Atoms

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 Physics Class XII, Chapter 12.

Why doesn't an atom collapse?

Every atom has a tiny positive nucleus and negative electrons circling it. Opposite charges attract, and a circling charge is an accelerating charge, which according to classical physics must radiate energy. Do the sum and an electron should crash into the nucleus in a fraction of a second. Yet the atoms in your body have lasted billions of years. This chapter is the story of how physicists discovered the nucleus, found that their own model was impossible, and fixed it with one bold idea: electrons can only occupy certain allowed orbits.

Thomson's plum pudding

J. J. Thomson had discovered the electron in 1897. In 1898 he suggested that an atom is a ball of uniformly spread positive charge with electrons embedded in it like seeds in a watermelon, just enough of them to make the atom neutral. This explained why atoms are neutral and roughly $10^{-10}$ m across. It could not explain why a hot, thin gas of atoms emits light only at a few sharp wavelengths, the line spectrum that acts as an element's fingerprint. Hot solids give a continuous rainbow. A gas of free atoms gives separate bright lines. Any correct model of the atom has to predict them.

The α-scattering experiment

Rutherford suggested, and his students Hans Geiger and Ernest Marsden carried out between 1908 and 1911, the experiment that ended the plum pudding. A narrow beam of 5.5 MeV α-particles (helium nuclei, charge $+2e$, mass about 7300 times that of an electron) was aimed at a very thin gold foil, about $2.1 \times 10^{-7}$ m thick, inside a vacuum chamber. A zinc sulphide screen with a microscope, which could rotate around the foil, recorded the flashes of light where α-particles landed.

Electrons are far too light to turn a heavy α-particle noticeably, much as a swarm of flies cannot stop a cricket ball. A diffuse positive pudding would only nudge an α-particle gently. Reversing one needs a single close encounter with a massive, positively charged object. So Rutherford concluded that all the positive charge and nearly all the mass sit in a tiny nucleus, with radius of the order of $10^{-15}$ to $10^{-14}$ m, which is $10^4$ to $10^5$ times smaller than the atom. If the nucleus were a 1 cm marble, the atom would be somewhere between 100 m and 1 km across. The electrons orbit that nucleus like planets around a sun. This is the nuclear (planetary) model. The number of α-particles detected falls steeply as the angle grows, in a way that agreed with Rutherford's Coulomb-force calculation over many orders of magnitude.

Impact parameter and distance of closest approach

The impact parameter $b$ is the perpendicular distance between the nucleus and the line along which the α-particle was originally heading. A large $b$ means a distant, gentle encounter and a small deflection. A small $b$ means a close encounter and a large deflection. For $b = 0$, a head-on collision, the α-particle is turned straight back ($\theta = 180^\circ$). The two are linked by $b = (Ze^2/(4\pi\varepsilon_0 K))\cot(\theta/2)$, where $K$ is the α-particle's kinetic energy.

In a head-on collision the α-particle slows down as the Coulomb repulsion grows, stops momentarily at the distance of closest approach $d$, then retreats. At that instant all its kinetic energy has become electrostatic potential energy, so

$$K = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d} \quad \Rightarrow \quad d = \dfrac{2Ze^2}{4\pi\varepsilon_0 K}$$

Because $e^2/4\pi\varepsilon_0 = 1.44 \text{ MeV fm}$ ($1 \text{ fm} = 10^{-15} \text{ m}$), you can do these sums in your head. The result is an upper limit for the nuclear radius, and it shows that $d$ is inversely proportional to $K$, so doubling the energy halves $d$.

Worked example: how close does an α-particle get? A 5.5 MeV α-particle is fired head-on at a gold nucleus ($Z = 79$). Find $d$.

The common wrong step: using charge $e$ instead of $2e$ for the α-particle, or $Z$ instead of $2Z$ in the numerator, which halves the answer. An α-particle carries two protons' charge. Also, an α-particle never touches the nucleus here, since d is larger than the sum of their radii. A more energetic 7.7 MeV α-particle gets closer, to about 30 fm, which shows the $1/K$ scaling.

