The History of Nuclear & Quantum Physics

1. The nucleus: the Geiger–Marsden–Rutherford experiment

By 1904, physicists knew that atoms contained tiny, negatively charged electrons — J. J. Thomson had found them using cathode rays. But since atoms are electrically neutral overall, there had to be positive charge somewhere too. Thomson's best guess was the "plum pudding" model: a blob of spread-out positive charge with electrons dotted through it, like fruit suspended in a pudding.

Fig. 1.1 Two competing models of the atom, before and after 1909.

In 1909, at Rutherford's laboratory in Manchester, two of his students, Hans Geiger and Ernest Marsden, fired a narrow beam of fast, positively charged alpha particles at an extremely thin sheet of gold foil (only a few hundred atoms thick). A screen coated in zinc sulfide, which produces a tiny flash of light whenever an alpha particle hits it, was used to detect the particles after they had passed through — or bounced off — the foil.

Fig. 1.2 The alpha-particle scattering apparatus (plan view).

In their own words

Rutherford later described the large-angle results as "quite the most incredible event that has ever happened to me... It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

The results were:

The overwhelming majority of alpha particles passed straight through the foil with little or no deflection — exactly as expected if the atom is mostly empty space.
A small fraction were deflected through moderate angles (more than about 10°).
An extremely small fraction — roughly 1 in 8000 — were deflected through more than 90°, and a few came almost straight back the way they came.

Rutherford's interpretation. Because most alpha particles passed straight through, most of the atom must be empty space. Because a few bounced back almost the way they came, there had to be something extremely small, extremely dense, and positively charged at the centre of the atom, able to repel a fast-moving positive alpha particle head-on. Rutherford called this the nucleus.

This was a genuine paradigm shift: the plum-pudding model was replaced by the nuclear model, in which almost all the mass and all the positive charge of an atom is concentrated in a nucleus roughly 10⁴–10⁵ times smaller than the atom itself, with electrons occupying the mostly-empty space around it.

Fire alpha particles at a plum-pudding atom and then at a nuclear atom, and see the difference in scattering pattern for yourself.

Open the PhET simulation ↗

Interactive simulation — open the online version of this workbook to launch it.

Simulation: Rutherford Scattering, PhET Interactive Simulations, University of Colorado Boulder — phet.colorado.edu.

Check your understanding

1Using the PhET simulation above, switch between the "Plum Pudding Atom" and "Rutherford Atom" screens. Describe one difference you observe in how the alpha particles scatter.

This is a descriptive, simulation-based question (compare the scattering patterns you observe) — check your wording against the description of the experiment in this section and with your teacher.

2In your own words (fewer than 100 words), explain why Rutherford concluded that atoms contain a small, positively charged, dense nucleus. Do not use a diagram.

This is a descriptive question (explain in your own words) — check your wording against the description of the experiment in this section and with your teacher.

3Suggest what would have happened in the Geiger–Marsden experiment if neutrons had been used instead of alpha particles. Explain your answer.

Neutrons carry no charge, so they would not be repelled electrically by the nucleus at all — almost none would be deflected by more than a small angle from occasional direct nuclear ("strong-force") collisions, so the large-angle scattering pattern that revealed the nucleus's charge and size would not appear in the same way.

2. Nuclear notation and isotopes

After Rutherford's discovery, it became clear that the nucleus itself is made of two kinds of particle: positively charged protons and uncharged neutrons, together called nucleons. Every atom of a given element has the same number of protons, but the number of neutrons can vary.

Particle	Relative mass	Relative charge	Location
proton	1	+1	nucleus
neutron	1	0	nucleus
electron	1/1840	−1	surrounding the nucleus

Key definitions.

Proton number, Z — the number of protons in the nucleus. This determines which element the atom is.
Nucleon number, A — the total number of protons and neutrons (also called the mass number).
Neutron number, N — the number of neutrons, where N = A − Z.

Fig. 2.1 Standard nuclide notation: ^A_ZX.

Two or more atoms of the same element (same Z) with different nucleon numbers A are called isotopes. They have identical chemical properties but different masses. For example, the three isotopes of hydrogen are:

¹₁H (hydrogen, no neutrons) · ²₁H (deuterium, 1 neutron) · ³₁H (tritium, 2 neutrons)

Worked example 2.1

A particular element has proton number 19.

