E.1 Structure of the atom

Set up the evidence

In the Geiger–Marsden–Rutherford experiment, alpha particles were directed at a thin gold foil and detected around the foil. Most particles passed through without deflection, some were deflected, and a very small number scattered backwards.

Infer the nuclear model

The results show that an atom is mostly empty space. The rare large deflections require a small, dense, positively charged nucleus that contains most of the atom’s mass; the positive charge cannot be spread uniformly through the whole atom.

Keep the conclusion qualitative

For SL, focus on linking each observation to the model: many undeflected particles imply empty space, while rare back-scattering implies a concentrated repulsive centre. Do not treat the experiment as evidence that electrons occupy fixed-radius orbits.

Common trap

Do not say that all alpha particles are deflected. The dominant observation is that most pass through essentially undeflected; the large-angle events are rare but decisive.

Read the symbol

Nuclear notation is written as ${}^{A}_{Z}X$. The chemical symbol $X$ identifies the element, the proton number $Z$ is written below, and the nucleon number $A$ is written above.

Count the nucleus

The nucleus contains $Z$ protons and $N=A-Z$ neutrons. The number of electrons is not encoded by $A$ and $Z$; for a neutral atom it equals $Z$, while an ion has gained or lost electrons.

Compare nuclides

Atoms of the same element have the same $Z$. Isotopes have the same $Z$ but different $A$, so they contain different numbers of neutrons.

Common trap

Do not use the electron count as the proton number for an ion, and do not confuse $A$ with the number of neutrons. Subtract $Z$ from $A$ to find the neutron number.

Emission lines

An excited gas emits light at particular frequencies, producing bright spectral lines rather than a continuous spread of frequencies. Each line corresponds to a permitted energy difference between atomic states.

Absorption lines

When continuous light passes through a cooler gas, the atoms remove the same frequencies they can emit. The resulting dark lines therefore occur at specific, repeatable wavelengths.

Infer discrete levels

Because only particular photon energies are emitted or absorbed, the atom’s energy states are discrete rather than continuous. The spectrum is evidence for quantized atomic energy levels.

Common trap

Do not treat every visible line as a separate element without considering transitions. A spectrum is evidence of allowed energy differences; the pattern, not simply the number of lines, carries the information.

Emission

When an electron moves from a higher atomic energy level to a lower one, the atom emits one photon. The photon energy equals the level difference: $E_\gamma=\Delta E$.

Absorption

An atom can absorb a photon only when its energy matches an allowed upward transition. The electron then moves to the higher level, so the spectrum records the same allowed energy differences in reverse.

Read a transition diagram

For each downward arrow, calculate the energy gap between its initial and final levels. A larger gap produces a higher-frequency photon and a shorter wavelength; a smaller gap produces a lower-frequency photon and a longer wavelength.

Common trap

Do not use the absolute energy of one level as the photon energy. A photon is associated with the difference between two levels, and emission requires a downward transition.

Worked example — hydrogen transition

A transition with $E_\gamma=1.89\,\mathrm{eV}$ has energy $(1.89)(1.60\times10^{-19})=3.02\times10^{-19}\,\mathrm{J}$. Hence $f=E/h=(3.02\times10^{-19})/(6.63\times10^{-34})=4.56\times10^{14}\,\mathrm{Hz}$ and $\lambda=c/f=6.58\times10^{-7}\,\mathrm{m}$.

Connect energy to a level gap

For an atomic transition, use $E_\gamma=|E_i-E_f|$. Take the magnitude of the energy difference, then convert units consistently before finding frequency or wavelength.

Predict the wavelength

A larger energy gap gives a higher-frequency photon and a shorter wavelength. Therefore the longest wavelength comes from the smallest non-zero transition energy.

Common trap

Do not carry a negative sign from bound-state energies into photon energy. The photon energy is positive and equals the magnitude of the level difference.

Treat a spectrum as a fingerprint

Each element has a characteristic set of emission and absorption wavelengths because its allowed energy differences are unique. The pattern can therefore identify the chemical species producing or absorbing the light.

Use comparison evidence

Record the observed spectral lines and compare their wavelengths or frequencies with laboratory spectra of known elements. Matching several characteristic lines supports the identification.

Apply it to stars

Light from a star can contain absorption lines produced by cooler gases in its atmosphere. Comparing those lines with known spectra reveals which elements are present, even when the source cannot be sampled directly.

