Structure of the Atom

Before Rutherford, the atom was pictured as a diffuse ball of positive charge with electrons dotted through it, so an alpha particle fired at it should barely be deflected.

In the Geiger-Marsden experiment, a narrow beam of alpha particles was fired at a very thin gold foil inside an evacuated chamber, and the scattered particles were detected as flashes on a movable fluorescent screen. Three observations carried the argument:

• Most alpha particles passed straight through, so the atom is mostly empty space.
• A small fraction was deflected through small angles, so a concentrated region of positive charge repels them.
• Roughly 1 in 8000 was deflected through more than $90^\circ$, so that region must be tiny, dense and contain almost all of the atom's mass.

The nuclear model that followed places a nucleus of diameter about $10^{-15}\ \text{m}$ inside an atom of diameter about $10^{-10}\ \text{m}$, with the electrons occupying the space outside it.

You must be able to describe the apparatus, pair each observation with its deduction, and quote this five-orders-of-magnitude difference in size between the atom and its nucleus.

A nuclide is a nucleus with a definite makeup of protons and neutrons, written $^{A}_{Z}X$. The proton number $Z$ (bottom) counts the protons and fixes which element $X$ is; the nucleon number $A$ (top) counts protons plus neutrons, so the number of neutrons is given by:

$$N = A - Z$$

For example, $^{14}_{6}\text{C}$ has $6$ protons and $14 - 6 = 8$ neutrons, and as a neutral atom it also has $6$ electrons.

Isotopes are nuclides with the same $Z$ but different $A$: carbon-12 and carbon-14 behave identically in chemical reactions because chemistry depends on the electrons, yet carbon-14 is radioactive while carbon-12 is stable.

The notation also makes nuclear equations easy to check, because in any nuclear process the total $A$ and the total $Z$ must balance on both sides.

You should be able to:

• read and write nuclide symbols
• count protons, neutrons and electrons for neutral atoms and ions
• pick out isotopes from a list of nuclides
• use conservation of $A$ and $Z$ to identify an unknown particle in a decay equation

Electrons in an atom can only occupy certain discrete energy levels; energies between the levels are simply not allowed. Energies are measured relative to a free electron, defined as $0\ \text{eV}$, so bound levels are negative: in hydrogen the ground state sits at $-13.6\ \text{eV}$ and the excited states crowd closer together as they approach zero.

The compelling evidence is line spectra. When a low-pressure gas is excited in a discharge tube, the light it emits contains only a few sharp, bright wavelengths (an emission spectrum); when white light passes through a cool sample of the same gas, exactly those wavelengths go missing, appearing as dark lines on a continuous background (an absorption spectrum).

A continuous range of electron energies would produce a continuous spectrum, so discrete lines directly demonstrate discrete levels. Because every element has its own ladder of levels, its spectrum is a fingerprint; astronomers identify the elements present in stars this way.

You must be able to sketch an energy-level diagram, explain why line spectra are evidence for quantized energy levels, and distinguish emission spectra from absorption spectra.

A transition between two levels involves exactly one photon whose energy equals the gap:

$$\Delta E = |E_f - E_i| = hf = \frac{hc}{\lambda}$$

In emission, an electron drops to a lower (more negative) level and the atom releases a photon; in absorption, an incoming photon is absorbed only if its energy matches a gap exactly, lifting the electron to a higher level. A bigger energy gap therefore means a higher frequency and a shorter wavelength.

In hydrogen, the drop from $-3.40\ \text{eV}$ to $-13.6\ \text{eV}$ releases a $10.2\ \text{eV}$ photon at about $122\ \text{nm}$, in the ultraviolet, while smaller drops between higher levels give visible and infrared lines.

Calculations usually require converting between electronvolts and joules using $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ before applying $E = hf$.

You must be able to compute the photon frequency or wavelength for a given transition, mark transitions as arrows on an energy-level diagram (downward for emission, upward for absorption), and count the number of different photon energies available from $n$ levels, which is $\frac{n(n-1)}{2}$.

Rutherford scattering offers two routes to nuclear size. First, for a head-on approach, all of the alpha particle's kinetic energy converts to electrostatic potential energy at the distance of closest approach $d$, where:

$$E_k = \frac{kq_1q_2}{d}$$

This gives an upper limit on the nuclear radius, not the radius itself.

Second, the classical analysis assumes the incoming alpha particle and the target nucleus are point-like charges interacting purely via electrostatic Coulomb repulsion. As the kinetic energy of the alpha particle increases, it penetrates deeper into the Coulomb barrier, getting closer to the nucleus.

When the alpha particle approaches to within a few femtometres of the nuclear surface (a centre-to-centre separation of the order of $10^{-14}\ \text{m}$ for a gold nucleus), the attractive strong nuclear force begins to act, drastically altering the scattering trajectory. Consequently, a graph of scattering intensity against energy deviates from classical predictions beyond a threshold energy, which provides a direct, non-destructive method for determining the physical size of the nucleus.

Such measurements support the empirical radius formula:

$$R = R_0 A^{1/3}$$

Here $R_0 \approx 1.2 \times 10^{-15}\ \text{m}$. Because $R^3 \propto A$, nuclear volume grows in proportion to nucleon number, so nuclear density is essentially constant across all nuclides, about $2.3 \times 10^{17}\ \text{kg m}^{-3}$.

You must be able to:

• calculate closest-approach distances
• apply the radius formula
• estimate nuclear density
• explain why high-energy scattering departs from the Coulomb model

Classically, an orbiting electron is accelerating, so it should radiate continuously, lose energy and spiral into the nucleus; atoms should not be stable at all. Bohr's resolution combines classical planetary orbits with quantization: only orbits in which the angular momentum satisfies the condition

$$mvr = \frac{nh}{2\pi}$$

for integer $n$ are allowed, and an electron in such an orbit does not radiate.

The mechanics comes from setting the electrostatic force as the centripetal force, $\frac{m v^2}{r} = \frac{ke^2}{r^2}$. The quantization condition is equivalent to fitting a whole number of de Broglie wavelengths around the orbit's circumference.

Solving these two equations simultaneously yields quantized orbital radii $r_n = \frac{n^2 h^2}{4\pi^2 k m e^2}$ and quantized total energies:

$$E_n = -\frac{13.6}{n^2}\ \text{eV}$$

These successfully explain why hydrogen atoms only emit or absorb light at the very specific frequencies observed in its spectral series.

The model works only for hydrogen and other one-electron systems and cannot predict the relative brightness of spectral lines, but it remains the standard quantitative bridge between classical orbits and full quantum theory.

You must be able to state the quantization postulate, carry out the two-equation derivation for $v$, $r$ or $E$, and calculate photon energies for transitions between Bohr levels.