12. Atoms

An atomic model earns its place by explaining evidence. Scattering tells us where charge is concentrated; spectra tell us which energy changes are possible. Constants in numerical examples are rounded classroom approximations.

Rutherford’s nuclear interpretation followed experiments in which Geiger and Marsden observed alpha-particle scattering. Bohr introduced quantised stationary states for hydrogen. Later matter-wave and quantum theories preserved the nuclear evidence while replacing literal classical electron orbits.

An atom is far too small to inspect with an ordinary microscope, so its structure must be inferred from evidence.

A successful atomic model must explain electrical neutrality, the distribution of positive and negative charge, the atom’s stability and the light it emits or absorbs. Each observation constrains the model differently. Explaining only one does not settle the whole structure.

In an early diffuse-charge model, electrons were embedded within spread-out positive charge. This picture could represent an electrically neutral atom, but it did not predict the observed frequency of rare, very large alpha-particle deflections. A later nuclear model concentrated positive charge and most mass in a tiny region.

The nuclear model explained scattering evidence but left important problems: why electrons did not continuously radiate away their energy in a classical orbit, and why atoms emitted particular spectral lines rather than every possible frequency. Bohr introduced a restricted set of allowed states to address these problems for hydrogen-like atoms.

A proposed atom is neutral but predicts no large-angle scattering and no discrete spectrum. Is neutrality enough to accept it?

Neutrality fixes total positive and negative charge. Scattering tests where charge and mass are concentrated. Spectra test allowed energy changes.

Result: No. It must account for all relevant evidence, not just its net charge.

1. What does electrical neutrality tell us about an atom?
2. Why are scattering and spectra complementary evidence?

Answers

1. Its total positive and negative charges balance.
2. Scattering probes spatial charge distribution; spectra probe energy differences.

A tiny fraction of strongly deflected projectiles can reveal a tiny, concentrated scattering centre.

In alpha-scattering experiments, a narrow beam of fast, positively charged particles meets a very thin metal foil. A surrounding detector records the directions of particles that emerge. The thin foil reduces multiple scattering, making a single close encounter a useful approximation for many events.

Most alpha particles pass with only a small deflection. This suggests that the atom’s positive charge does not fill its entire volume at high density. Rare large deflections require a strong repulsive force during a close encounter. Concentrating positive charge and most mass into a small nucleus explains both results together.

For an alpha particle approaching a positive nucleus, electrostatic repulsion changes the direction of momentum. A distant path receives a small impulse; a close path can turn through a large angle. The impact parameter is the perpendicular separation between the initial straight-line direction and the nuclear centre. It is not generally the closest-approach distance.

For a head-on encounter with a heavy, initially stationary nucleus, the alpha particle is momentarily at rest at the turning point. Neglect nuclear recoil. Energy conservation gives $K=k(2q_{\rm e})(Zq_{\rm e})/r_{\min}$. Thus $r_{\min}=2kZq_{\rm e}^2/K$. A larger incident kinetic energy permits a smaller closest approach.

An alpha particle of kinetic energy $6.0\,\mathrm{MeV}$ approaches a nucleus with $Z=79$ head-on. Use $kq_{\rm e}^2=1.44\,\mathrm{MeV\,fm}$ and neglect recoil.

At closest approach, the initial kinetic energy has become electrostatic potential energy.

$r_{\min}=2\times79\times1.44/6.0\,\mathrm{fm}$.

Result: $r_{\min}=37.92\,\mathrm{fm}$. This is a turning distance, not a direct measurement of nuclear radius.

In a classical circular-orbit picture, electrostatic attraction supplies centripetal force: $mv^2/r=kq_{\rm e}^2/r^2$ for hydrogen. The electron has inward acceleration. Classical electrodynamics predicts radiation from an accelerating charge, so such an orbit would lose energy and shrink.

A classical orbit therefore does not by itself explain a stable atom. This failure motivated a change in the allowed-state rules, not the abandonment of the nuclear evidence. Modern quantum mechanics does not assign the electron a tiny planet-like trajectory.

1. How does the head-on turning distance change when incident kinetic energy doubles?
2. What is the central classical stability problem?

