Atoms

12.1 Introduction — Thomson's model and why it failed

• By 1897 J. J. Thomson's discharge-tube experiments had shown that atoms of every element contain identical negatively charged electrons, yet the atom as a whole is neutral. So an atom must also carry positive charge.
• Thomson's plum-pudding (watermelon) model, 1898: the positive charge is spread uniformly through the whole volume of the atom, with the electrons embedded in it like seeds in a watermelon.
• Consequences of the model: the mass distribution is nearly continuous, the electrons sit in stable electrostatic equilibrium, and the electric field inside is weak everywhere — so a fast $\alpha$-particle could only ever be deflected through a very small angle.
• Killed by the Geiger-Marsden experiment: the observed large-angle scattering (1 in 8000 beyond $90^\circ$) is impossible if the positive charge is diffuse.
• CBSE's own comparison: the atomic size is no different in the two models ($\sim10^{-10}$ m); in Thomson's model electrons are in stable equilibrium, in Rutherford's they always experience a net force; a classical atom based on Rutherford's model is doomed to collapse; the mass distribution is nearly continuous in Thomson's and highly non-uniform in Rutherford's; the positively charged part carries most of the mass in both models.

12.2 The Geiger-Marsden ($\alpha$-scattering) apparatus, in words

• Source: a ${}^{214}_{83}\text{Bi}$ radioactive source emitting $\alpha$-particles of $5.5$ MeV. An $\alpha$-particle is a helium nucleus ${}^{4}_{2}\text{He}$, charge $+2e$, mass $6.64\times10^{-27}$ kg.
• Collimator: the emitted particles pass through a channel bored in lead bricks, which absorbs everything travelling in the wrong direction and leaves a narrow parallel beam.
• Target: a very thin gold foil, thickness $2.1\times10^{-7}$ m. Thin, so that an $\alpha$-particle suffers at most one scattering on the way through; gold, because it is heavy ($Z=79$, nucleus about $50$ times heavier than an $\alpha$-particle, so it stays effectively stationary) and rolls into very thin sheets.
• Detector: a zinc sulphide (ZnS) screen with a microscope, mounted so that it can be rotated about the foil to any angle $\theta$. Each $\alpha$-particle striking the screen produces a tiny flash of light — a scintillation — which is counted through the microscope.
• The whole apparatus sits in an evacuated (vacuum) chamber, so that air molecules do not scatter or absorb the beam.
• Measurement made: the number of scintillations per unit time as a function of the scattering angle $\theta$.

What was observed

• Most of the $\alpha$-particles passed straight through, undeviated.
• About $0.14\%$ were scattered by more than $1^\circ$.
• About $1$ in $8000$ was deflected by more than $90^\circ$; a few came almost straight back ($\theta\approx180^\circ$).
• The count $N(\theta)$ falls very steeply with angle, following $N(\theta)\propto 1/\sin^4(\theta/2)$.

What it proves

12.2 Rutherford's nuclear (planetary) model

• All the positive charge $+Ze$ and almost all the mass are concentrated in a tiny central nucleus of radius $10^{-15}$ to $10^{-14}$ m; the atom's radius is $10^{-10}$ m, i.e. $10^4$ to $10^5$ times larger.
• Electrons revolve around the nucleus in circular orbits, the Coulomb attraction supplying the centripetal force, just as planets orbit the sun under gravity.
• Results worth memorising: $r=\dfrac{e^2}{4\pi\varepsilon_0 mv^2}$; $K=\dfrac{e^2}{8\pi\varepsilon_0 r}$; $U=-\dfrac{e^2}{4\pi\varepsilon_0 r}$; $E=-\dfrac{e^2}{8\pi\varepsilon_0 r}$. Negative $E$ means bound.

Impact parameter and trajectory

• Impact parameter $b$: the perpendicular distance of the initial velocity vector of the $\alpha$-particle from the centre of the nucleus, measured while the particle is still far away.
• The trajectory is a hyperbola with the nucleus at the outer focus, because the force is an inverse-square repulsion.
• Small $b$ gives large $\theta$; $b=0$ (head-on) gives $\theta=180^\circ$; large $b$ gives $\theta\approx0^\circ$. Relation: $b=\dfrac{Ze^2\cot(\theta/2)}{4\pi\varepsilon_0 E}$.
• Because only a few particles come back, only a few have very small $b$ — so the target must be tiny. This is how the experiment gives an upper limit on the nuclear size.

Distance of closest approach

• Definition: the distance from the nucleus at which the whole of the kinetic energy of the head-on $\alpha$-particle has been converted into electrostatic potential energy, so that it momentarily stops.
• $r_0=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{K}$, so $r_0\propto\dfrac{1}{K}$ and $r_0\propto$ charge of the projectile.
• Double the kinetic energy, $r_0$ halves. Replace the $\alpha$-particle by a proton of the same energy, $r_0$ halves (charge $+e$ instead of $+2e$).
• Numbers CBSE has used: $7.7$ MeV on gold gives $r_0=3.0\times10^{-14}$ m $=30$ fm; $5.12$ MeV gives $44.4$ fm; $2.56\times10^{-12}$ J gives $14.2$ fm.
• $r_0$ always comes out larger than the true nuclear radius (about $7$ fm for gold) because the $\alpha$-particle turns round without ever touching the nucleus — both are treated as point charges. So $r_0$ is an upper limit on the nuclear size, not the size itself.