Identify the element.
Its most common isotope has a nucleon number of 39. State the number of protons, neutrons and electrons in a neutral atom of this isotope.
Write the full nuclide symbol for this isotope.

Answer:
a) Potassium (K).
b) 19 protons, 20 neutrons (39 − 19), 19 electrons (the atom is neutral).
c) ³⁹₁₉K

Check your understanding

4Consider a neutral atom.

What do you get when you change the number of protons in an atom?
What do you get when you change the number of neutrons in an atom?
What do you get when you change the number of electrons in an atom?

a) a different element — changing the number of protons (Z) changes which element it is; b) a different isotope of the same element — changing the number of neutrons keeps Z (and so the element) the same, but changes the nucleon number A and the mass; c) an ion — changing the number of electrons means the atom is no longer electrically neutral, so it becomes a charged ion (positive if electrons are removed, negative if electrons are added).

5Chlorine, Cl, has proton number 17. Its two most common isotopes are chlorine-35 and chlorine-37. Write the full nuclide symbol for each, and state the number of neutrons in each.

³⁵₁₇Cl (18 neutrons); ³⁷₁₇Cl (20 neutrons).

6An ion is formed from a nitrogen atom (Z = 7, A = 14) by removing three electrons. State the number of protons, neutrons and electrons in this ion, and its overall charge.

7 protons, 7 neutrons, 4 electrons; overall charge = +3.

3. How big is a nucleus? Radius, density, and closest approach

The Geiger–Marsden results can be analysed quantitatively using conservation of energy. As an alpha particle travels directly towards a nucleus, it slows down as its kinetic energy is converted into electric potential energy. At the instant it is momentarily at rest — the distance of closest approach, r — all of its initial kinetic energy has been transferred. This gives a maximum possible radius of a nucleus:

initial kinetic energy of alpha particle = electric potential energy at closest approach
E_k = k·q₁q₂ / r

Worked example 3.1

An alpha particle (charge +2e) with kinetic energy 6.4 MeV is fired head-on at a gold nucleus (charge +79e). Determine the closest distance it can approach the nucleus.

Answer:
E_k = k q₁q₂ / r, so r = k q₁q₂ / E_k
r = (8.99 × 10⁹) × (2 × 1.60 × 10⁻¹⁹) × (79 × 1.60 × 10⁻¹⁹) / (6.4 × 10⁶ × 1.60 × 10⁻¹⁹)
r ≈ 3.6 × 10⁻¹⁴ m

Because the alpha particle in this calculation gets so close without deviating from the inverse-square law, this method gives an upper limit for the nuclear radius (the real gold nucleus turns out to be a little smaller).

A separate line of evidence — probing nuclei with high-energy electrons instead of alpha particles — leads to an empirical rule connecting the radius R of a nucleus to its nucleon number A:

R = R₀A^1/3 where R₀ = 1.20 × 10⁻¹⁵ m (the "Fermi radius")

Worked example 3.2 (HL)

Show that the radius of a lead-206 nucleus (²⁰⁶₈₂Pb) is about 7.1 × 10⁻¹⁵ m, and use it to estimate the nuclear density.

Answer:
R = R₀A^1/3 = (1.20 × 10⁻¹⁵) × 206^1/3 = 7.1 × 10⁻¹⁵ m ✓

ρ = m/V ≈ Au / (4/3 πR³) = (206 × 1.661 × 10⁻²⁷) / (4/3 π (7.1 × 10⁻¹⁵)³) ≈ 2.3 × 10¹⁷ kg m⁻³

Why this matters: nuclear density is (almost) constant. Because both mass and volume scale with A (mass ∝ A, and V ∝ R³ ∝ A), the A cancels out of the density calculation. Every nucleus — hydrogen or uranium — has approximately the same density, about 2–3 × 10¹⁷ kg m⁻³. This is roughly 10¹⁴ times the density of everyday solids, and is comparable only to the density of a neutron star.

Check your understanding

7Estimate the radius of an oxygen-16 nucleus (¹⁶₈O).

R = (1.20×10⁻¹⁵)(16)^1/3 ≈ 3.0×10⁻¹⁵ m.

8A gold nucleus has a measured radius of 6.98 × 10⁻¹⁵ m. Using R = R₀A^1/3, show this is consistent with A = 197.