Rutherford's atom is unstable

For an electron of charge $-e$ circling a nucleus of charge $+Ze$, the Coulomb attraction supplies the centripetal force:

$$\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2} = \dfrac{mv^2}{r}$$

so the kinetic energy is $\text{KE} = Ze^2/(8\pi\varepsilon_0 r)$ and the potential energy is $\text{PE} = -Ze^2/(4\pi\varepsilon_0 r)$. The total energy is

$$E = \text{KE} + \text{PE} = -\dfrac{Ze^2}{8\pi\varepsilon_0 r}$$

So $\text{KE} = -E$ and $\text{PE} = 2E = -2\,\text{KE}$. The total is negative because the electron is bound: you must supply |E| to free it. These relations hold for every Bohr orbit, and a question that gives you the kinetic energy K can be answered immediately: total energy $-K$, potential energy $-2K$.

The theory then fails twice. A circling electron is accelerating, and classical electromagnetism says an accelerating charge radiates, so the electron would lose energy, spiral inward and hit the nucleus in a tiny fraction of a second (a classical estimate gives about $10^{-11}$ s). And since the orbit would shrink continuously, the radiation would come out at all frequencies, a continuous spectrum, whereas real atoms give sharp lines. Both are failures of classical physics, and both are fixed by Bohr's postulates.

Bohr's three postulates

In 1913 Niels Bohr kept Rutherford's planetary picture but added three rules that classical physics does not allow.

Postulate 1: stationary orbits. An electron can revolve only in certain special orbits, and while it does so it does not radiate energy. Each such orbit has a fixed total energy. Postulate 2: quantised angular momentum. The allowed orbits are those where the electron's angular momentum is an integral multiple of $h/2\pi$:

$$L = mvr = \dfrac{nh}{2\pi} = n\hbar, \qquad n = 1, 2, 3, \ldots$$

Postulate 3: frequency condition. An electron jumping between two stationary states emits or absorbs one photon whose energy is the difference of the two energies:

$$h\nu = E_i - E_f$$

If the electron drops to a lower state ($E_i > E_f$), a photon is emitted. If it rises, a photon of exactly that energy must be absorbed.

Here $n$ is the principal quantum number. Postulate 1 repairs the instability problem and postulate 3 explains the line spectrum, since only certain energy differences, hence only certain wavelengths, exist.

The numbers that come out

Combine Coulomb's force balance with $L = n\hbar$. For hydrogen (and for any single-electron ion with nuclear charge Ze) you get these results.

The energies for hydrogen come out as $E_1 = -13.6 \text{ eV}$, $E_2 = -3.40 \text{ eV}$, $E_3 = -1.51 \text{ eV}$, $E_4 = -0.85 \text{ eV}$, $E_5 = -0.54 \text{ eV}$, and $E_\infty = 0$. The higher the orbit, the less negative (higher) the energy, and the closer together the levels get. The ionisation energy is the energy needed to take the electron from the ground state to $n = \infty$: 13.6 eV for hydrogen. The excitation energy to $n = 2$ is 10.2 eV, not 3.4 eV. Don't confuse the energy of a level with the gap between levels.

Hydrogen-like ions have one electron but a bigger nuclear charge: He⁺ ($Z = 2$), Li²⁺ ($Z = 3$). Their radii shrink by $1/Z$, and their energies grow by $Z^2$, which is why He⁺ has a ground state of −54.4 eV.

Worked example: a hydrogen electron in $n = 3$. Find its radius, speed, kinetic, potential and total energy, and the energy needed to ionise it from this level.

The common wrong step: calling the kinetic energy negative, or taking 13.6 eV as the ionisation energy from every level. 13.6 eV is the ionisation energy from the ground state only. An electron already in $n = 3$ needs far less.

Worked example: a hydrogen-like ion. Find the ground-state energy and radius of He⁺, and the energy and wavelength of the photon emitted when it falls from $n = 2$ to $n = 1$.

The common wrong step: forgetting $Z^2$, which would give 10.2 eV, the hydrogen value.

Why only these orbits? De Broglie's answer

Bohr simply asserted that $L = n\hbar$. In 1923 Louis de Broglie explained it. An electron has a wavelength $\lambda = h/mv$. A wave that circles an orbit must meet itself crest to crest after one trip, otherwise it cancels itself out on repeated laps. That means the circumference holds a whole number of wavelengths:

$$2\pi r_n = n\lambda = \dfrac{nh}{mv_n} \quad \Rightarrow \quad m v_n r_n = \dfrac{nh}{2\pi}$$

This is exactly Bohr's second postulate. Allowed orbits are standing matter waves around the nucleus, like the standing waves on a plucked guitar string.