Two limitations of the Rutherford model

• Instability: a revolving electron is accelerating, so by classical electromagnetic theory it must radiate continuously, spiral inward and fall into the nucleus in about $10^{-8}$ s. Atoms are in fact stable.
• Spectrum: as the electron spirals in, its frequency of revolution changes continuously, so the emitted radiation would be a continuous spectrum. Atoms actually emit sharp, discrete line spectra characteristic of the element.

12.3 Atomic spectra

• Dense matter (solids, liquids, dense gases) emits a continuous spectrum, because radiation comes from atoms interacting with their neighbours.
• A rarefied gas excited by an electric discharge emits only certain discrete wavelengths, because the atoms are far apart and radiate individually. This is an emission line spectrum: bright lines on a dark background.
• Absorption spectrum: pass white light through the cool gas and the same wavelengths appear as dark lines on a continuous bright background.
• Every element has its own fixed set of lines — a fingerprint used to identify the gas.
• Balmer's empirical formula (1885) for the visible hydrogen lines: $\dfrac{1}{\lambda}=R\left[\dfrac{1}{2^2}-\dfrac{1}{n^2}\right]$, $n=3,4,5,\dots$ Generalised by Rydberg to $\dfrac{1}{\lambda}=R\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]$.

12.4 Bohr's three postulates (learn the wording)

• First (stationary orbits): an electron in an atom can revolve in certain stable orbits without emitting radiant energy, contrary to classical electromagnetic theory. Each such state has a definite total energy and is called a stationary state.
• Second (quantisation of angular momentum): the electron revolves only in those orbits for which the angular momentum is an integral multiple of $h/2\pi$, i.e. $L=mvr=\dfrac{nh}{2\pi}$, $n=1,2,3,\dots$
• Third (frequency condition): an electron may jump from one non-radiating orbit to another of lower energy, emitting a single photon whose energy equals the difference of the two state energies: $h\nu=E_i-E_f$. Absorption is the same process reversed: $E_i+h\nu=E_f$.
• Postulate one removes the stability problem; postulate three removes the continuous-spectrum problem.

Standard Bohr results

12.4.1 The energy-level diagram, described in words

• Draw a vertical energy axis marked in eV, with $0$ eV at the top and increasingly negative values downward.
• Horizontal lines mark the allowed states: $n=1$ far down at $-13.6$ eV, $n=2$ at $-3.40$ eV, $n=3$ at $-1.51$ eV, $n=4$ at $-0.85$ eV, $n=5$ at $-0.54$ eV, $n=6$ at $-0.38$ eV, converging on $n=\infty$ at $0.00$ eV.
• The lines crowd closer and closer together as $n$ increases, because $E_n\propto1/n^2$; everything above $0$ eV is a continuum of free-electron states, usually shaded.
• Downward arrows between lines are emission transitions, upward arrows absorption. Label the arrow with the photon energy $\lvert \Delta E\rvert$ or with the wavelength.
• Group the arrows by their lower level: all arrows ending on $n=1$ are the Lyman series, on $n=2$ the Balmer series, on $n=3$ the Paschen series, on $n=4$ the Brackett series, on $n=5$ the Pfund series.
• The single longest arrow shown, $n=\infty\to1$ (or the reverse), is the ionisation of hydrogen, $13.6$ eV.

12.5 Line spectra of hydrogen

• Photon energy $h\nu_{if}=E_{n_i}-E_{n_f}$. Since $n_i$ and $n_f$ are integers, only discrete frequencies appear — the line spectrum.
• Bohr's theory reproduces Rydberg's formula and predicts $R=\dfrac{me^4}{8\varepsilon_0^2h^3c}$, in excellent agreement with the measured $1.097\times10^7$ m$^{-1}$. Bohr received the 1922 Nobel Prize for this.

• Longest wavelength in a series = smallest energy gap = transition from $n_1+1$. Shortest wavelength (series limit) = largest gap = transition from $\infty$.
• Number of lines emitted when an atom de-excites from level $n$: $N=\dfrac{n(n-1)}{2}$. Exciting hydrogen to $n=4$ therefore gives $6$ lines: $3$ Lyman ($4\to1,3\to1,2\to1$), $2$ Balmer ($4\to2,3\to2$) and $1$ Paschen ($4\to3$).
• Electron-beam bombardment: a beam of energy $E$ excites hydrogen to the highest $n$ with $13.6\left[1-1/n^2\right]\le E$. A $12.75$ eV beam reaches exactly $n=4$; a $12.5$ eV beam gives $n\approx3.5$, so the atom is excited only to $n=3$ and only Lyman and Balmer lines follow.

12.6 de Broglie's explanation of the second postulate

• de Broglie (1923): matter has wave nature, $\lambda=h/p$; verified for electrons by Davisson and Germer (1927).
• Treat the orbiting electron not as a point particle but as a wave spread around the whole orbit. Like a wave on a plucked string, only those wavelengths survive that form a standing wave; all others interfere destructively with themselves after one circuit and die away.
• Standing-wave condition on a circle: the circumference must contain a whole number of wavelengths, $2\pi r_n=n\lambda$.
• Put $\lambda=h/mv_n$: $\;2\pi r_n=\dfrac{nh}{mv_n}$, i.e. $mv_nr_n=\dfrac{nh}{2\pi}$ — Bohr's second postulate, now explained rather than assumed.
• The diagram to draw: a circle with a smooth wavy line drawn along it, closing on itself after exactly $n$ crests-and-troughs (NCERT Fig. 12.8 shows $n=4$, four wavelengths round the circumference).
• Quantised orbits and quantised energies are therefore consequences of the wave nature of the electron.