Structure of Atom

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 2.

If the nucleus were a marble, how big would the atom be?

The nucleus has a radius of about 10⁻¹⁵ m and the atom about 10⁻¹⁰ m, a ratio of 100,000. Put a marble on the centre spot of a stadium and the atom would stretch out about a kilometre, with a few electrons somewhere in the empty space between. Almost all of the mass sits in the marble. How did anyone discover this, when no one has ever seen an atom? By firing particles at atoms, watching what bounced back, and building a model that explains it. This chapter follows that chain of models, and every one of them breaks for a reason you can state in one line.

The particles inside the atom

In the 1800s experiments with electricity passing through gases at very low pressure revealed the first subatomic particle. When a high voltage is applied across two electrodes in a sealed tube with the air pumped down, a stream of particles flows from the negative electrode (cathode) towards the positive one (anode). These cathode rays travel in straight lines when there are no fields, make a fluorescent screen glow where they hit it, and bend towards the positive plate in an electric field, which shows they are negatively charged. Most importantly, the rays behave the same whatever the cathode metal or the gas in the tube. So the particle is a universal part of all matter.

In 1897 J.J. Thomson measured the charge-to-mass ratio of these particles by adjusting electric and magnetic fields until the beam went straight again. He found

$$\dfrac{e}{m_e} = 1.758820 \times 10^{11}\ \text{C kg}^{-1}$$

and called the particle the electron. R.A. Millikan's oil drop experiment then measured the charge itself. He balanced the weight of a charged oil droplet against an electric force, and found that every charge was a whole number multiple of $e = 1.602 \times 10^{-19}$ C. Combining the two results gives the electron's mass.

Worked example: mass of the electron. Using $e = 1.602 \times 10^{-19}$ C and $e/m_e = 1.7588 \times 10^{11}$ C kg⁻¹, find $m_e$.

Positive particles came from tubes with a perforated cathode. Rays streamed backwards through the holes, and these canal rays were positive ions. The lightest ones came from hydrogen and were identified as the proton. The neutron was found in 1932 by James Chadwick, when beryllium bombarded with α-particles gave off a penetrating neutral radiation made of particles slightly heavier than protons.

A proton is about 1,836 times heavier than an electron, so the mass of an atom is essentially the mass of its protons and neutrons.

Beyond the textbook. For the exam: the neutron is a stable particle that sits in the nucleus with the protons. In reality: a neutron alone in free space is unstable. It decays into a proton, an electron and an antineutrino in about 15 minutes on average. Inside most nuclei it is stabilised by the forces holding the nucleus together. Why it matters: this is why neutrons were found so late, and why a nucleus needs the right mix of protons and neutrons to last.

From the plum pudding to the nuclear atom

Once electrons were found, the question was how they sit inside a neutral atom. Thomson (1898) pictured a sphere of positive charge with electrons embedded in it like seeds in a watermelon, or plums in a pudding. It explained neutrality, but a scattering experiment destroyed it.

Rutherford's students Geiger and Marsden fired α-particles (doubly charged helium nuclei) at a very thin gold foil about 100 nm thick, and watched where each one landed on a fluorescent screen. If the plum pudding were right, every particle would sail through with a tiny nudge. What happened instead:

Rutherford (1911) therefore proposed the nuclear model: a tiny positive nucleus of radius about 10⁻¹⁵ m, with electrons revolving around it like planets round the Sun, held by electrostatic attraction. The model failed for a reason you must remember. An electron moving in a circle is constantly accelerating, and a charge that accelerates must radiate energy. It would lose energy continuously and spiral into the nucleus in a fraction of a second. The model also said nothing about how the electrons are arranged or how much energy each has.

Beyond the textbook. For the exam: the nucleus is a tiny, dense region at the centre with almost all of the atom's mass. In reality: nuclear matter is so dense, about 10¹⁷ kg per cubic metre, that a matchbox full of it would weigh around a billion tonnes. Why it matters: it puts the "empty space" claim in perspective. An atom is mostly empty only in the sense that the electrons occupy a huge volume compared with the nucleus.

Atomic number, mass number and isotopes

The atomic number $Z$ is the number of protons in the nucleus, and it identifies the element. A neutral atom has $Z$ electrons. The mass number $A$ is the total number of protons and neutrons, so the number of neutrons is $A - Z$. An atom is written as $^{A}_{Z}\text{X}$.

Worked example: counting particles. Find the number of protons, neutrons and electrons in ¹⁴₆C, ³⁵₁₇Cl⁻, ²⁷₁₃Al³⁺ and ⁵⁶₂₆Fe²⁺.

