E.1 Structure of the atom | IB DP Physics

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This topic starts with the most famous experiment in physics and ends with the model that replaced the one it established. The through-line is evidence: every claim here was forced on physicists by something they measured.

🎯What you need to be able to do

Use nuclear notation $ ^{A}_{Z}\text{X} $, and state what protons, neutrons and electrons contribute in mass and charge.
Describe the nuclear model of the atom, including its scale — and state the limitation that eventually broke it.
Describe the Rutherford–Geiger–Marsden experiment and explain what each observation shows.
Distinguish emission from absorption spectra, and explain both in terms of quantized energy levels.
Use $ E = hf $ to relate a photon’s energy to the gap between two levels.
HLUse $ R = R_0A^{1/3} $, and show that nuclear density is the same for every nucleus.
HLFind the distance of closest approach of an alpha particle from energy conservation.
HLUse the Rydberg formula, and describe the Bohr model with its successes and its limitations.

🧩What everything is made of

All matter, living or otherwise, is built from about a hundred different kinds of atom. Atoms of a single kind form an element, each with a name and a chemical symbol — H for hydrogen, O for oxygen — and the full list is the periodic table. Every atom is made of just three things: protons, neutrons and electrons.

\[ ^{A}_{Z}\text{X} \]

$Z$ is the atomic number, the number of protons, and it is what decides which element you have. $A$ is the nucleon number (or mass number), the number of protons plus neutrons. So the neutron count is $ A - Z $, which is worth writing down as a formula because it is asked for constantly.

Nuclear notation set out around a chemical symbol X, with the nucleon number A written at the upper left and the atomic number Z at the lower left. A is labelled as the number of protons plus neutrons and Z as the number of protons, which fixes the element. Beside it a worked instance shows uranium-235 written with 235 at the upper left and 92 at the lower left, giving 92 protons, 143 neutrons and, in a neutral atom, 92 electrons. Beneath, a table gives the relative mass and charge of each particle: proton mass 1 charge plus 1, neutron mass 1 charge neutral, electron mass negligible charge minus 1. — Because $Z$ alone fixes the element, the symbol and $Z$ carry the *same* information — which is why $Z$ is often left off entirely, and why an exam question can ask you to justify writing either one.

✏️Worked example 1 — reading and writing the notation

A nucleus of one form of uranium contains 92 protons and 143 neutrons. Write it in nuclear notation, and state how many electrons the neutral atom has.

Atomic number. $ Z = 92 $, the proton count — and 92 is uranium, so the symbol is U.

Nucleon number. $ A = 92 + 143 = 235 $.

\[ ^{235}_{\ 92}\text{U} \]

A neutral atom has as many electrons as protons, so 92 electrons.

$A$ is always the bigger number, because it counts the protons and the neutrons. If you have written the smaller one on top, you have them the wrong way round. The only nucleus for which they are equal is ordinary hydrogen, $ ^{1}_{1}\text{H} $, which has no neutron at all.

⚛️The nuclear model, and how empty it is

In the nuclear model, a very small central nucleus containing the protons and neutrons — collectively the nucleons — is surrounded by electrons arranged in different energy levels. Essentially all the mass and all the positive charge is in the nucleus; the electrons supply all the negative charge and almost none of the mass.

The numbers are what make this strange. An atom is about $ 10^{-10} $ m across and its nucleus about $ 10^{-15} $ m — a factor of 100 000. Cube that and the nucleus occupies roughly one part in $ 10^{15} $ of the atom’s volume. The vast majority of an atom is nothing at all.

The scale of an atom. A large faint sphere represents the whole atom, about ten to the minus ten metres across, with electrons shown as diffuse clouds rather than dots, the caption noting that their exact positions are not known but that different orbitals correspond to different energy levels. A magnified inset shows the nucleus at the centre, about ten to the minus fifteen metres across, containing protons and neutrons packed together. A scale note states that the atom is a hundred thousand times wider than its nucleus, so the nucleus occupies about one part in ten to the fifteen of the atom's volume, and that if the nucleus were a marble the atom would be a kilometre wide. — Electrons are drawn as clouds, not dots, because their exact positions are not known — only the regions where they are likely to be found. Those regions are the energy levels that the rest of this page is about.

