Atoms

The chapter runs in one continuous chain of reasoning, and it pays to see the chain before the details. First comes the experiment: alpha-particle scattering off gold foil, and what its results force us to conclude about where an atom's mass and charge live (the nuclear model, the impact parameter, and the distance of closest approach). Next comes the classical crisis: why a Rutherford atom, treated with ordinary electromagnetism, cannot be stable, and why it cannot explain the sharp emission and absorption lines seen in real atomic spectra. Then comes Bohr's fix: three postulates that quantise the electron's orbit, from which we derive the radius $r_n$, the energy $E_n$, and the full hydrogen spectrum organised into the Lyman, Balmer, Paschen, Brackett and Pfund series. Finally comes the deeper "why": de Broglie's matter-wave picture, which shows Bohr's quantisation rule is really a standing-wave condition — and a candid look at where the Bohr model succeeds and where it runs out of road.

Rutherford's nuclear model and the scattering experiment

Geiger and Marsden aimed a collimated beam of 5.5 MeV alpha particles (helium nuclei, charge $+2e$) at a gold foil only $2.1\times10^{-7}\ \text{m}$ thick and counted scattered particles with a rotatable zinc-sulphide screen. The overwhelming majority passed through with little or no deflection; only about 0.14% scattered by more than 1°, and roughly 1 in 8000 turned back by more than 90°. A diffuse, Thomson-style positive charge could never produce that occasional violent rebound — only a concentrated charge could supply the large repulsive Coulomb force needed. Rutherford therefore proposed that essentially all of an atom's positive charge and mass sits in a nucleus roughly $10^{-15}$ to $10^{-14}\ \text{m}$ across, while the atom itself is about $10^{-10}\ \text{m}$ across — the electrons orbit at a distance ten thousand to a hundred thousand times the nuclear radius, so an atom is almost entirely empty space. Since the gold foil is thin, each alpha particle can be treated as scattering off a single stationary nucleus (gold is about 50 times more massive than an alpha particle). The Coulomb repulsion between the alpha particle (charge $2e$) and the nucleus (charge $Ze$) is

where: $r$ is the instantaneous centre-to-centre separation, $Z$ is the atomic number of the target ($Z = 79$ for gold), and $\varepsilon_0$ is the permittivity of free space. This force acts along the line joining the two particles and grows sharply as $r$ shrinks, which is exactly what is needed to turn a fast alpha particle around if it gets close enough. The path an individual alpha particle follows is fixed by its impact parameter $b$, the perpendicular distance from the nucleus to the alpha particle's original straight-line path (had there been no force). A small $b$ sends the particle deep into the repulsive field, producing large-angle scattering, even a full rebound ($\theta \approx \pi$) for a direct, head-on approach. A large $b$ leaves the particle only weakly deflected ($\theta \approx 0$). Because only a small fraction of incoming particles have a small enough $b$ to rebound, only a small fraction of the atom's cross-sectional area is occupied by the nucleus — confirming that the nucleus is minute compared with the atom.

Stable orbits in the Rutherford picture, before Bohr

Treating the Rutherford atom purely classically, the electrostatic attraction between the electron (charge $-e$) and nucleus (charge $+e$, for hydrogen) supplies the centripetal force that keeps the electron in a circular orbit of radius $r$:

where: $m$ is the electron mass, $v$ is its orbital speed, and $r$ is the orbit radius. Rearranging gives $r = e^2/(4\pi\varepsilon_0 m v^2)$, linking radius and speed for any classical circular orbit — but so far nothing picks out which radius is allowed; classically, any $r$ would do. The kinetic energy is $K = \tfrac{1}{2}mv^2 = e^2/(8\pi\varepsilon_0 r)$, the potential energy is $U = -e^2/(4\pi\varepsilon_0 r)$, and adding them gives the total energy

The negative sign says the electron is bound: energy must be supplied to pull it free to $r \to \infty$, where $E \to 0$. This classical picture is a necessary stepping stone, but it is silent on why only certain orbits and energies actually occur in nature — that requires Bohr's quantum postulates, taken up next.