Answers

1. It is halved in the fixed-heavy-nucleus Coulomb model.
2. An orbiting accelerating electron would radiate energy and spiral inward.

A line spectrum records selected energy differences, not a continuous range of electron energies.

A hot dense source can produce a broad continuous spectrum. A low-pressure excited gas instead often shows separated bright emission lines. Passing broadband light through a cooler gas can produce dark absorption lines. A spectrometer separates wavelengths so these patterns can be measured.

An atom can absorb a photon when its energy matches an allowed upward energy change. It can emit a photon during an allowed downward transition: $h\nu=E_i-E_f$ for emission, where $E_i>E_f$. Energy levels are properties of the atom; spectral lines are differences between pairs of levels.

Each type of atom has a characteristic set of possible transitions. This makes spectra useful for identifying materials at a distance. The intensity of a line also depends on populations and transition probabilities, so absence of a bright line does not always mean the associated element is absent.

Two allowed states differ by $2.48\,\mathrm{eV}$. Find the emitted wavelength using $hc=1240\,\mathrm{eV\,nm}$.

Use the energy difference as the photon energy, not either absolute state energy.

$\lambda=hc/\Delta E$.

Result: $\lambda=500\,\mathrm{nm}$.

1. Why is a spectrum made of lines instead of every wavelength?
2. Can several pairs of levels generate the same energy difference?

Answers

1. Only certain energy differences are allowed for the emitting atomic system.
2. Yes; a line position alone need not uniquely identify a single pair without additional information.

Bohr kept the nucleus but replaced the unrestricted classical orbit with a set of allowed stationary states.

• Allowed stationary states do not radiate continuously in the model.
• For a circular hydrogen orbit, angular momentum satisfies $mvr=n\hbar=nh/(2\pi)$, with positive integer n.
• Radiation is emitted or absorbed only when the atom changes state, with photon energy equal to the state-energy difference.

Combine the force equation $mv^2/r=kq_{\rm e}^2/r^2$ with $mvr=n\hbar$. Substituting $v=n\hbar/(mr)$ gives $n^2\hbar^2/(mr^3)=kq_{\rm e}^2/r^2$. Solving gives $r_n=n^2\hbar^2/(mkq_{\rm e}^2)=a_0n^2$ for hydrogen.

The kinetic energy is $K=kq_{\rm e}^2/(2r)$ and potential energy, with zero at infinite separation, is $U=-kq_{\rm e}^2/r$. Therefore $E=K+U=-kq_{\rm e}^2/(2r)$. Negative total energy means the electron is bound relative to the separated proton and electron.

$$r_n=a_0n^2,\qquad E_n=-\frac{13.6}{n^2}\,\mathrm{eV}\quad\text{(hydrogen, approximate)}$$

For a one-electron ion of nuclear charge Z, $r_n=a_0n^2/Z$ and $E_n=-13.6Z^2/n^2\,\mathrm{eV}$ in the simple fixed-nucleus treatment. The ground state has n=1. Excited states have larger n and less negative energy. The ionisation limit is E=0.

Find the radius and energy of hydrogen in n=3. Use $a_0=0.0529\,\mathrm{nm}$. How much energy is needed to ionise it from this state?

The radius grows as n squared. The energy has the reciprocal-square dependence with a negative sign. Ionisation raises the energy to zero.

Result: $r_3=0.4761\,\mathrm{nm}$; $E_3=-1.5111\,\mathrm{eV}$; ionisation energy $=1.51111\,\mathrm{eV}$.

Bohr’s model reproduces the main hydrogen spectrum and gives useful one-electron energy scales. It does not correctly describe general many-electron atoms, all line intensities, fine splitting or a modern electron probability distribution. Its circular paths are historical model elements, not literal modern quantum trajectories.

1. Why is the bound-state energy negative?
2. What is the energy ratio E₂/E₁ for hydrogen in this model?

Answers

1. The reference zero is separated electron and nucleus at rest infinitely far apart.
2. $0.25$. Both energies are negative.

Every hydrogen spectral series groups transitions with the same final principal quantum number.