This model contains the seed of its own destruction, and that is the point. An electron going round a nucleus is continually changing direction, so it is accelerating — and accelerating charges are known to radiate energy. An orbiting electron should therefore lose energy continuously and spiral into the nucleus in a fraction of a second. Atoms plainly do not do this. The nuclear model is right about where the mass and charge are and wrong about what the electrons are doing, and resolving that is what quantum theory was built to do.

🔬The evidence: Rutherford, Geiger and Marsden

The nuclear model is strange enough that it needs good evidence, and the best of it comes from one experiment. Positive alpha particles were fired at a very thin gold foil, about $ 10^{-8} $ m thick, in a vacuum, and a movable detector counted how many arrived at each angle.

The expectation, on the then-current model of the atom as a diffuse blob of positive charge, was that almost everything would pass more or less straight through. What they found was:

Most alpha particlespassed straight through, barely deviated

Somewere deflected through large angles

About 1 in 8000came back — deflected through more than 90°

$On the left, the apparatus: a source of alpha particles behind lead screens produces a narrow beam that strikes a thin gold foil target in a vacuum, with a movable detector on an arm that can be swung round to count particles arriving at any angle theta. Most particles are shown passing straight through, some deviated through a large angle, and about one in eight thousand repelled straight back. On the right, the atomic explanation: a stream of alpha particles approaches a row of gold atoms, each drawn with a tiny positive nucleus at its centre; particles passing far from any nucleus continue almost undeviated, those passing close are deflected through large angles by electrostatic repulsion, and one heading almost directly at a nucleus is turned back the way it came. A note says the diagram is not to scale and that only a minute fraction are scattered.$

Each observation forces a separate conclusion, and exam questions ask for them as a matched pair — so learn them as pairs, not as a list.

Most pass straight throughthe atom is mostly empty space

A few deflect a lotthe positive charge is concentrated in a tiny volume

Some come backthat concentration is also very massive — far heavier than an alpha particle

The mathematics went further than the qualitative picture. The number of particles deflected through any given angle matched the prediction of an inverse-square law of repulsion from a point-like nucleus — which is Coulomb’s law from D.2, turning up as the explanation for a scattering pattern.

🌈Emission and absorption spectra

Give an element enough energy and it emits light. Split that light with a prism or a diffraction grating and you do not get a continuous band of colour. You get a handful of sharp lines at particular wavelengths — an emission spectrum — and which lines you get depends on the element.

Now do the reverse. Shine light containing all frequencies through the same element as a cool gas, and exactly those same wavelengths are missing from what comes out, leaving dark lines on a continuous background. That is an absorption spectrum.

Two experimental arrangements and their results. In the emission set-up, a sample of hot gas emits light that passes through a slit and then a prism, producing a few bright coloured lines on a dark background. In the absorption set-up, a source of all frequencies shines through a cool sample of the same gas and then through the slit and prism, producing a continuous coloured band interrupted by dark lines. Beneath, the emission and absorption spectra of hydrogen are drawn one above the other on a common wavelength scale from 400 to 700 nanometres, showing that the bright lines of the first fall at exactly the same wavelengths as the dark lines of the second. — The bright lines and the dark lines are at **exactly the same wavelengths**. That is the single most important fact about these two spectra, and it is what tells you they have a common cause.

Because the pattern is unique to each element, a spectrum is a fingerprint. The yellow-orange glow of a street lamp is sodium announcing itself; the dark lines in sunlight tell you which elements are in the Sun’s outer layers, without going there.

