
$p$	3	6
$d$	5	10
$f$	7	14

Aufbau Principle (HL)

Electrons fill orbitals from lowest to highest energy. The order is:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, …

Example (HL): Iron ( $Z = 26$ ): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$ or $[\mathrm{Ar]\,4s^2 3d^6$ .

Worked Example 4 (HL): Write the full electron configuration and orbital diagram for chromium ( $Z = 24$ ). (Hint: chromium is an exception.)

Expected: $[\mathrm{Ar]\,4s^2 3d^4$ . However, chromium is an exception: $[\mathrm{Ar]\,4s^1 3d^5$ .

This is because a half-filled $d$ subshell ( $3d^5$ ) is more stable than a partially filled one ( $3d^4$ ). Similarly, copper ( $Z = 29$ ) is $[\mathrm{Ar]\,4s^1 3d^{10}$ rather than $[\mathrm{Ar]\,4s^2 3d^9$ .

Hund’s Rule (HL)

When filling degenerate orbitals (same energy), electrons occupy separate orbitals with parallel Spins before pairing.

Example: The $2p$ subshell of nitrogen ( $Z = 7$ ):

$\underset{\uparrow}{\boxed{2p_x}} \quad \underset{\uparrow}{\boxed{2p_y}} \quad \underset{\uparrow}{\boxed{2p_z}}$

Three unpaired electrons with parallel spins.

Pauli Exclusion Principle (HL)

No two electrons in an atom can have the same set of four quantum numbers. An orbital holds at most Two electrons with opposite spins.

Quantum Numbers (HL)

Quantum number	Symbol	Values	Describes
Principal	$n$	$1, 2, 3, \ldots$	Energy level / shell
Angular momentum	$\ell$	$0$ to $n-1$	Subshell ( $s=0, p=1, d=2$ )
Magnetic	$m_\ell$	$-\ell$ to $+\ell$	Orbital orientation
Spin	$m_s$	$+\frac{1}{2}, -\frac{1}{2}$	Electron spin

Worked Example 5 (HL): List all possible quantum numbers for a $3p$ electron.

$n = 3$$\ell = 1$ (p subshell), $m_\ell = -1, 0, +1$$m_s = +\frac{1}{2}$ or $-\frac{1}{2}$ .

There are 6 possible combinations: 3 orbitals $\times$ 2 spins = 6 electrons (matches the maximum For the $p$ subshell).

Atomic Emission Spectra (HL)

Hydrogen Spectrum

When hydrogen atoms absorb energy, electrons are excited to higher energy levels. When they return To lower levels, they emit photons:

\Delta E = E_{\mathrm{higher} - E_{\mathrm{lower} = hf = \frac{hc}{\lambda}

Balmer series (transitions to $n = 2$ ): visible light.

Lyman series (transitions to $n = 1$ ): ultraviolet.

Paschen series (transitions to $n = 3$ ): infrared.

Derivation of the energy levels of hydrogen:

The energy of an electron in the $n$ Th level of hydrogen is:

$E_n = -\frac{13.6}{n^2} \mathrm{ eV$

This is derived from the Bohr model, which postulates that the angular momentum of the electron is Quantised: $mvr = n\hbar$ . Solving for the allowed radii and energies gives the above formula.

Example (HL): Find the wavelength of the photon emitted when an electron in hydrogen drops from $n = 4$ to $n = 2$ .

\Delta E = 13.6\left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times \frac{3}{16} = 2.55\mathrm{ eV

\lambda = \frac{hc}{\Delta E} = \frac{1240\mathrm{ eV nm}{2.55\mathrm{ eV} = 486\mathrm{ nm

This is the blue-green line in the Balmer series.

Worked Example 6 (HL): A photon of wavelength $97.2\mathrm{ nm$ is emitted from a hydrogen atom. Identify the transition involved.