A^1/3 = 6.98×10⁻¹⁵/1.20×10⁻¹⁵ = 5.82 ⟹ A ≈ 197 ✓ (matches gold).

9An alpha particle of kinetic energy 4.0 MeV is fired head-on at a copper nucleus (Z = 29). Calculate the closest distance of approach.

r = kq₁q₂/E_k ≈ 2.1×10⁻¹⁴ m.

10Explain why nuclear densities are all approximately equal, regardless of the size of the nucleus.

Mass ∝ A and volume ∝ R³ ∝ A, so A cancels in ρ = m/V — density is independent of nucleon number and (almost) the same for every nucleus.

4. When Coulomb's law breaks down: the strong nuclear force

Rutherford's calculations assumed that the only force acting between the alpha particle and the nucleus is the repulsive electric (Coulomb) force. This assumption matches experimental scattering angles extremely well — provided the alpha particles do not have too much energy.

If alpha particles of increasingly high energy are used, they can get close enough to the nucleons that a second, much stronger, but extremely short-range force starts to act: the strong nuclear force. Once this happens, the scattering pattern no longer matches the simple inverse-square-law prediction — this deviation from "Rutherford scattering" is itself evidence for the strong force.

Combined nuclear force vs. separation

Separation, r = 1.50 fm

Coulomb force: …
Strong force: …
Net force: …

Positive = repulsive (pushing the alpha particle away); negative = attractive (pulling it in). Dashed lines show the two individual contributions; the solid line is their sum, the net force actually felt by the particle. This is an illustrative model of the correct qualitative shape, not an exact nuclear force law.

Fig. 4.1 The Coulomb force (dashed, always repulsive, 1/r²) and the strong nuclear force (dashed, repulsive at extremely short range, attractive at intermediate range, negligible beyond ~4 fm) combine to give the net force (solid) on an alpha particle approaching a nucleus. Drag the slider to see how the balance shifts with distance.

Key idea. A very energetic alpha particle can get close enough to a nucleus to feel the attractive strong nuclear force as well as the repulsive electric force. When this happens, the scattering can no longer be explained by Coulomb's law alone — this is the experimental signature that the strong force exists.

Check your understanding

11Explain why deviations from the Rutherford scattering pattern only appear at high alpha-particle energies, not low energies.

Only high-energy alpha particles get close enough to a nucleus (small distance of closest approach) to enter the very short range (≲ 3×10⁻¹⁵ m) over which the strong force acts; low-energy particles never get that close, so Coulomb's law alone still fits.

12Suggest why the strong nuclear force must be short-range, given that ordinary atoms (with nuclei made of many protons) do not collapse or fly apart.

Because the strong force is short-range, each nucleon effectively only interacts with its near neighbours rather than every other nucleon in the nucleus — this stops nuclei from either collapsing (it turns repulsive at extremely short range) or growing without limit, and it has no effect at all at atomic/molecular distances.

5. Evidence for energy levels: emission and absorption spectra

The simple picture of electrons orbiting a nucleus (rather like planets orbiting a star) turns out to be seriously incomplete. Orbiting satellites can have any orbital energy — a continuous range. Electrons in atoms cannot: they can only exist with certain very precise, separated (discrete) energies, called atomic energy levels. The lowest of these is the ground state.

The evidence for this comes from studying the light that atoms give out or take in.

When a gas is excited (heated, or given energy by an electric current), it emits light only at certain specific frequencies. Viewed through a prism or diffraction grating, this appears as a series of bright, separate lines on a dark background: an emission spectrum.
When white light (a continuous spectrum) is passed through a cool gas, the gas absorbs light at exactly those same frequencies. This produces dark lines on an otherwise continuous, bright spectrum: an absorption spectrum.

Key idea. Every line in an emission or absorption spectrum corresponds to an electron transition between two discrete energy levels. Since different elements have different sets of energy levels, every element produces its own unique "fingerprint" of spectral lines. This is why spectra can be used to identify the chemical composition of a distant star, a nebula, or an unknown sample in a lab.

Live simulation: build your own line spectrum

This tool (Foothill College AstroSims) lets you add individual coloured emission lines onto a continuous spectrum and see the resulting pattern, just as astronomers use spectrometers to read the chemical fingerprint of starlight.

Interactive simulation — open the online version of this workbook to launch it.