The common wrong step: changing the number of protons for an ion. Only electrons are gained or lost. $Z$ never changes in a chemical reaction.

Pause and predict: ¹²C, ¹³C and ¹⁴C are isotopes of carbon. Why do they react in the same way?

Check your answer: Chemical behaviour depends on the number and arrangement of electrons, and all three have 6 electrons. The extra neutrons change the mass, and so physical properties such as density and rates of diffusion, but not the chemistry.

Light as a wave

To go beyond Rutherford, scientists studied how atoms absorb and emit light. Light is an electromagnetic wave: oscillating electric and magnetic fields, at right angles to each other and to the direction of travel, moving at $c = 3.0 \times 10^8$ m s⁻¹ in vacuum with no medium needed. A wave is described by its wavelength $\lambda$ (distance between crests), frequency $\nu$ (crests passing per second, in Hz) and wavenumber $\bar{\nu}$ (waves per unit length).

$$c = \nu\lambda \qquad \bar{\nu} = \dfrac{1}{\lambda}$$

The electromagnetic spectrum runs from long, low frequency waves to short, high frequency ones. Visible light is only a thin slice, from about 380 nm (violet) to 780 nm (red).

Worked example: frequency, wavenumber and photon energy. Green light has $\lambda = 500$ nm. Find its frequency and wavenumber, and the energy of one photon and of one mole of photons.

The common wrong step: forgetting to convert nanometres to metres before using $c$ in m s⁻¹. That is an error of $10^9$.

Energy comes in packets: Planck and the photoelectric effect

Classical wave theory predicted that a hot object should emit infinite energy at short wavelengths, which is not what happens. In 1900 Planck solved this by proposing that energy is emitted or absorbed only in discrete packets called quanta. The energy of one quantum is proportional to the frequency, where $h = 6.626 \times 10^{-34}$ J s is Planck's constant.

$$E = h\nu = \dfrac{hc}{\lambda}$$

The idea explains why a heated iron bar glows dull red, then orange, then white. Hotter objects emit more high frequency quanta, which carry more energy each.

The decisive test was the photoelectric effect. When light of high enough frequency falls on a clean metal surface, electrons are knocked out. Three facts defied wave theory, and Einstein (1905) explained all of them by treating light as a stream of photons, each of energy $h\nu$.

Energy conservation for one photon and one electron gives Einstein's equation.

$$h\nu = h\nu_0 + \tfrac{1}{2}m_e v^2 \qquad \text{so} \qquad KE = h(\nu - \nu_0)$$

Worked example: photoelectric numerical. Light of frequency $1.5 \times 10^{15}$ Hz falls on a metal whose threshold frequency is $5.0 \times 10^{14}$ Hz. Find the kinetic energy and the speed of an ejected electron.

The common wrong step: using $h\nu$ alone as the kinetic energy. Part of the photon's energy, $h\nu_0$, is spent in just freeing the electron from the metal.

Pause and predict: Light of frequency above $\nu_0$ is made twice as bright. What happens to the kinetic energy of each electron and to the number of electrons?

Check your answer: The kinetic energy of each electron stays the same, because the photon energy has not changed. The number of electrons doubles, because there are twice as many photons.

Beyond the textbook. For the exam: light shows both wave and particle behaviour, and the photoelectric effect is the particle evidence. In reality: the same photon logic explains why sunburn comes from ultraviolet light and not from a very bright red lamp. UV photons carry enough energy each to break bonds in skin cells, but red photons do not, however many arrive. Solar cells and digital camera sensors also depend on photons freeing electrons in a material. Why it matters: it is the same argument, the energy per photon matters, not the total brightness.

Atomic spectra: the fingerprint of an element

Heat a solid and it glows with a continuous spectrum, every colour merging into the next. Heat a gas of free atoms, or pass a high voltage through it, and it emits light only at a few sharp wavelengths, a line spectrum. Each element has its own pattern, so a spectrum acts as a fingerprint. Absorption spectra show dark lines at exactly the same wavelengths, because an atom absorbs only what it can emit.

The hydrogen spectrum is the simplest. All its lines fit one formula discovered by Rydberg, where $n_2 > n_1$ and the Rydberg constant is $109677$ cm⁻¹.

$$\bar{\nu} = 109677\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\ \text{cm}^{-1}$$

Each value of $n_1$ gives a family of lines called a series.

Worked example: a Balmer line. Find the wavelength of the line in the Balmer series that arises from $n_2 = 3$ to $n_1 = 2$.