💡Why they are lines: the energy levels are quantized

Electrons in an atom are bound to the nucleus — they cannot escape without being given energy, and if one is given enough to leave, the atom is left positive and is said to be ionized. The crucial discovery is that a bound electron cannot have just any energy. It may occupy only certain particular energies: they are quantized.

When an electron moves between two levels it must absorb or emit exactly the difference, and it does so as a single packet of light called a photon:

\[ E = hf \qquad h = 6.63 \times 10^{-34}\ \text{J s} \]

and since $ c = f\lambda $, the same relation can be written $ E = hc/\lambda $.

Absorption and emission are the same transition run in opposite directions, which is exactly why the two spectra have lines at the same wavelengths.

A photon’s energy is a DIFFERENCE, never a level. If the levels are at −3.4 eV and −1.5 eV, the photon carries 1.9 eV — not 3.4, not 1.5, and certainly not −1.9. Take the magnitude of the gap. The negative signs belong to the levels (an electron is bound, so its energy is below the zero at infinity, exactly as in D.1) and they cancel when you subtract.

✏️Worked example 2 — from a transition to a colour

In hydrogen the $ n = 4 $ level lies at −0.85 eV and the $ n = 2 $ level at −3.40 eV. Find the wavelength of the photon emitted when an electron falls from $n = 4$ to $n = 2$, and say what colour it is.

The gap. $ \Delta E = 3.40 - 0.85 = 2.55 $ eV. Convert to joules: $ 2.55 \times 1.60\times10^{-19} = 4.08\times10^{-19} $ J.

\[ f = \frac{E}{h} = \frac{4.08\times10^{-19}}{6.63\times10^{-34}} = 6.15\times10^{14}\ \text{Hz} \]

\[ \lambda = \frac{c}{f} = \frac{3.00\times10^{8}}{6.15\times10^{14}} = 4.9\times10^{-7}\ \text{m} = 487\ \text{nm} \]

That is in the blue-green part of the visible spectrum.

The accepted value is 486.1 nm, and the 1 nm difference is instructive. It is not an error in the method — it comes from the rounding in “−0.85” and “13.6”, which are themselves rounded versions of more precise numbers. Spectroscopy is one of the most precise measurements in all of physics, so a model that got only three significant figures right would not have convinced anyone. Keep more digits when a question is about agreement with experiment rather than about the method.

HLHow big is a nucleus, and how dense?

More massive nuclei are larger, and detailed analysis of scattering data shows they behave like hard spheres packed at an essentially constant density. That single fact fixes how the radius grows with the nucleon number:

\[ R = R_0A^{1/3} \qquad R_0 \approx 1.2 \times 10^{-15}\ \text{m} = 1.2\ \text{fm} \]

On the left, a graph of nuclear radius in femtometres against nucleon number A from 0 to 250, showing a curve that rises steeply at first and then flattens, following the cube root of A; carbon-12, iron-56 and uranium-238 are marked on it at 2.7, 4.6 and 7.4 femtometres. On the right, three nuclei drawn to scale as spheres of closely packed nucleons, with the caption that although uranium-238 has about twenty times the nucleons of carbon-12 it is less than three times the radius, because volume goes as the cube of the radius. A note states that the number of nucleons per unit volume is the same for all three, so every nucleus has the same density of about two times ten to the seventeen kilograms per cubic metre. — The cube root is the whole content of the formula: **a nucleus with eight times the nucleons has only twice the radius.** Volume goes as $R^{3}$, and volume is what is proportional to $A$.

Because the volume is $ V = \tfrac{4}{3}\pi R^{3} = \tfrac{4}{3}\pi R_0^{3}A $, the nucleons per unit volume is $ A/V = 3/(4\pi R_0^{3}) $ — and $A$ has cancelled. Multiply by the mass of a nucleon ($ \approx 1.7\times10^{-27} $ kg) and you get the density of every nucleus:

\[ \rho = \frac{3m}{4\pi R_0^{3}} \approx 2 \times 10^{17}\ \text{kg m}^{-3} \]

That is a genuinely absurd number. A teaspoon of it would have a mass of around a billion tonnes. The only macroscopic objects in the universe with this density are neutron stars, which is where E.5 ends up.