Simulation: Foothill College AstroSims — Spectroscopy Demonstrator (opens the full version in a new tab).

Explore real spectra — atomic-spectra.net

Browse the real emission-line spectrum of any element in the periodic table.

Open atomic-spectra.net ↗

External resource: atomic-spectra.net. Opens in a new tab.

Check your understanding

13Using the AstroSims tool above (or atomic-spectra.net), compare the emission spectra of hydrogen and helium. Sketch or describe two differences.

This is a descriptive, simulation-based question — see the key idea box in this section for the underlying physics; check your specific comparison with your teacher.

14Explain how an astronomer could use the absorption spectrum of a star to determine which elements are present in its outer atmosphere.

This is a descriptive question — see the key idea box in this section, which explains why each element's absorption lines act as a unique fingerprint; check your wording with your teacher.

15Explain why the existence of line spectra (rather than continuous, "rainbow" spectra) is evidence that atomic energy levels are discrete rather than continuous.

Discrete spectral lines correspond to discrete photon frequencies (E = hf); since only specific frequencies appear, only specific energy differences — and therefore only discrete energy levels — can exist within the atom.

6. Photons and atomic transitions

When an electron in an atom drops from a higher energy level to a lower one, the atom emits a single "packet" of electromagnetic energy called a photon. Conversely, an atom can absorb a photon and jump from a lower level to a higher one — but only if the photon carries exactly the right amount of energy to bridge the gap between the two levels.

energy of one photon: E = hf (h = Planck's constant = 6.63 × 10⁻³⁴ J s)

Because atomic energy levels are discrete, the differences between them are also discrete — so only certain photon energies (and therefore only certain frequencies, since E = hf) can be emitted or absorbed by a given atom. This is exactly why line spectra exist.

Physically, each energy level corresponds to an electron occupying one of the atom's circular "shells" around the nucleus (this is the Bohr model — see Section 7). But the emission/absorption diagrams used throughout this topic, like Fig. 6.2 below, draw each level as a flat horizontal line rather than a circle — and the line spectra you explored in Section 5 are themselves just another picture of the very same spacing. Fig. 6.1 shows how all three pictures (shells → flat lines → spectral lines) are connected.

Fig. 6.1 The circular electron shells are then shown as a horizontal "sliver" to save space and make the model as clear as possible.

Fig. 6.2 Emission transitions (downward, coloured) release photons of different energies; absorption (upward, dashed) requires a photon of exactly the right energy. Notice the gaps between levels shrink higher up, as in a real atom (E ∝ −1/n²).

Worked example 6.1

An electron in an atom drops from an energy level of −1.20 × 10⁻¹⁸ J to a level of −3.06 × 10⁻¹⁸ J. Calculate the frequency of the photon emitted, and state the part of the electromagnetic spectrum it belongs to.

Answer:
ΔE = (−1.20 × 10⁻¹⁸) − (−3.06 × 10⁻¹⁸) = 1.86 × 10⁻¹⁸ J
E = hf ⟹ f = E/h = (1.86 × 10⁻¹⁸)/(6.63 × 10⁻³⁴) = 2.81 × 10¹⁵ Hz
This frequency lies in the ultraviolet part of the spectrum.

Check your understanding

16A photon of frequency 5.0 × 10¹⁴ Hz is absorbed by an atom. Calculate the energy gained by the atom, in both joules and electronvolts.

E = hf = 3.32×10⁻¹⁹ J = 2.07 eV.

17An atom has just four energy levels. State the maximum possible number of different spectral lines (transitions) it could produce.

6 possible transitions (4 choose 2).

18Explain why a photon can only be absorbed by a particular atom if its energy exactly matches the gap between two of that atom's energy levels — rather than being absorbed partially.

Because atomic energy levels are discrete, an electron can only ever be in one exact level or another — there is no "in-between" state to absorb partial energy into, so the photon energy must exactly bridge two allowed levels.