$\Delta E = \frac{1240}{97.2} = 12.76 \mathrm{ eV$

$12.76 = 13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

This is a large energy, suggesting a transition to $n = 1$ (Lyman series):

$12.76 = 13.6\left(1 - \frac{1}{n_i^2}\right)$

$\frac{12.76}{13.6} = 1 - \frac{1}{n_i^2}$

$\frac{1}{n_i^2} = 1 - 0.938 = 0.062$

$n_i^2 = 16.1 \implies n_i = 4$

Transition: $n = 4 \to n = 1$ (Lyman series).

Periodic Trends (OL/HL)

Atomic Radius

Decreases across a period (increasing nuclear charge pulls electrons closer). Increases down a group (additional shells).

Worked Example 7 (OL): Arrange the following in order of increasing atomic radius: $\mathrm{Na$ $\mathrm{Mg$$\mathrm{Al$$\mathrm{K$ .

$\mathrm{Al < \mathrm{Mg < \mathrm{Na < \mathrm{K$ .

$\mathrm{Na$$\mathrm{Mg$ And $\mathrm{Al$ are in period 3: radius decreases from left to right. $\mathrm{K$ is in period 4 (below $\mathrm{Na$ ) and has a much larger radius.

Ionisation Energy (OL/HL)

The energy required to remove the outermost electron from a gaseous atom.

Trends:

Increases across a period (stronger nuclear attraction).
Decreases down a group (outer electrons are further from the nucleus and shielded).

Anomalies: Between groups 2 and 13 (e.g., Mg to Al) and groups 15 and 16 (e.g., P to S), there Are slight dips because the $s$ -subshell is full and the new electron enters a higher energy $p$ -subshell, or because pairing begins.

Example (OL): Explain why the first ionisation energy of aluminium is lower than that of Magnesium.

Magnesium has the electron configuration $[\mathrm{Ne]\,3s^2$ — a filled $3s$ subshell. Aluminium Has $[\mathrm{Ne]\,3s^2 3p^1$ . The $3p$ electron in aluminium is at a higher energy level and is Shielded by the $3s$ electrons, so it is easier to remove.

Worked Example 8 (HL): Explain why the first ionisation energy of sulfur is lower than that of Phosphorus.

Phosphorus: $[\mathrm{Ne]\,3s^2 3p^3$ — each $3p$ orbital has one electron (half-filled subshell, Stable). Sulfur: $[\mathrm{Ne]\,3s^2 3p^4$ — one $3p$ orbital has two electrons. The pairing energy In sulfur’s $3p^4$ configuration makes the fourth electron slightly easier to remove than Phosphorus’s third $3p$ electron.

Electronegativity (OL/HL)

The ability of an atom to attract bonding electrons.

Increases across a period.
Decreases down a group.

Fluorine is the most electronegative element (3.98 on the Pauling scale).

Summary of Periodic Trends

Property	Across a period	Down a group
Atomic radius	Decreases	Increases
First ionisation energy	Increases (with anomalies)	Decreases
Electronegativity	Increases	Decreases
Metallic character	Decreases	Increases
Melting point (metals)	Increases	Decreases

Mass Spectrometry (HL)

Principle

Atoms/molecules are ionised and then separated by their mass-to-charge ratio ( $m/z$ ).

Steps

Vaporisation: sample is vaporised.
Ionisation: atoms are ionised (electron impact or electrospray).
Acceleration: ions are accelerated by an electric field.
Deflection: ions are deflected by a magnetic field (lighter ions are deflected more).
Detection: ions hit a detector, producing a signal.

Interpreting Mass Spectra

The peak at the highest $m/z$ value corresponds to the molecular ion ( $M^+$ ). Other peaks represent Fragments.

Isotope Abundance from Mass Spectra

Worked Example 9 (HL): The mass spectrum of an element shows peaks at $m/z = 52$ (83%) and $m/z = 54$ (17%). Calculate the relative atomic mass.

$A_r = \frac{83 \times 52 + 17 \times 54}{100} = \frac{4316 + 918}{100} = \frac{5234}{100} = 52.34$

This element is chromium.