For an emission from upper level $n_i$ to lower level $n_f$, subtract the Bohr energies: $\Delta E=13.6(1/n_f^2-1/n_i^2)\,\mathrm{eV}$. Combining with $hc/\lambda=\Delta E$ gives the Rydberg form.

$$\frac1\lambda=R_{\rm H}\left(\frac1{n_f^2}-\frac1{n_i^2}\right),\qquad n_i>n_f$$

Within a series, increasing the upper level makes the wavelength approach a minimum called the series limit. The energy spacing between high levels gets smaller, so neighbouring lines crowd together near that limit. A transition from infinity means a limiting state at zero binding energy; it is not an electron starting at an actual infinity-sized circular ring in the laboratory.

Calculate the hydrogen n=3 to n=2 photon energy and wavelength using $13.6\,\mathrm{eV}$ and $hc=1240\,\mathrm{eV\,nm}$.

$\Delta E=13.6(1/4-1/9)$.

The final state is n=2, so this belongs to the Balmer series.

$\lambda=1240/\Delta E$ in nm.

Result: $\Delta E=1.888889\,\mathrm{eV}$ and $\lambda\approx656.47\,\mathrm{nm}$.

If an excited atom cascades through several levels, each photon corresponds to one step. Total emitted photon energy equals the complete energy decrease, but the energy can be divided among different photons. For N distinct levels, at most $N(N-1)/2$ different pairs exist; not every possible line must be populated or allowed in every physical experiment.

1. Which series includes transitions ending at n=3?
2. Why do high-n lines become crowded?

Answers

1. Paschen.
2. Adjacent high-n levels have increasingly small energy separations.

An integer number of wavelengths around a closed orbit reproduces Bohr’s angular-momentum rule.

In a standing-wave interpretation of the historical circular model, the wave must join smoothly after one turn. Its circumference must contain a whole number of wavelengths: $2\pi r=n\lambda$. Insert de Broglie’s $\lambda=h/(mv)$ to obtain $2\pi r=nh/(mv)$ and hence $mvr=nh/(2\pi)=n\hbar$. A non-integer fit would not have the same phase after a complete turn.

This argument connects matter waves with quantisation and makes Bohr’s rule less arbitrary. It is still a model-level argument. A full quantum treatment solves for allowed wavefunctions and predicts probability densities rather than requiring a real electron to travel around a circular track.

For the nth Bohr orbit, $\lambda_n=2\pi r_n/n=2\pi a_0n$ for hydrogen. The radius grows as n squared while the wavelength along the idealised orbit grows as n. The speed decreases as the level number rises.

Use $a_0=0.0529\,\mathrm{nm}$ to find the de Broglie wavelength associated with n=2 in the circular Bohr model.

First use $r_2=4a_0$. The circumference contains two wavelengths, so $\lambda_2=2\pi r_2/2=4\pi a_0$.

Result: $\lambda_2=0.66476\,\mathrm{nm}$.

1. What condition makes the wave phase repeat after one turn?
2. Does a modern atomic orbital describe a fixed classical track?

Answers

1. $2\pi r=n\lambda$, with integer n.
2. No. It describes a quantum state and its spatial probability structure.

• A heavy-nucleus head-on Coulomb model gives $r_{\min}=2kZq_{\rm e}^2/K$.
• The simple hydrogen Bohr model gives $r_n=a_0n^2$ and $E_n=-13.6/n^2\,\mathrm{eV}$.
• A transition photon has $h\nu=|E_i-E_f|$.
• Spectral series use $1/\lambda=R_{\rm H}(1/n_f^2-1/n_i^2)$.
• The circular standing-wave condition $2\pi r=n\lambda$ motivates $mvr=n\hbar$.

• A turning distance in a repulsive Coulomb encounter is not automatically a nuclear radius.
• An energy-level diagram uses height to represent energy, not the electron’s physical position.
• Bohr’s model is historically important and quantitatively useful in a limited domain; modern atomic orbitals are not planetary tracks.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

12.1 Give two independent observations an atomic model should explain.

12.2 Why is a very thin foil used in a simple scattering experiment?

12.3 Distinguish impact parameter from closest approach.

12.4 For a head-on alpha encounter with $Z=47$, $K=5\,\mathrm{MeV}$ and $kq_{\rm e}^2=1.44\,\mathrm{MeV\,fm}$, find turning distance.