7. The Bohr model of the hydrogen atomHL

In 1913, the Danish physicist Niels Bohr proposed a bold new rule for the hydrogen atom: an electron can only orbit the nucleus in certain allowed orbits, and while it stays in one of these orbits, it does not radiate energy (even though classical physics said an accelerating charge always should). Bohr justified this by proposing that the electron's angular momentum is quantized — it can only take specific values:

mvr = nh / 2π (n = 1, 2, 3, ... — the "principal quantum number")

Combining this quantization rule with Coulomb's law and circular motion allows the energy of each allowed orbit (n = 1, 2, 3, ...) to be calculated. Remarkably, the result matches the hydrogen line spectrum almost perfectly:

energy levels of hydrogen: E = −13.6 / n² (eV)

Fig. 7.1 Energy levels of hydrogen. The gaps between levels shrink rapidly as n increases (E = −13.6/n² eV), converging on the ionisation limit at n = ∞.

Worked example 7.1

Calculate the energy of the n = 4 level of hydrogen, in eV and in joules. Hence calculate the frequency of the photon emitted when an electron falls from n = 4 to n = 2.

Answer:
E₄ = −13.6/4² = −0.850 eV = −0.850 × 1.60 × 10⁻¹⁹ = −1.36 × 10⁻¹⁹ J
E₂ = −13.6/2² = −3.40 eV
ΔE = (−0.850) − (−3.40) = 2.55 eV = 4.08 × 10⁻¹⁹ J
f = ΔE/h = (4.08 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 6.15 × 10¹⁴ Hz (visible light)

Worked example 7.2

Calculate the angular momentum of an electron in the n = 2 orbit of hydrogen.

Answer:
L = nh/2π = (2 × 6.63 × 10⁻³⁴)/(2π) = 2.11 × 10⁻³⁴ kg m² s⁻¹

Why the Bohr model matters — and where it fails

Bohr's model was the first to correctly predict the numerical energy levels of hydrogen from a quantization rule, rather than just fitting a curve to data. But it only works well for hydrogen (and other one-electron ions); it fails for atoms with more than one electron, and it does not explain why angular momentum should be quantized. That deeper explanation had to wait for the discovery of the wave nature of matter — the subject of Section 11.

Check your understanding

19Calculate the energy (in eV) of the n = 5 level of hydrogen.

E₅ = −13.6/25 = −0.544 eV.

20Determine the frequency of the photon emitted when an electron in a hydrogen atom falls from n = 3 to the ground state (n = 1). State which part of the electromagnetic spectrum this photon belongs to.

ΔE = 13.6 − 1.51 = 12.1 eV ⟹ f ≈ 2.92×10¹⁵ Hz (ultraviolet — part of the Lyman series).

21Calculate the ionisation energy of a hydrogen atom that is already in its n = 2 state (i.e. the energy needed to remove the electron completely, starting from n = 2 rather than the ground state).

Ionisation energy from n = 2 is 3.40 eV (= 0 − (−3.40 eV)).

22Determine the angular momentum of an electron in the n = 3 orbit of hydrogen, (a) in terms of h/π and (b) in SI units.

a) L = 3h/2π = 1.5(h/π); b) L = 3.17×10⁻³⁴ kg m² s⁻¹.

Optional extra — PhET: Build an Atom

Not required, but useful for visualising how electrons are organised into shells around a nucleus, and how this connects to isotope and ion notation from Section 2.

Open the simulation ↗

Simulation: PhET Interactive Simulations, University of Colorado Boulder — phet.colorado.edu.

8. The photoelectric effect

By the late 1800s, light was firmly believed to be a wave. But in 1887, Heinrich Hertz noticed something odd: shining ultraviolet light onto a clean metal surface could cause it to emit electrons — the photoelectric effect, and the ejected electrons are called photoelectrons.

Careful experiments revealed three observations that a wave model of light simply could not explain:

There is no time delay — photoelectrons are emitted the instant light hits the surface, however dim the light.
For a given metal, photoelectrons are only emitted if the light's frequency is above a certain minimum value, the threshold frequency, f₀ — no matter how intense the light is below that frequency.
Increasing the intensity of light above the threshold frequency increases the number of photoelectrons per second, but not their individual maximum kinetic energy.

Einstein's explanation (1905). Einstein proposed that light itself is quantized into individual photons, each carrying energy E = hf. A single photon transfers all of its energy to a single electron, instantaneously — which explains why there is no time delay, and why the effect depends on frequency (the energy per photon) rather than intensity (the number of photons per second).

Play with the live simulation below: it models a beam of monochromatic light photons hitting a metal cathode, releasing photoelectrons that travel towards a collector plate.

Try changing only the wavelength slider first, then only the metal, then only the intensity — notice which sliders change whether electrons are emitted at all, and which just change how many.