Common Pitfalls

Confusing mass number with atomic number — mass number includes neutrons.
Electron configuration — 4s fills before 3d but loses electrons before 3d when forming ions.
Ionisation energy anomalies — explain using subshell stability (filled/half-filled).
Relative atomic mass — use percentage abundances, not numbers of atoms.
Orbital diagrams — apply Hund’s rule correctly.
Transition metal electron configurations — remember the exceptions: Cr is $[\mathrm{Ar]\,4s^1 3d^5$ and Cu is $[\mathrm{Ar]\,4s^1 3d^{10}$ .
Ion formation — when a transition metal forms an ion, the $4s$ electrons are lost BEFORE the $3d$ electrons, even though $4s$ fills first.
Spectral lines — emission spectra have dark background with bright lines; absorption spectra have bright background with dark lines.

Practice Questions

Ordinary Level

Describe the structure of an atom, naming the three subatomic particles and their properties.
Write the electron configuration for potassium ( $Z = 19$ ).
Calculate the relative atomic mass of neon from its isotopes: $^{20}\mathrm{Ne$ (90.5%) and $^{22}\mathrm{Ne$ (9.5%).
State the trends in atomic radius and ionisation energy across a period.

Higher Level

Write the full electron configuration and orbital diagram for chromium ( $Z = 24$ ). (Hint: chromium is an exception.)
Explain why the first ionisation energy of sulfur is lower than that of phosphorus.
A photon of wavelength $97.2\mathrm{ nm$ is emitted from a hydrogen atom. Identify the transition involved.
The relative atomic mass of boron is 10.81. If boron has two isotopes, $^{10}\mathrm{B$ and $^{11}\mathrm{B$ Calculate their percentage abundances.

Extended Questions

Explain why the first ionisation energy of neon is much higher than that of sodium.
Write the electron configuration of $\mathrm{Fe^{2+}$ and $\mathrm{Fe^{3+}$ Explaining why $\mathrm{Fe^{3+}$ is more stable.
Calculate the energy of a photon with wavelength $121.6\mathrm{ nm$ and identify the hydrogen transition that produces it.
Explain why the second ionisation energy of sodium ( $4562 \mathrm{ kJ/mol$ ) is much higher than the first ( $496 \mathrm{ kJ/mol$ ).
The mass spectrum of silicon shows three peaks at $m/z = 28$ 29, 30 with relative abundances 92.2%, 4.7%, and 3.1%. Calculate the relative atomic mass of silicon.
Explain the significance of the line spectrum of hydrogen in the development of atomic theory, referencing the Bohr model.

Successive Ionisation Energies

Successive ionisation energies provide evidence for electron shell structure. When all electrons in A shell have been removed, the next ionisation energy increases significantly.

Evidence for Shell Structure

For sodium ( $Z = 11$ ):

Ionisation	Energy (kJ/mol)	Jump?
1st	496	—
2nd	4562	—
3rd	6912	—
4th	9544	—
5th	13354	—
6th	16610	—
7th	20117	—
8th	25496	—
9th	28932	—
10th	141362	Large jump
11th	159076	—

The large jump between the 9th and 10th ionisation energies indicates that the 10th electron is Being removed from an inner shell (closer to the nucleus, less shielded). This confirms the electron Configuration 2, 8, 1.

Using Successive Ionisation Energies to Identify Elements

Worked Example 10 (HL): The successive ionisation energies of an element are:

$738, 1451, 7733, 10540 \mathrm{ kJ/mol$

Identify the element and explain your reasoning.

The large jump between the 2nd and 3rd ionisation energies indicates that the first two electrons Are in the outer shell and the third is in an inner shell. The element has 2 valence electrons, Which corresponds to Group 2. The low values ( $738, 1451$ ) are consistent with magnesium ( $Z = 12$ Electron configuration $2, 8, 2$ ).