The lines of each series crowd together towards a limit as $n_2 \to \infty$. For the Lyman series, that limit is $\lambda = 1/109677$ cm $= 91.2$ nm. If electrons fall from level $n$ to lower levels in many atoms, the number of different lines is $\dfrac{n(n-1)}{2}$, so 6 lines from $n = 4$.

Beyond the textbook. For the exam: every element has a unique line spectrum. In reality: this is how we know what stars are made of. Helium was identified from dark lines in sunlight in 1868, almost 30 years before it was found on Earth, and its name comes from helios, the Greek word for Sun. Why it matters: sharp lines are evidence that electrons in atoms can have only certain energies, which is the idea Bohr builds on next.

Bohr's model of hydrogen

Rutherford's atom could not explain why spectra are lines. In 1913 Niels Bohr kept the nucleus-and-orbit picture but added three rules.

From these, the energy and radius of the $n$th orbit of a one electron species (H, He⁺, Li²⁺, Be³⁺) with atomic number $Z$ come out as

$$E_n = -2.18 \times 10^{-18}\,\dfrac{Z^2}{n^2}\ \text{J atom}^{-1} \qquad r_n = 52.9\,\dfrac{n^2}{Z}\ \text{pm}$$

The radius of the first orbit of hydrogen, 52.9 pm, is the Bohr radius. Energies are negative because the electron is bound: the zero of energy is an electron at rest infinitely far away ($n = \infty$). The lowest level, $n = 1$, is the ground state, and anything higher is an excited state. Energy rises with $n$, since $-1/n^2$ increases towards zero, and the levels crowd together at high $n$. Note that $Z^2$ is on top and $n^2$ is below in the energy formula, and the reverse way round for the radius.

Worked example: Bohr formulas. Find (a) the energy of the $n = 2$ orbit of He⁺, (b) the radius of the first orbit of Li²⁺, (c) the ionisation energy of hydrogen per mole, and (d) the wavelength of the light emitted when hydrogen falls from $n = 4$ to $n = 2$.

Pause and predict: Which has the larger radius in its second orbit, H or He⁺?

Check your answer: Hydrogen. Since $r_n \propto n^2/Z$, the $n = 2$ orbit has radius $52.9 \times 4/1 = 211.6$ pm in H, but only $52.9 \times 4/2 = 105.8$ pm in He⁺. The doubled nuclear charge pulls the electron in closer.

Bohr's model matched the hydrogen spectrum perfectly, but it fails beyond that.

Matter has waves too, and certainty has limits

Light behaves as both wave and particle. In 1924 de Broglie asked whether matter could do the same. His answer was that every moving particle has a wavelength.

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

The electron in a Bohr orbit can only exist if its wave closes smoothly on itself, which brings back Bohr's quantised angular momentum rule. Matter waves were confirmed in 1927 when electrons were diffracted by a crystal.

Worked example: why only tiny things behave as waves. Compare the wavelength of an electron moving at $1.0 \times 10^6$ m s⁻¹ with that of a 1.0 kg ball moving at 10 m s⁻¹.

Beyond the textbook. For the exam: electrons behave as waves, and the wave character is significant only for microscopic particles. In reality: electron microscopes are built on this. Electrons accelerated by high voltage have wavelengths far smaller than visible light (hundreds of nanometres), so they can show details a light microscope cannot resolve. Why it matters: it is the working proof that de Broglie's wavelength is a real property and not a mathematical trick.

Wave nature has a second consequence. In 1927 Heisenberg showed that it is impossible to know both the exact position and the exact momentum of a microscopic particle at the same time.

$$\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi} \qquad \text{or} \qquad \Delta x \cdot \Delta v \geq \dfrac{h}{4\pi m}$$

The more precisely you pin down where an electron is, the more uncertain its speed becomes. For an electron located within $\Delta x = 10^{-10}$ m (one atom), $\Delta v \geq \dfrac{6.626 \times 10^{-34}}{4\pi \times 9.109 \times 10^{-31} \times 10^{-10}} \approx 5.8 \times 10^{5}$ m s⁻¹, which is huge. For a 1 mg dust grain located within 1 µm, the same formula gives about $5 \times 10^{-23}$ m s⁻¹, which is utterly negligible. So uncertainty matters for electrons and not for cricket balls. It also kills Bohr's picture, because a definite orbit means knowing both position and velocity at every moment.