Simulation by Dr Dan Jones (drjonesphysics.com). Try changing only the wavelength slider first, then only the metal, then only the intensity — notice which sliders change whether electrons are emitted at all, and which just change how many.

Units

Cathode Metal

Wavelength (700 nm)

Intensity (50%)

Stopping Voltage (0.0 V)

Calculations

Ammeter

Digital Ammeter

0.000 nA

Lab Observation

Wavelength too long (f < f₀)

Calculations

Photon Energy (hf): 1.77 eV

Work Function (Φ): 2.28 eV

Max KE (hf - Φ): 0.00 eV

Stopping Potential (V_s): 0.00 V

Interactive simulation modelled on Dr Dan Jones's photoelectric effect demonstration — drjonesphysics.com/pe-effect.

Interactive simulation — open the online version of this workbook to launch it.

Simulation by Dr Dan Jones (drjonesphysics.com).

Simulation tasks

23Select Sodium as the cathode metal. Slowly increase the wavelength slider from 150 nm upwards. Record the wavelength at which photoelectrons stop being emitted (use the "Lab Observation" panel).

Sodium's threshold wavelength is λ₀ = hc/Φ ≈ 544 nm — emission should stop close to this value.

24With the wavelength fixed at a value that does produce photoelectrons, turn the intensity slider down to 0% and then up to 100%. Describe what changes on the canvas, and what does not change.

The rate of photon/electron arrivals (and hence the ammeter current) increases, but the maximum kinetic energy (and stopping voltage) of each electron is unchanged.

25Switch the cathode metal to Gold (a much larger work function) without changing the wavelength. What happens, and why?

Gold's work function (5.10 eV) is much larger than sodium's (2.28 eV); if the photon energy is now below 5.10 eV, no photoelectrons are emitted at all, however intense the light.

26The threshold frequency of zinc is 1.04 × 10¹⁵ Hz. Calculate the corresponding threshold wavelength, and state whether visible light (400–700 nm) could ever release photoelectrons from zinc.

λ₀ = c/f₀ ≈ 289 nm (UV) — shorter than the whole visible range (400–700 nm), so visible light can never release photoelectrons from zinc.

9. Einstein's photoelectric equation

Different electrons within the same metal surface need different amounts of energy to escape, depending on how tightly bound they are. But there is a well-defined minimum amount of energy needed to free the most loosely-bound electron from the surface of a given metal — this is called the work function, Φ.

The photoelectric effect is really just a statement of conservation of energy. A single photon hands over all of its energy to a single electron, and that energy has to go somewhere: some of it is "spent" doing the minimum amount of work needed to free the electron from the metal — the work function, Φ (the photoelectric-effect equivalent of the ionisation energy of an atom) — and whatever is left over becomes the electron's kinetic energy:

energy of photon = work function (ionisation energy) + kinetic energy of electron

For the most loosely-bound electrons, this uses up the smallest possible share of the photon's energy on escaping, leaving the largest possible share as kinetic energy — so this gives the maximum possible kinetic energy of a photoelectron:

hf = Φ + E_max ⟺ E_max = hf − Φ

If hf < Φ: no photoelectrons are emitted, regardless of intensity.
If hf = Φ: this defines the threshold frequency, f₀ = Φ/h, with photoelectrons emitted with zero kinetic energy.
If hf > Φ: photoelectrons are emitted with a range of kinetic energies, up to the maximum E_max.

This can be tested experimentally by measuring the stopping voltage, V_s — the minimum reverse potential difference needed to just stop the fastest photoelectrons from reaching a collector. Since work done = charge × p.d., E_max = eV_s, so a graph of stopping voltage against frequency should be a straight line:

eV_s = hf − Φ ⟺ V_s = (h/e)f − Φ/e

Reading the graph. On a graph of V_s (or E_max in eV) against frequency: the gradient gives Planck's constant, h (divided by e, if plotting V_s); the x-intercept gives the threshold frequency, f₀; and the y-intercept gives −Φ/e.

Worked example 9.1

Radiation of wavelength 4.20 × 10⁻⁷ m is incident on a metal surface with a work function of 1.90 eV.

Calculate the energy of one photon of this radiation, in eV.
Determine whether the photoelectric effect occurs.
Calculate the maximum kinetic energy of the emitted photoelectrons, in eV and in joules.
Calculate the stopping voltage needed to stop these photoelectrons.