Wave-Particle Duality of Electrons (HL)

De Broglie Wavelength

Louis de Broglie proposed that all matter has wave properties:

$\lambda = \frac{h}{mv} = \frac{h}{p}$

Where $h$ is Planck’s constant, $m$ is mass, and $v$ is velocity.

For an electron in an orbital, the standing wave condition requires that an integer number of Wavelengths fit around the orbit: $n\lambda = 2\pi r$ .

This explains why only certain energy levels are allowed: only orbits that satisfy the standing wave Condition are stable.

Heisenberg Uncertainty Principle

$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$

It is impossible to simultaneously know both the exact position and exact momentum of an electron. This is why we use probability distributions (orbitals) rather than fixed orbits to describe Electron positions.

Quantum Mechanical Model (HL)

Orbitals as Probability Distributions

An orbital is a region of space where there is a high probability ( 90%) of finding an Electron. Each orbital is described by a wave function $\psi$ And $|\psi|^2$ gives the probability Density.

Shapes of Orbitals

s orbitals: Spherical. All $s$ orbitals are spherical but increase in size with $n$ .
p orbitals: Dumbbell-shaped. Three orientations: $p_x$$p_y$$p_z$ .
d orbitals: Clover-shaped (five orientations).

Radial Distribution Functions

The radial distribution function shows the probability of finding an electron at a given distance From the nucleus. For the $1s$ orbital, the most probable distance equals the Bohr radius ( $a_0 = 52.9 \mathrm{ pm$ ). For the $2s$ orbital, there is a node (zero probability) at a certain Distance.

Electron Configuration of Ions (HL)

Transition Metal Ions

When transition metals form ions, the $4s$ electrons are removed before the $3d$ electrons.

Example: $\mathrm{Fe$ : $[\mathrm{Ar]\,4s^2 3d^6$

$\mathrm{Fe^{2+}$ : $[\mathrm{Ar]\,3d^6$ (two $4s$ electrons removed)

$\mathrm{Fe^{3+}$ : $[\mathrm{Ar]\,3d^5$ (three electrons removed: two from $4s$ One from $3d$ )

Note: $\mathrm{Fe^{3+}$ has a half-filled $3d$ subshell, making it particularly stable.

Common Ion Configurations

Ion	Electron configuration
$\mathrm{Na^+$	$[\mathrm{Ne]$ or $1s^2 2s^2 2p^6$
$\mathrm{Ca^{2+}$	$[\mathrm{Ar]$
$\mathrm{O^{2-}$	$[\mathrm{Ne]$
$\mathrm{Cl^-$	$[\mathrm{Ar]$
$\mathrm{Cu^+$	$[\mathrm{Ar]\,3d^{10}$
$\mathrm{Zn^{2+}$	$[\mathrm{Ar]\,3d^{10}$

Isoelectronic Series

Ions with the same number of electrons are isoelectronic.

Example: $\mathrm{O^{2-}$$\mathrm{F^-$$\mathrm{Ne$$\mathrm{Na^+$$\mathrm{Mg^{2+}$ all have 10 Electrons ( $[\mathrm{Ne]$ configuration).

Trend in ionic radius: For isoelectronic ions, the radius decreases as nuclear charge increases:

$\mathrm{O^{2-} > \mathrm{F^-> \mathrm{Ne > \mathrm{Na^+ > \mathrm{Mg^{2+}$

The Periodic Table in Detail

Blocks of the Periodic Table

Block	Subshell being filled	Groups	Examples
s-block	$s$	1, 2	$\mathrm{Na$$\mathrm{Mg$
p-block	$p$	13-18	$\mathrm{Al$$\mathrm{Cl$$\mathrm{Ar$
d-block	$d$	3-12	$\mathrm{Fe$$\mathrm{Cu$
f-block	$f$	Lanthanides, Actinides	$\mathrm{U$$\mathrm{Ce$

Periodic Trends in Detail: Atomic Radius

Across a period: Nuclear charge increases by one proton per element, but shielding increases Only slightly (electrons in the same shell do not shield effectively). The effective nuclear charge ( $Z_{\mathrm{eff}$ ) increases, pulling electrons closer.