Atomic spectra: the fingerprint that classical physics could not explain

Rarefied atomic gases excited electrically (as in a neon sign) emit light only at certain sharp, discrete wavelengths — a bright-line emission spectrum unique to each element, a kind of fingerprint. When white light instead passes through a cooler gas, the same wavelengths are missing from the transmitted light, appearing as dark lines against a continuous background — the absorption spectrum. Hydrogen's emission spectrum, the simplest and most studied, shows a beautifully regular pattern of lines that Balmer captured empirically in 1885 for the visible series — long before anyone knew why the pattern held. A classical, continuously radiating spiralling electron could never produce such sharp, discrete lines; it would radiate a smear of every frequency as it spiralled inward. Explaining this fingerprint is the real target of Bohr's postulates.

Bohr's postulates and the stability problem

An electron orbiting the nucleus is constantly accelerating (centripetally), and classical electromagnetic theory insists an accelerating charge continuously radiates electromagnetic energy. A Rutherford-model electron should therefore lose energy steadily, its orbit shrinking as it spirals inward, until it crashes into the nucleus in a fraction of a second — matter as we know it could not exist. Worse, as the electron spirals its frequency of revolution changes continuously, so classically the emitted light should form a continuous spectrum, not the sharp lines actually observed. Rutherford's model could locate the nucleus but could not explain why atoms are stable or why their spectra are discrete. Bohr's 1913 resolution was to accept Rutherford's nucleus but overrule classical electrodynamics inside the atom with three postulates. Bohr's first postulate: an electron can revolve in certain stable orbits, called stationary states, without radiating energy at all, contrary to classical prediction. Bohr's second postulate: only those orbits are allowed for which the electron's angular momentum is an integral multiple of $h/2\pi$ — the angular momentum is quantised:

where: $L = mvr$ is the orbital angular momentum, $h = 6.63\times10^{-34}\ \text{J s}$ is Planck's constant, and $n$ is the principal quantum number, a positive integer labelling the allowed orbit. This is a purely quantum rule with no classical justification at the time — de Broglie later supplied one, covered below. Bohr's third postulate: an electron can jump from a higher-energy stationary state to a lower one, emitting a single photon whose energy equals exactly the energy difference between the two states:

where: $\nu$ is the photon frequency, and $E_i > E_f$ are the initial and final state energies. Absorption is the reverse process, $E_i + h\nu = E_f$, with the atom absorbing a photon of exactly the right energy to jump up. Because $n$ is restricted to integers, only certain energy differences — hence only certain frequencies — are possible, and that is precisely the origin of the sharp line spectrum.

Orbit radius and energy of the hydrogen atom

Combining Bohr's quantisation condition with the classical force balance (worked in full in the first derivation below) gives the radius of the $n$-th orbit,

where: $a_0$ is the Bohr radius, the radius of the smallest ($n=1$) orbit, and $n = 1, 2, 3, \ldots$ labels successive orbits, which grow as $n^2$. Substituting $r_n$ back into the energy expression gives the quantised total energy,

where: the constant $13.6\ \text{eV}$ is fixed entirely by the electron mass, charge, $\varepsilon_0$ and $h$ — it is not put in by hand. The lowest energy state, $n = 1$, is the ground state, with $E_1 = -13.6\ \text{eV}$; states with $n > 1$ are excited states, with energies closer and closer to zero (unbound) as $n$ grows. The minimum energy to free the ground-state electron entirely ($n \to \infty$, $E_\infty = 0$) is the ionisation energy, $13.6\ \text{eV}$ — one of the best-tested numbers in all of physics.

The hydrogen line spectrum and the Rydberg formula

When an electron falls from a higher level $n_i$ to a lower level $n_f$ (with $n_f < n_i$), the emitted photon's energy is $E_{n_i} - E_{n_f}$, and using $E_n = -13.6\ \text{eV}/n^2$ together with $E = hc/\lambda$ gives the wavelength of every possible hydrogen line:

where: $R$ is the Rydberg constant, $n_f$ is the fixed lower level that a given series converges to, and $n_i > n_f$ ranges over all higher levels. Fixing $n_f$ groups the infinitely many possible transitions into named series: $n_f = 1$ gives the Lyman series (ultraviolet), $n_f = 2$ gives the Balmer series (partly visible — this is the series Balmer found empirically in 1885), $n_f = 3$ gives the Paschen series, $n_f = 4$ the Brackett series, and $n_f = 5$ the Pfund series (all infrared). Within any one series, the lines crowd closer together as $n_i \to \infty$, converging on a series limit at $1/\lambda = R/n_f^2$, the wavelength corresponding to ionisation from that level.