✏️HLWorked example 3 — two nuclei, one density

Find the radius of an iron-56 nucleus and of a uranium-238 nucleus, and show that their densities are the same. Take $ R_0 = 1.2 $ fm and the mass of a nucleon as $ 1.66\times10^{-27} $ kg.

Radii. $ R_{\text{Fe}} = 1.2 \times 56^{1/3} = 1.2 \times 3.83 = 4.6 $ fm, and $ R_{\text{U}} = 1.2 \times 238^{1/3} = 1.2 \times 6.20 = 7.4 $ fm.

Iron. $ V = \tfrac{4}{3}\pi(4.59\times10^{-15})^{3} = 4.05\times10^{-43} $ m$^{3}$; mass $ = 56 \times 1.66\times10^{-27} = 9.30\times10^{-26} $ kg; so $ \rho = 2.3\times10^{17} $ kg m$^{-3}$.

Uranium. $ V = 1.72\times10^{-42} $ m$^{3}$; mass $ = 3.95\times10^{-25} $ kg; so $ \rho = 2.3\times10^{17} $ kg m$^{-3}$.

Identical, to every figure carried. That is not a coincidence to be remarked on — it is built into $ R = R_0A^{1/3} $, which was written down because the density is constant. Uranium has 4.25 times the nucleons of iron and 4.25 times the volume, so the ratio is unchanged. If your two answers differ, you have made an arithmetic slip, not a discovery.

HLHow close can an alpha particle get?

An alpha particle fired at a nucleus is repelled, slows down, stops, and comes back. At the moment it is stationary all of its kinetic energy has become electric potential energy, and that is the whole calculation:

\[ E_k = \frac{q_1q_2}{4\pi\varepsilon_0 r} \qquad\Longrightarrow\qquad r = \frac{q_1q_2}{4\pi\varepsilon_0 E_k} \]

Several alpha particles approaching a nucleus along different lines. Those passing to one side follow curved hyperbolic paths and are deflected through various angles, while one aimed directly at the nucleus travels in, slows, stops at a distance r and returns straight back along the same path. Beside it, an energy bar chart at three moments: far away the particle has all kinetic energy and no potential energy; part way in it has some of each; at the closest approach it is momentarily stationary and has all potential energy and no kinetic energy. The note states that the alpha particle never actually touches the nucleus because it does not have enough energy. — The alpha particle never reaches the nucleus — it runs out of energy first. The distance at which that happens is an *upper bound* on the nuclear radius, which is how the size of a nucleus was first estimated.

✏️HLWorked example 4 — closest approach to a gold nucleus

An alpha particle ($ Z = 2 $) with 4.2 MeV of kinetic energy is fired directly at a gold nucleus ($ Z = 79 $). Find its distance of closest approach. ($ \varepsilon_0 = 8.85\times10^{-12} $, $ e = 1.6\times10^{-19} $ C.)

Convert the energy. $ E_k = 4.2\times10^{6} \times 1.6\times10^{-19} = 6.72\times10^{-13} $ J.

The two charges. $ q_1 = 2e $ and $ q_2 = 79e $.

\[ r = \frac{(2 \times 1.6\times10^{-19})(79 \times 1.6\times10^{-19})}{4\pi \times 8.85\times10^{-12} \times 6.72\times10^{-13}} = 5.4\times10^{-14}\ \text{m} \]

Compare that with the nucleus itself. A gold nucleus has $ R = 1.2 \times 197^{1/3} = 7.0 $ fm $ = 7.0\times10^{-15} $ m, so the alpha particle stops nearly eight times further out than the nuclear surface. It never gets near enough to touch, which is exactly why ordinary Rutherford scattering sees only the Coulomb repulsion and never the strong nuclear force. Push the energy high enough and that stops being true — see below.