Down a group: Each new shell is further from the nucleus. Although $Z$ increases, the additional Inner shells provide significant shielding, so the outer electrons are less tightly held.

Periodic Trends in Detail: Ionisation Energy

The first ionisation energy increases across a period with two notable dips:

Group 2 to 13 (e.g., Be to B): The $2s$ subshell is full in Be. The new electron in B enters the $2p$ subshell, which is higher in energy and more shielded, making it easier to remove.
Group 15 to 16 (e.g., N to O): N has a half-filled $2p^3$ configuration (stable). In O, the fourth $2p$ electron pairs with another, introducing electron-electron repulsion that makes it easier to remove.

Comparison of Group 1 and Group 17 Properties

Property	Group 1 (Alkali metals)	Group 17 (Halogens)
Atomic radius	Large, increases down group	Small, increases down group
First IE	Low, decreases down group	High, decreases down group
Reactivity	Increases down group	Decreases down group
Electronegativity	Low	High
Type of element	Metal	Non-metal
Bonding	Metallic/ionic	Covalent
Typical ion	$+1$	$-1$

Advanced: Effective Nuclear Charge (HL)

The effective nuclear charge ( $Z_{\mathrm{eff}$ ) experienced by an electron is:

$Z_{\mathrm{eff} = Z - S$

Where $Z$ is the actual nuclear charge and $S$ is the shielding constant (Slater’s rules).

Slater’s Rules (simplified):

Electrons in the same group (ns, np) shield each other by 0.35 (except 1s, which shields by 0.30).
Electrons in the (n-1) shell shield by 0.85.
Electrons in lower shells shield by 1.00.

Worked Example 11 (HL): Calculate $Z_{\mathrm{eff}$ for a $3p$ electron in chlorine.

$\mathrm{Cl$ : $1s^2 2s^2 2p^6 3s^2 3p^5$ . $Z = 17$ .

For a $3p$ electron:

Other electrons in the same group ( $n = 3$ ): 7 electrons $\times 0.35 = 2.45$
Electrons in $n = 2$ : 8 electrons $\times 0.85 = 6.80$
Electrons in $n = 1$ : 2 electrons $\times 1.00 = 2.00$

$S = 2.45 + 6.80 + 2.00 = 11.25$

$Z_{\mathrm{eff} = 17 - 11.25 = 5.75$

This relatively high $Z_{\mathrm{eff}$ explains chlorine’s high electronegativity and small atomic Radius.

X-ray Spectra and Moseley’s Law (HL)

Henry Moseley discovered that the frequency of characteristic X-rays emitted by an element is Related to its atomic number:

$\sqrt{f} = a(Z - b)$

Where $a$ and $b$ are constants. This established that atomic number (not atomic mass) is the Fundamental property that determines an element’s identity, leading to the modern periodic table.

Applications of Atomic Structure

Lasers

Lasers rely on stimulated emission: an excited electron is stimulated to drop to a lower energy Level by an incoming photon of exactly the right energy, emitting a second photon with the same Wavelength, phase, and direction.

Conditions for laser action:

Population inversion (more atoms in excited state than ground state)
Metastable state (long-lived excited state)
Optical cavity (mirrors to reflect photons back and forth)

Fluorescence

When atoms or molecules absorb UV light, electrons are excited to higher energy levels. They then Return to lower levels in steps, emitting visible light. This is the principle behind fluorescent Lamps and biological staining.

Electron Microscopy

Electron microscopes use beams of electrons (which have much shorter wavelengths than visible light) To achieve much higher resolution than optical microscopes. Transmission electron microscopy (TEM) And scanning electron microscopy (SEM) are two common types.