De Broglie's explanation of quantised angular momentum

Bohr's second postulate looked like a rule pulled from nowhere until Louis de Broglie, in 1923, pointed out that an electron, like any particle, has an associated wavelength $\lambda = h/p$. Picture the electron's circular orbit as a string bent into a loop: just as a plucked string only sustains a standing wave when a whole number of wavelengths fits along its length, an electron's orbit is stable only if the orbit's circumference holds a whole number of de Broglie wavelengths, so the wave meets itself in phase after one full trip and does not cancel itself out — exactly reproducing Bohr's quantisation rule, as the third derivation below shows in full.

Where the Bohr model stops working

The model is remarkably successful for hydrogenic atoms — hydrogen itself and any ion with a single electron orbiting a nucleus of charge $+Ze$, such as singly ionised helium — because these are the only systems the three postulates were built to handle. It fails for any atom with two or more electrons, such as neutral helium, because the postulates account only for the nucleus-electron force and completely ignore electron-electron repulsion, which is not negligible when charges and distances are comparable. The model also correctly predicts which frequencies hydrogen emits but has nothing to say about their relative intensities — why some spectral lines are bright and others faint. Both limitations were later resolved only by full quantum mechanics, in which an electron's state is not a classical orbit but a probability distribution, and a complete description needs four quantum numbers, not just $n$.

Board level - orbital radius and speed of the electron from the ionisation energy

It is found experimentally that 13.6 eV of energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the speed of the electron in the ground state of hydrogen.

Step 1. Convert the given energy to joules: $E = -13.6\ \text{eV} = -13.6\times1.6\times10^{-19}\ \text{J} = -2.2\times10^{-18}\ \text{J}$ (negative, since it is a bound state).

Step 2. Use $E = -\dfrac{e^2}{8\pi\varepsilon_0 r}$, rearranged for $r$: $r = -\dfrac{e^2}{8\pi\varepsilon_0 E} = \dfrac{(9\times10^9)(1.6\times10^{-19})^2}{2\times(2.2\times10^{-18})} = 5.3\times10^{-11}\ \text{m}$.

Step 3. Find the speed from the force-balance relation $v = \sqrt{\dfrac{e^2}{4\pi\varepsilon_0 mr}}$, or equivalently $v = e/\sqrt{4\pi\varepsilon_0 mr}$, using $m = 9.1\times10^{-31}\ \text{kg}$: $v = 2.2\times10^{6}\ \text{m/s}$. Answer: $r = 5.3\times10^{-11}\ \text{m}$ (the Bohr radius) and $v = 2.2\times10^{6}\ \text{m/s}$.

Board level - kinetic and potential energy in the ground state

The ground-state energy of the hydrogen atom is $-13.6\ \text{eV}$. Find the kinetic and potential energies of the electron in this state.

Step 1. Recall the general relations for a Coulomb-bound circular orbit: $K = -E$ and $U = 2E$ (since $E = K + U$ and $K = -U/2$, so $U = -2K$ and $E = K - 2K = -K$).

Step 2. Apply them with $E = -13.6\ \text{eV}$: $K = -E = 13.6\ \text{eV}$, and $U = 2E = -27.2\ \text{eV}$.

Step 3. Check consistency: $K + U = 13.6 + (-27.2) = -13.6\ \text{eV} = E$, which matches. Answer: $K = 13.6\ \text{eV}$ and $U = -27.2\ \text{eV}$.

Competitive level (JEE / NEET) - wavelength and frequency of the photon absorbed for a ground-to-n=4 transition

A hydrogen atom initially in the ground state ($n=1$) absorbs a photon and is excited to the $n=4$ level. Determine the wavelength and frequency of the absorbed photon.

Step 1. Find the energies of the two levels: $E_1 = -13.6\ \text{eV}$ and $E_4 = -13.6/16 = -0.85\ \text{eV}$.