HLWhere Rutherford scattering breaks down

Rutherford’s analysis models the scattering as pure Coulomb repulsion between the alpha particle and the nucleus, and at moderate energies it predicts the measured intensities at each angle very accurately. At high energies it stops working.

Fix the detector at one angle and increase the alpha energy, and the measured intensity follows the Rutherford prediction until, at around 27.5 MeV, it departs from it. The reason is the previous section in reverse: more energy means a closer approach, and eventually the alpha particle gets close enough for the strong nuclear force (E.3) to start acting. Once a second force is involved, a model built on only one of them must fail.

This is a nice piece of scientific reasoning to be able to reproduce: the point at which a model breaks tells you something real. Here the breakdown energy is a measure of how close you have to get before the strong force appears, and therefore of the size of the nucleus. Probing finer detail needs a different tool again — high-energy electrons, which do not feel the strong force at all.

HLThe hydrogen spectrum and the Rydberg formula

Hydrogen’s emission spectrum was measured long before anyone could explain it. In 1885 a Swiss schoolteacher, Johann Balmer, noticed that the visible wavelengths fitted a formula. Those lines are the Balmer series, and they turned out to be one of several such families, all described by a single expression — the Rydberg formula:

\[ \frac{1}{\lambda} = R_H\left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right) \qquad R_H = 1.097\times10^{7}\ \text{m}^{-1} \]

with $m$ any whole number larger than $n$. Each value of $n$ gives a whole series, named after whoever found it:

Lyman, $ n = 1 $ultraviolet

Balmer, $ n = 2 $visible

Paschen, $ n = 3 $infrared

Brackett, $ n = 4 $ · Pfund, $ n = 5 $further into the infrared

An energy level diagram for hydrogen with levels drawn at minus 13.6, minus 3.40, minus 1.51, minus 0.85 and minus 0.54 electronvolts for n equals 1 to 5, crowding together towards zero at the top which is marked as the ionisation limit. Downward arrows are grouped into three families: those ending on n equals 1 are labelled the Lyman series and marked ultraviolet, those ending on n equals 2 are the Balmer series and marked visible, and those ending on n equals 3 are the Paschen series and marked infrared. A note explains that the series ending on the lowest level involve the biggest energy gaps and therefore the shortest wavelengths, which is why only the Balmer series falls in the visible range. — Which series a line belongs to is set by *where it lands*, not where it starts. Transitions down to $n=1$ cross the biggest gaps, so the Lyman series is the most energetic and lies in the ultraviolet.

HLThe Bohr model

Niels Bohr took the planetary picture of hydrogen and added a restriction that had no justification at the time but produced the right answer. He postulated:

First postulatean electron in a stable orbit does not radiate, and only orbits whose angular momentum is a whole multiple of $ h/2\pi $ are allowed: $ m_evr = nh/2\pi $

Second postulatean electron moving between allowed orbits emits or absorbs the energy difference as a photon

Setting the electrostatic attraction equal to the centripetal force and substituting the quantized angular momentum gives the allowed radii, and hence the allowed energies:

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

On the left, the Bohr picture of hydrogen: a nucleus with three allowed circular orbits drawn round it, labelled n equals 1, 2 and 3, with the radii increasing as n squared, and a dashed circle between two of them marked as not allowed. An electron is shown jumping from the third orbit to the second and emitting a photon. On the right, two columns headed what it got right and what it could not do. The successes are that the predicted energies match the measured hydrogen wavelengths and that the Rydberg constant can be calculated from other known constants. The failures are that it fails for atoms with more than one electron, that the angular momentum postulate has no theoretical justification, that classical theory still says the orbiting electron should radiate, and that it accounts for neither the relative intensities of lines nor their fine structure. — The agreement with hydrogen is genuinely impressive — and the model is still wrong. Being able to say *both* of those things about the same model is what the question “evaluate the Bohr model” is asking for.