Summary: Atomic Structure Key Relationships

Concept	Formula/Relationship	Key Idea
Bohr model energy	$E_n = -13.6/n^2 \mathrm{ eV$	Quantised energy levels
Photon energy	$E = hf = hc/\lambda$	Emitted/absorbed photons
De Broglie	$\lambda = h/mv$	Wave-particle duality
Heisenberg	$\Delta x \cdot \Delta p \geq h/4\pi$	Measurement limits
Effective nuclear charge	$Z_{\mathrm{eff} = Z - S$	Shielding effects
Shell capacity	$2n^2$	Maximum electrons per shell
Moseley’s law	$\sqrt{f} = a(Z - b)$	Atomic number defines element

Practice Questions (Extended)

The successive ionisation energies of an element are: $578, 1817, 2745, 11578, 14842, 18379 \mathrm{ kJ/mol$ . Identify the element and explain the pattern.
Calculate the de Broglie wavelength of an electron travelling at $2.0 \times 10^6 \mathrm{ m/s$ . (Mass of electron $= 9.11 \times 10^{-31} \mathrm{ kg$ , $h = 6.63 \times 10^{-34} \mathrm{ J s$ .)
Explain why potassium ( $Z = 19$ ) is placed after argon ( $Z = 18$ ) in the periodic table, despite argon having a filled $3p$ subshell and potassium having an electron in the $4s$ subshell.
Calculate $Z_{\mathrm{eff}$ for a valence electron in sodium and in potassium. Use your results to explain why potassium is more reactive than sodium.
The first four ionisation energies of an element X are: $1090, 2350, 4620, 6220 \mathrm{ kJ/mol$ . To which group does X belong? Explain your reasoning.
Draw the shape of a $2p_z$ orbital and a $3d_{xy}$ orbital, labelling the axes.
Explain the difference between a continuous spectrum and a line spectrum. Why do hot gases produce line spectra but hot solids produce continuous spectra?
Calculate the wavelength of light emitted when an electron in a hydrogen atom transitions from $n = 5$ to $n = 2$ . In which spectral series does this line appear?

History of Atomic Theory

Key Developments

Scientist	Contribution	Date
Dalton	Atomic theory: all matter is made of atoms	1808
Thomson	Discovered the electron (cathode ray experiments)	1897
Rutherford	Nuclear model: gold foil experiment showed dense positive nucleus	1911
Bohr	Quantised energy levels for hydrogen	1913
de Broglie	Wave-particle duality	1924
Heisenberg	Uncertainty principle	1927
Schrodinger	Wave equation for electrons	1926

Rutherford’s Gold Foil Experiment

Most alpha particles passed straight through the gold foil, but a few were deflected at large angles And some bounced back. This led to the conclusion that:

Most of the atom is empty space (most particles pass through)
The nucleus is very small, dense, and positively charged (few particles deflected)
Electrons orbit the nucleus (to account for the overall neutral atom)

Thomson’s Plum Pudding Model

Before Rutherford, Thomson proposed that the atom was a sphere of positive charge with electrons Embedded in it (like plums in a pudding). The gold foil experiment disproved this model because the Large-angle deflections could not be explained by a diffuse positive charge.

Limitations of the Bohr Model

The Bohr model successfully explained the hydrogen spectrum but failed for multi-electron atoms Because:

It treats electrons as particles in fixed orbits, not as standing waves
It cannot explain the fine structure of spectral lines (splitting due to spin-orbit coupling)
It violates the Heisenberg uncertainty principle (specifying both orbit radius and momentum)

The quantum mechanical model (Schrodinger equation) overcomes these limitations by describing Electrons as probability waves.

Worked Examples

Example 1: Applying chemical principles

A typical problem in atomic structure involves identifying the relevant chemical concepts, writing appropriate equations, and performing calculations with correct units and significant figures.

Solution approach:

Identify the relevant chemical species and their states
Write balanced chemical equations where applicable
Apply the correct formulae and show working
Give the final answer with appropriate units and significant figures

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