Step 2. The absorbed photon supplies exactly the energy difference: $\Delta E = E_4 - E_1 = -0.85 - (-13.6) = 12.75\ \text{eV} = 12.75\times1.6\times10^{-19} = 2.04\times10^{-18}\ \text{J}$.

Step 3. Get the frequency from $\Delta E = h\nu$: $\nu = \dfrac{2.04\times10^{-18}}{6.63\times10^{-34}} = 3.08\times10^{15}\ \text{Hz}$.

Step 4. Get the wavelength from $c = \nu\lambda$: $\lambda = \dfrac{3\times10^{8}}{3.08\times10^{15}} = 9.7\times10^{-8}\ \text{m} = 97\ \text{nm}$. This lies in the ultraviolet, consistent with it belonging to the Lyman series ($n_f = 1$). Answer: $\lambda \approx 97\ \text{nm}$ and $\nu \approx 3.08\times10^{15}\ \text{Hz}$. Timing note: about ninety seconds. The fast route is to compute $E_n = -13.6/n^2$ for both levels first, subtract to get $\Delta E$ in eV, convert to joules only once, then chain through $\nu = \Delta E/h$ and $\lambda = c/\nu$ — do not convert eV to joules twice or round $\Delta E$ too early, since a rounding slip at 12.75 eV shifts the final wavelength by several nanometres. Recognising the series (Lyman, since $n_f=1$) is a free sanity check: Lyman lines must fall in the UV, so a wavelength above 400 nm would flag an arithmetic error.

Students routinely confuse the size ordering: the nucleus (about $10^{-15}$ m) is ten thousand to a hundred thousand times smaller than the atom (about $10^{-10}$ m), not merely ten or a hundred times — this is why an atom is described as almost entirely empty space. Another frequent slip is treating $n$ in $E_n = -13.6/n^2$ as a continuous variable; it is strictly a positive integer, and $E_n$ is negative for every bound state, becoming zero (not positive) only in the unbound limit $n \to \infty$. Many students also forget that the photon energy in a transition equals $E_i - E_f$ (initial minus final, for emission) and mix up which state is $n_i$ and which is $n_f$, especially when computing absorption, where the photon energy is $E_f - E_i$ with $f > i$. It is also easy to muddle the series names: Lyman terminates at $n_f = 1$ and lies in the ultraviolet, Balmer terminates at $n_f = 2$ and is the one with visible lines, and Paschen, Brackett and Pfund (terminating at $n_f = 3, 4, 5$) are all infrared — mixing up which series lands where on the electromagnetic spectrum is a common exam error. Finally, remember the Bohr model's scope: it applies only to hydrogenic (single-electron) systems, so trying to apply $E_n = -13.6\ Z^2/n^2\ \text{eV}$ or the raw hydrogen formulas directly to neutral helium or any multi-electron atom gives wrong answers, since electron-electron repulsion is not included anywhere in the model.

• State what the Geiger-Marsden alpha-scattering experiment observed and explain why the results forced the conclusion of a small, dense, positively charged nucleus.
• Define the impact parameter and explain qualitatively how it controls the scattering angle in Rutherford scattering.
• Derive the distance of closest approach for a head-on alpha-particle collision using conservation of energy.
• State Bohr's three postulates in your own words and explain what problem each one solves.
• Derive the Bohr-model expressions for orbit radius $r_n$ and total energy $E_n$, and quote the numerical values $a_0 = 5.3\times10^{-11}$ m and $E_1 = -13.6$ eV.
• Explain why the total energy of a bound electron is negative and connect this to the ionisation energy.
• Use $\Delta E = E_i - E_f = h\nu$ to compute the wavelength or frequency of a photon emitted or absorbed in a transition between two named levels.
• Identify the Lyman, Balmer, Paschen, Brackett and Pfund series by their final quantum number and their region of the electromagnetic spectrum.
• Explain de Broglie's standing-wave argument for why angular momentum is quantised in units of $h/2\pi$.
• State at least two limitations of the Bohr model and why each one arises from the assumptions built into the three postulates.

Ask the tutor: if you would like to work through the wavelength or energy of any specific hydrogen transition, or see the full algebra behind any of the three derivations above, just ask and we can step through it together.