Common trap

Do not identify an element from one broad colour alone. The evidence is the set of matching spectral lines and their wavelengths.

Worked example — gold-197

For $A=197$, $R=(1.20\times10^{-15})(197)^{1/3}=6.98\times10^{-15}\,\mathrm{m}$. Since volume is proportional to $R^3\propto A$ while nuclear mass is also approximately proportional to $A$, the model predicts approximately constant nuclear density.

Infer the density

Nuclear volume scales as $R^3$, so $V\propto A$. Since nuclear mass is approximately proportional to $A$, the mass per unit volume is approximately constant across nuclei.

Scale carefully

If $A$ changes by a factor of $k$, radius changes by $k^{1/3}$, not by $k$. The density remains approximately unchanged in this model.

Common trap

Do not assume a nucleus with eight times the nucleon number has eight times the radius. It has twice the radius and approximately the same density.

Start with Rutherford scattering

At moderate energies, alpha-particle scattering can be modelled as electrostatic repulsion from a concentrated positive nucleus. The predicted deflections follow the Rutherford picture.

Read the high-energy deviation

At sufficiently high alpha-particle energies, the particles can approach more closely and the observed scattering departs from the electrostatic prediction. This provides evidence that the nucleus has a finite size and that a short-range strong interaction becomes relevant.

State what the evidence supports

The deviation is evidence about the nuclear scale and the interaction at very small separation. It is not evidence about the size of the alpha particle or the weak force.

Common trap

Do not continue applying pure Coulomb scattering after the experiment has entered the regime where the alpha particle probes the nuclear force.

Use energy conservation

For a head-on alpha particle, the initial kinetic energy is converted into electric potential energy as the particle approaches the positive nucleus. At the turning point, the radial kinetic energy is zero.

Set the energies equal

At the turning point the radial kinetic energy is zero, so the initial kinetic energy equals the electric potential energy of the repulsive alpha-particle–nucleus system. Both positive charges must be included.

Worked example — alpha particle toward gold

For $E_k=5.0\,\mathrm{MeV}=8.0\times10^{-13}\,\mathrm{J}$, $q_\alpha=2e$ and $q_N=79e$, $r_{\min}=k(2e)(79e)/E_k=4.5\times10^{-14}\,\mathrm{m}$. This is a turning-point distance, not automatically the nuclear radius.

Check the turning point

At closest approach the alpha particle has momentarily stopped moving toward the nucleus, then reverses. A larger initial kinetic energy gives a smaller closest-approach distance.

Common trap

Do not use the charge of gold alone: the interaction contains both $q_\alpha$ and $q_{nucleus}$. Also do not leave energy in MeV while using $k$ in SI units.

Worked example — fifth level

For $n=5$, $E_5=-13.6/5^2=-0.544\,\mathrm{eV}$. In joules this is $(-0.544)(1.60\times10^{-19})=-8.70\times10^{-20}\,\mathrm{J}$. Keep the negative sign for the bound level; use a positive energy difference for a photon.

Find a transition energy

For a transition between levels, calculate $\Delta E=|E_i-E_f|$. Emission occurs for a downward transition and absorption for an upward transition; the photon then obeys $E_\gamma=hf=hc/\lambda$.

Compare levels

The gaps are not equally spaced. A transition involving low $n$ can have a larger energy difference than one involving high $n$, so compare the actual level values rather than relying on the visual spacing of an unscaled sketch.

Common trap

Do not omit the negative sign while identifying the level, but do use the positive magnitude of the difference when calculating photon energy.

Worked example — $n=4$

For $n=4$, $L=4h/(2\pi)=4(6.63\times10^{-34})/(2\pi)=4.22\times10^{-34}\,\mathrm{kg\,m^2\,s^{-1}}$. An intermediate value is not an allowed Bohr-orbit angular momentum.

Connect quantization to energy

Only selected radii, speeds and total energies are allowed. The electron cannot occupy an intermediate orbit energy in this model, which explains discrete atomic levels and line spectra.

Use ratios efficiently

For hydrogen, combining the quantization condition with the electrostatic circular-orbit model gives $r_n\propto n^2$ and $v_n\propto 1/n$. If $r_2/r_1=4$, then $v_2/v_1=1/2$.

Common trap

Do not say that quantization fixes only the radius. The condition restricts angular momentum and leads to discrete allowed energies; do not treat $mvr$ as an arbitrary continuous value.