Line spectra and the hydrogen series

When a gas of hydrogen is excited, electrons jump up and then fall back, and each fall emits a photon. Because the levels are discrete, only certain wavelengths come out. Pass white light through cool hydrogen instead, and the same wavelengths are absorbed, leaving dark lines on the continuous background. The wavelengths of the emission lines are given by

$$\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) \quad \text{with } R = 1.097 \times 10^7 \text{ m}^{-1}$$

where $n_f$ is the level the electron lands on and $n_i$ is the level it starts from. Which $n_f$ you land on defines the series:

The pattern is that the further down the electron lands, the larger the energy jump and the shorter the wavelength. Lyman lies deepest, so it holds the shortest wavelengths. The shortest-wavelength limits arrange in the order Lyman < Balmer < Paschen < Brackett < Pfund. An electron starting from level $n$ can fall by different routes, giving up to $n(n - 1)/2$ different lines.

Worked example: the Balmer lines. An electron in $n = 4$ falls to $n = 2$. Find the photon's wavelength, and count the lines possible from $n = 4$.

The common wrong step: writing $(1/n_i^2 - 1/n_f^2)$, which gives a negative wavelength. The larger n must come second in the bracket.

Worked example: absorption. A hydrogen atom in its ground state absorbs a photon of 12.75 eV. To which level does it go, and by how much does its angular momentum change?

The common wrong step: assuming a photon of more than 10.2 eV can be absorbed to some level in between. A photon is absorbed only if its energy exactly equals the gap between two levels (a photon of 10.5 eV would pass straight through). Only such exact gaps are absorbed.

Pause and predict: a beam of photons of energy 9.0 eV is shone on hydrogen atoms in the ground state. Does any atom get excited?

Check your answer: no. The smallest gap from $n = 1$ is 10.2 eV, so a 9.0 eV photon cannot lift the electron to any allowed level. The atoms are transparent to it.

Pause and predict: in which orbit is the electron fastest, and in which is it slowest to complete a lap?

Check your answer: fastest in $n = 1$ ($v \propto 1/n$), and its lap time is the shortest there ($T \propto n^3$). The higher the orbit, the slower the electron and the longer each lap.

What Bohr's model gets wrong

The model works for hydrogen and hydrogen-like ions (He⁺, Li²⁺), single-electron systems where only the nucleus attracts the electron. It fails for atoms with two or more electrons, because it ignores the repulsion between electrons, and even helium is beyond it. It also cannot predict the relative intensities of the spectral lines. These shortcomings led to the full quantum mechanics of Schrödinger and Heisenberg.

Beyond the textbook. For the exam: the electron moves in a circle with $L = n\hbar$, and the ground state has $L = \hbar$. In reality: modern quantum mechanics shows there are no definite orbits, only probability clouds, and the ground state of hydrogen has zero orbital angular momentum, not $\hbar$. The Bohr energy levels, however, turn out to be correct for hydrogen. The most probable distance of the electron from the nucleus in the ground state is still $a_0$. Why it matters: Bohr's pictures are a useful tool for energy-level questions, and the exam asks for exactly that.

Beyond the textbook. For the exam: the Balmer series is the visible series of hydrogen. In reality: the red line at 656 nm, the first Balmer line, is the reason glowing hydrogen clouds in space, such as the Orion Nebula, look pink-red in photographs. Astronomers use these lines to identify hydrogen and to measure how fast distant galaxies recede. The shifts in the line wavelengths give the speed. Why it matters: the same formula that explains a lab discharge tube reaches across the universe.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Dual Nature of Radiation and Matter supplies photon energy $h\nu$ and the de Broglie wavelength that turn Bohr's postulates into physics.
• Next: Nuclei goes inside the nucleus that Rutherford discovered here: its size, its binding energy and its radioactivity.
• Related: Electric Charges and Fields is where Coulomb's law comes from, and Electromagnetic Waves explains why an accelerating charge radiates, the very reason classical atoms collapse.