Answer:
a) f = c/λ = (3.00×10⁸)/(4.20×10⁻⁷) = 7.14×10¹⁴ Hz; E = hf = (6.63×10⁻³⁴)(7.14×10¹⁴) = 4.73×10⁻¹⁹ J = 2.96 eV
b) Yes — photon energy (2.96 eV) > work function (1.90 eV).
c) E_max = hf − Φ = 2.96 − 1.90 = 1.06 eV = 1.70×10⁻¹⁹ J
d) V_s = E_max/e = 1.06 V

Check your understanding

27The threshold frequency of a metal is 5.5 × 10¹⁴ Hz. Calculate its work function in eV.

Φ = hf₀ ≈ 2.28 eV.

28Light of frequency 9.0 × 10¹⁴ Hz produces photoelectrons with a maximum kinetic energy of 1.75 eV from a certain metal. Calculate the work function of the metal and its threshold frequency.

Φ = hf − E_max ≈ 1.98 eV; f₀ = Φ/h ≈ 4.78×10¹⁴ Hz.

29Sketch a fully labelled graph of E_max against frequency for a metal, showing clearly where h, f₀ and Φ can be read from the graph. (Describe your sketch in words if you cannot draw here.)

Straight line, gradient = h (or h/e for a V_s–f graph), x-intercept = f₀, y-intercept = −Φ (or −Φ/e).

30Explain why increasing the intensity of light incident on a metal surface increases the photoelectric current but not the maximum kinetic energy of the photoelectrons.

More intensity means more photons per second (more photoelectrons per second, higher current), but each photon still carries the same energy hf, so E_max = hf − Φ is unchanged.

10. Compton scattering

The photoelectric effect was strong, but not universally convincing, evidence for photons — sceptics could still try to patch up the wave theory. In 1923, Arthur Compton provided much more direct evidence by firing high-frequency X-rays at a block of graphite and measuring the X-rays that scattered off the loosely-bound outer electrons in the carbon atoms.

A wave theory of light predicts that scattered radiation should keep the same wavelength as the incident radiation. Compton found instead that the scattered X-rays had a longer wavelength (lower energy) than the incident X-rays — and that the size of this shift depended on the scattering angle, θ, but not on the wavelength or intensity of the incident beam.

Why this proves light is particle-like. Compton explained his results by treating the interaction as a "collision" between a particle photon (with momentum p = h/λ) and an individual electron, obeying conservation of momentum and energy — exactly like two billiard balls colliding. Since the photon transfers some of its energy (and momentum) to the electron, the scattered photon must have lower energy, and therefore a longer wavelength than before. This only makes sense if light is made of particle-like photons.

Fig. 10.1 Compton scattering: an incoming X-ray photon transfers some of its energy and momentum to an electron. The scattered photon emerges at angle θ with a longer wavelength (lower energy) than it arrived with; the electron recoils.

Conservation of momentum.

momentum of initial photon = momentum of electron + momentum of final photon

p_i = p_e + p_f (a vector equation — the electron and the scattered photon fly off in different directions, so their momenta add like vectors, not like plain numbers)

Before the collision, the electron is at rest, so it has zero momentum — all of the momentum belongs to the incoming photon. After the collision, the electron is recoiling, so it now has some momentum of its own. That momentum cannot come from nowhere: it has been transferred directly from the photon. Since total momentum is conserved, whatever momentum the electron gains, the photon must lose. A photon's momentum is p = h/λ, so if the photon's momentum decreases, its wavelength λ must increase. This is why the scattered photon always comes out with a longer wavelength than it went in with — not because of an equation to memorise, but because the electron had to get its momentum from somewhere, and the photon is the only place it could have come from.

shift in wavelength: λ_f − λ_i = Δλ = (h / m_ec)(1 − cos θ)

where θ is the scattering angle and h/(m_ec) = 2.426 × 10⁻¹² m is called the Compton wavelength.

Worked example 10.1

X-rays of wavelength 1.00 × 10⁻¹⁰ m are Compton-scattered through an angle of 60°. Calculate the wavelength of the scattered X-rays.