The quantum mechanical model: orbitals

The answer is a probability description. Schrödinger's equation gives, for each allowed energy of the electron, a wave function $\psi$. By itself $\psi$ has no direct meaning, but its square, $\psi^2$, gives the probability density of finding the electron at a point. A region of space with high probability of finding the electron, with a definite shape and energy, is an atomic orbital. This is the key distinction to remember: an orbit is a fixed circular path (Bohr), while an orbital is a three dimensional region described by a wave function.

Each orbital is labelled by three quantum numbers, and an electron in it needs a fourth.

Counting follows directly. A subshell has $2l + 1$ orbitals, a shell has $n^2$ orbitals, and a shell holds $2n^2$ electrons, since each orbital holds two.

Subshells contain 1 (s), 3 (p), 5 (d) and 7 (f) orbitals.

Worked example: allowed quantum numbers. Are the sets (i) $n = 3, l = 3, m_l = 0$ and (ii) $n = 2, l = 1, m_l = -1$ allowed? How many orbitals does the 4f subshell have?

Shapes and nodes. An s orbital is a sphere. A p orbital has two lobes along an axis (a dumbbell), so there are three, $p_x$, $p_y$ and $p_z$, at right angles. Four of the five d orbitals have four lobes ($d_{xy}$, $d_{yz}$, $d_{xz}$ and $d_{x^2-y^2}$), and $d_{z^2}$ has two lobes along $z$ with a ring around the middle. A node is a place where the probability of finding the electron is zero. The number of nodes follows from $n$ and $l$.

Energy of orbitals. In hydrogen, energy depends only on $n$, so 2s and 2p have the same energy (they are degenerate). In an atom with more than one electron, the electrons shield each other from the nucleus, and the energy now depends on $n$ and $l$. An s electron spends more time close to the nucleus than a p electron of the same shell, so it feels a stronger pull and sits lower: $ns < np < nd < nf$.

Filling the orbitals: Aufbau, Pauli and Hund

Three rules fill the orbitals of an atom in its ground state.

Why does 4s fill before 3d? Check $(n + l)$: 4s is $4 + 0 = 4$ while 3d is $3 + 2 = 5$. When two orbitals tie, as 3d and 4p do at 5, the smaller $n$ fills first.

Chromium and copper are the famous exceptions. Moving one electron from 4s to 3d gives a half filled (d⁵) or fully filled (d¹⁰) subshell, which is extra stable for two reasons: symmetrical distribution of charge, and exchange energy. Parallel spin electrons in degenerate orbitals can swap places, which lowers energy, and a d⁵ set has 10 possible exchanges against only 6 for d⁴.

Worked example: configurations of ions. Write the configuration of Fe²⁺, Fe³⁺, Cu⁺ and Cu²⁺, and give the number of unpaired electrons in each.

The common wrong step: removing 3d electrons first, because 3d fills later. In an ion, 4s electrons leave first.

Beyond the textbook. For the exam: 4s fills before 3d, but when a transition metal forms a cation, the 4s electrons are removed first. In reality: both are explained by one fact. Once electrons start entering 3d, the 3d orbitals are pulled down below 4s in energy, so 4s is the outer, higher energy shell and gives up its electrons first. In scandium to zinc the filling order is a bookkeeping rule for neutral atoms and not a statement that 4s is always lower. Why it matters: this is why Fe³⁺ is more stable than Fe²⁺, since it reaches the half filled 3d⁵ arrangement.

Pause and predict: What is the electronic configuration of Mn²⁺, and why is it unusually stable?

Check your answer: Mn is [Ar] 3d⁵ 4s², so Mn²⁺ is [Ar] 3d⁵. Losing the two 4s electrons leaves a half filled d subshell with five unpaired electrons, which is the same stabilising arrangement seen in Cr and Fe³⁺.

Exam traps

Formula sheet

Constants: $h = 6.626 \times 10^{-34}$ J s, $c = 3.0 \times 10^{8}$ m s⁻¹, $m_e = 9.109 \times 10^{-31}$ kg, $N_A = 6.022 \times 10^{23}$ mol⁻¹.

60-second revision

Where this chapter connects

• Before this: Some Basic Concepts of Chemistry introduced the atom as indivisible and the mole as a count of atoms, and isotopes explain why atomic masses are averages.
• Next: Classification of Elements and Periodicity in Properties turns the electron configurations you have just learned into the shape of the periodic table.
• Forward link: Chemical Bonding and Molecular Structure overlaps the orbitals from this chapter to form bonds. In Physics, Atoms and Dual Nature of Radiation and Matter return to the Bohr model and the photoelectric effect.