Two things follow immediately from that expression, and both are worth stating in an answer. The energy is negative, which means the electron is bound — trapped by the proton. And it goes as $ 1/n^{2} $, so the levels crowd together as $n$ rises and converge on zero, which is the ionisation limit: 13.6 eV supplied to a ground-state electron frees it completely.

Bohr’s expression for $ 1/\lambda $ has exactly the form of the Rydberg formula, and the Rydberg constant can be calculated from $m_e$, $e$, $\varepsilon_0$, $c$ and $h$ — constants measured in completely unrelated experiments. That agreement is why the model was taken seriously. But it fails for every atom with more than one electron, its central postulate has no justification, and it still cannot explain why the orbiting electron does not radiate. The modern account replaces orbits with wavefunctions, which is where E.2 goes.

🔭See it happen

PhET, Rutherford Scattering lets you fire alpha particles at a nucleus and vary their energy and the target’s atomic number — watch the closest approach shrink as you wind the energy up. Then PhET, Models of the Hydrogen Atom puts the competing models side by side and shows you the spectrum each one predicts, with the real spectrum alongside for comparison. It is the fastest way to see why Bohr’s model was believed and why it was not enough.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A nucleus contains 26 protons and 30 neutrons. Write it in nuclear notation, and state which of the three symbols $A$, $Z$ and X carry the same information.

$ Z = 26 $, which is iron, so the symbol is Fe. $ A = 26 + 30 = 56 $, giving $ ^{56}_{26}\text{Fe} $. $Z$ and X carry the same information: the atomic number fixes the element and the element fixes the atomic number, so writing both is strictly redundant — which is why you often see just $ ^{56}\text{Fe} $. $A$ is the one that adds something, because it tells you the neutron count, $ A - Z $.

2. Explain the significance of the negative values of atomic energy levels.

The zero of energy is taken to be a free electron at rest, infinitely far from the nucleus. A bound electron has less energy than that, so its energy is negative. The magnitude of a level is therefore the energy that must be supplied to remove that electron from the atom entirely. The more negative the level, the more tightly bound the electron — exactly the same convention, and the same reasoning, as gravitational potential in D.1.

3. Four levels of an atom lie at −0.85, −1.51, −3.40 and −13.6 eV. Find the longest and the shortest wavelength that transitions between them can produce.

The longest wavelength comes from the smallest energy gap, which is between the top two levels: $ \Delta E = 1.51 - 0.85 = 0.66 $ eV $ = 1.06\times10^{-19} $ J, giving $ \lambda = hc/E = 1.9\times10^{-6} $ m — about 1900 nm, in the infrared. The shortest comes from the largest gap, from the top level down to the bottom: $ \Delta E = 13.6 - 0.85 = 12.75 $ eV $ = 2.04\times10^{-18} $ J, giving $ \lambda = 9.7\times10^{-8} $ m — about 97 nm, in the ultraviolet. Note that the biggest gap gives the shortest wavelength; getting that inversion the wrong way round is the usual error.

4. Determine how many different spectral lines transitions between those four levels can produce.

Six. Every distinct pair of levels gives one line, and the number of pairs from four levels is $ \binom{4}{2} = (4 \times 3)/2 = 6 $. Listing them: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1. It does not matter that a cascade like 4→3→1 involves two steps — each step is one of the six.

5. Explain how an absorption spectrum arises in the light from a star, and outline what can be deduced from it.

The star’s hot, dense interior emits a continuous spectrum. On the way out, that light passes through the cooler, less dense gas of the star’s outer atmosphere. Electrons in those atoms absorb photons whose energies exactly match gaps between their energy levels, removing those particular wavelengths from the beam and leaving dark lines. What can be deduced: the wavelengths of the missing lines identify which elements are present, because the level pattern is unique to each element; the relative darkness of the lines indicates how much of each element there is; and any overall shift of the whole pattern gives the star’s speed towards or away from us by the Doppler effect (C.5).