12.5 Why does doubling alpha kinetic energy reduce its head-on turning distance?

12.6 Why is a classical electron orbit unstable under classical electrodynamics?

12.7 State the three main rules in Bohr’s model.

12.8 Find r₂/r₁ and r₄/r₁ for hydrogen.

12.9 Find the hydrogen n=4 energy.

12.10 Find the ionisation energy from hydrogen n=2.

12.11 Find energy released in hydrogen n=4 to n=2.

12.12 Using $hc=1240\,\mathrm{eV\,nm}$, find the wavelength in the preceding transition.

12.13 Why is the ground-state energy negative?

12.14 Identify the series for transitions ending at n=1, n=2 and n=3.

12.15 Find the Balmer series-limit wavelength using $R=1.097\times10^7\,\mathrm{m^{-1}}$.

12.16 Why does a series have a minimum wavelength?

12.17 For a one-electron ion with Z=2, find ground-state energy and radius using a₀=0.0529 nm.

12.18 Can the elementary Bohr formulas be applied unchanged to a neutral multi-electron atom?

12.19 A model atom has five accessible levels. How many distinct pairs of levels exist?

12.20 Explain why a cascade can emit several photons but conserve the full transition energy.

12.21 An absorption line has photon energy $2.0\,\mathrm{eV}$. Does the atomic energy rise or fall, and by how much?

12.22 A circular matter-wave model has circumference $1.2\,\mathrm{nm}$ and contains three wavelengths. Find wavelength.

12.23 Derive mvr=nħ from the standing-wave circumference condition.

12.24 Why should a classroom atom diagram state that it is not to scale?

Answers

12.1 Examples: large-angle alpha scattering and discrete atomic spectra; neutrality and stability are additional requirements.

12.2 It reduces multiple encounters so a single-scattering interpretation is more useful.

12.3 Impact parameter is the initial perpendicular offset from the nuclear centre; closest approach is the minimum actual separation along the bent path.

12.4 $r_{\min}=2Z(1.44)/K=27.072\,\mathrm{fm}$, neglecting recoil.

12.5 Energy conservation gives turning distance inversely proportional to the initial kinetic energy.

12.6 The accelerating charge radiates, loses energy and cannot remain in the same orbit.

12.7 Stationary allowed states do not radiate continuously; angular momentum is an integer multiple of ħ; transitions exchange photons with energy equal to the state-energy difference.

12.8 $4$ and $16$.

12.9 $E_4=-0.85\,\mathrm{eV}$.

12.10 $3.4\,\mathrm{eV}$.

12.11 $13.6(1/4-1/16)=2.55\,\mathrm{eV}$.

12.12 $486.27\,\mathrm{nm}$.

12.13 Energy is referenced to an infinitely separated, stationary electron and nucleus; a bound system has lower energy.

12.14 Lyman, Balmer and Paschen, respectively.

12.15 $\lambda=4/R=364.63\,\mathrm{nm}$.

12.16 The largest energy jump into its fixed final level comes from the ionisation limit.

12.17 $E_1=-54.4\,\mathrm{eV}$ and $r_1=0.02645\,\mathrm{nm}$.

12.18 No. Electron interactions and the quantum state structure require a more complete treatment.

12.19 $5(5-1)/2=10$. This is a pair count, not proof that every line is allowed or populated.

12.20 The step energies add by cancellation of intermediate state energies, giving the same total initial-minus-final energy.

12.21 It rises by 2.0 eV in the absorbing transition.

12.22 $0.4\,\mathrm{nm}$.

12.23 Use $2\pi r=n\lambda$ and $\lambda=h/(mv)$; rearrange to $mvr=nh/(2\pi)$.

12.24 The nucleus and electron distribution occupy vastly different length scales, which cannot usually both be shown clearly in one small diagram.