6. State the three main observations of the Rutherford–Geiger–Marsden experiment and the conclusion each one supports.

Most alpha particles passed straight through, barely deviated → the atom is mostly empty space. A small number were deflected through large angles → there is a concentration of positive charge in a very small volume, repelling them. About 1 in 8000 was turned back through more than 90° → that concentration must also be very massive compared with an alpha particle, or it would simply have been knocked aside instead. A fourth point earns credit if asked for the mathematics: the number scattered at each angle matched an inverse-square law of repulsion, confirming a point-like Coulomb source.

7. HLEstimate the radius of a lead-208 nucleus, and state how it compares with carbon-12.

$ R = R_0A^{1/3} = 1.2\times10^{-15} \times 208^{1/3} = 1.2\times10^{-15} \times 5.93 = 7.1\times10^{-15} $ m, or 7.1 fm. For carbon-12, $ R = 1.2 \times 12^{1/3} = 2.7 $ fm. So lead has 17 times the nucleons of carbon but only 2.6 times the radius — because $ R \propto A^{1/3} $, and $ 17^{1/3} = 2.6 $.

8. HLShow that every nucleus has approximately the same density, and comment on its size.

Volume $ V = \tfrac{4}{3}\pi R^{3} = \tfrac{4}{3}\pi R_0^{3}A $, so $ V \propto A $. The mass is also proportional to $A$, since it is $A$ nucleons each of mass $ \approx 1.7\times10^{-27} $ kg. Density is mass over volume, so $A$ cancels and the density is the same for every nuclide: \[ \rho = \frac{3m}{4\pi R_0^{3}} = \frac{3 \times 1.7\times10^{-27}}{4\pi(1.2\times10^{-15})^{3}} \approx 2\times10^{17}\ \text{kg m}^{-3} \] This is around $ 10^{14} $ times the density of water. Nothing on Earth comes close; the only objects with it are neutron stars, which are essentially nuclear matter on an astronomical scale.

9. HLAlpha particles of mass $ 6.7\times10^{-27} $ kg travelling at $ 2.0\times10^{6} $ m s$^{-1}$ are fired at gold nuclei ($ Z = 79 $). Calculate how close they get.

First the kinetic energy: $ E_k = \tfrac{1}{2}mv^{2} = \tfrac{1}{2} \times 6.7\times10^{-27} \times (2.0\times10^{6})^{2} = 1.34\times10^{-14} $ J. At the closest approach all of it has become electric potential energy: \[ r = \frac{q_1q_2}{4\pi\varepsilon_0E_k} = \frac{(2 \times 1.6\times10^{-19})(79 \times 1.6\times10^{-19})}{4\pi \times 8.85\times10^{-12} \times 1.34\times10^{-14}} = 2.7\times10^{-12}\ \text{m} \] Note how far out that is — about 400 times the radius of the gold nucleus — because these alphas are slow. Compare the 4.2 MeV alphas of Worked example 4, which reach $ 5.4\times10^{-14} $ m.

10. HLOutline two successes and two limitations of the Bohr model.

Successes: it predicts the energy levels of hydrogen as $ E_n = -13.6/n^{2} $ eV, and the wavelengths that follow agree with the measured hydrogen spectrum to high precision; and it yields an expression of exactly the Rydberg form, allowing $R_H$ to be calculated from $m_e$, $e$, $\varepsilon_0$, $c$ and $h$ — constants measured in entirely unrelated experiments — which again matches. Limitations: it fails to give the correct spectra for any atom or ion with more than one electron; the postulate that angular momentum is quantized in units of $h/2\pi$ has no theoretical justification within the model; classical theory still predicts that an orbiting (accelerating) electron must radiate and spiral inwards, which the model simply asserts does not happen; and it accounts for neither the relative intensities of the spectral lines nor their fine structure.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

HyperPhysics — the Rutherford experiment, hydrogen spectrum and nuclear size
The Physics Hypertextbook — atomic structure and spectra