Answer:
Δλ = (2.426×10⁻¹²)(1 − cos 60°) = (2.426×10⁻¹²)(0.500) = 1.213×10⁻¹² m
λ_f = λ_i + Δλ = (1.00×10⁻¹⁰) + (1.213×10⁻¹²) = 1.012×10⁻¹⁰ m

Check your understanding

31Calculate the change in wavelength for X-rays Compton-scattered through an angle of 90°.

Δλ = (2.426×10⁻¹²)(1 − cos 90°) = 2.43×10⁻¹² m.

32At what scattering angle is the wavelength shift (a) a maximum, and (b) zero? Explain both answers physically.

Maximum shift at θ = 180° (photon bounces straight back, transferring the most momentum); zero shift at θ = 0° (photon continues undeviated, no momentum transferred).

33Explain why Compton scattering is considered stronger evidence for the particle nature of light than the photoelectric effect.

Compton scattering can only be explained by treating light as particles colliding with electrons and obeying conservation of momentum and energy — a wave model predicts no wavelength shift at all, so the shift is direct, quantitative, and inescapable evidence, whereas photoelectric-effect observations were initially argued about for longer.

34State two reasons why Compton scattering is not observed with visible light.

(1) the wavelength shift is a fixed ~2.4×10⁻¹² m regardless of incident wavelength — utterly negligible compared to a visible wavelength of ~5×10⁻⁷ m; (2) visible photons have energies comparable to the binding energy of outer electrons, so those electrons cannot be treated as free, stationary particles as the simple Compton model requires.

11. The wave nature of matter

The photoelectric effect and Compton scattering both show that light — long thought to be a wave — sometimes behaves like a stream of particles. In 1924, the French physicist Louis de Broglie asked the reverse question: could matter, long thought to be made of particles, sometimes behave like a wave?

de Broglie wavelength of a moving particle: λ = h / p (p = momentum = mv)

This bold hypothesis was confirmed in 1927 by the Davisson–Germer experiment: a beam of electrons fired at a nickel crystal produced a diffraction pattern, just like X-rays diffracting from the regularly-spaced atoms in a crystal lattice. Diffraction — constructive and destructive interference spreading a wave out around obstacles — is a behaviour unique to waves, so this was direct evidence that electrons have wave-like properties.

Wave–particle duality. Put together, the photoelectric effect, Compton scattering, and electron diffraction show that all particles have wave-like properties, and all electromagnetic waves have particle-like (photon) properties. Neither the "pure wave" nor "pure particle" picture is complete on its own — this is wave–particle duality. In practice, the wave nature of matter is only noticeable for objects with extremely small momentum (because h is so tiny): it is significant for electrons, but utterly undetectable for a thrown ball or a moving car.

Worked example 11.1

Electrons are accelerated from rest through a potential difference of 3.0 kV.

Calculate the kinetic energy gained, in joules.
Calculate the speed of the electrons.
Calculate their de Broglie wavelength.

Answer:
a) E_k = qV = (1.60×10⁻¹⁹)(3000) = 4.80×10⁻¹⁶ J
b) E_k = ½mv² ⟹ v = √(2E_k/m) = √(2×4.80×10⁻¹⁶ / 9.11×10⁻³¹) = 3.25×10⁷ m s⁻¹
c) p = mv = (9.11×10⁻³¹)(3.25×10⁷) = 2.96×10⁻²³ kg m s⁻¹
λ = h/p = (6.63×10⁻³⁴)/(2.96×10⁻²³) = 2.24×10⁻¹¹ m

Check your understanding

35A proton has a de Broglie wavelength of 1.2 × 10⁻¹³ m. Calculate its momentum, and hence its speed (assume non-relativistic motion).

p = h/λ ≈ 5.53×10⁻²¹ kg m s⁻¹; v = p/m ≈ 3.3×10⁶ m s⁻¹.

36A proton and an electron travel at the same speed. Which one has the longer de Broglie wavelength? Explain your answer.

The electron — since λ = h/(mv), a smaller mass at the same speed gives a larger wavelength.

37A tennis ball of mass 0.058 kg travels at 25 m s⁻¹. Estimate its de Broglie wavelength and explain why the wave nature of the ball is never observed.

λ ≈ 4.6×10⁻³⁴ m — many orders of magnitude smaller than any aperture or obstacle that could diffract it, so no wave effects can ever be observed.

38Outline how the Davisson–Germer experiment provided evidence for de Broglie's hypothesis.

Electrons fired at a nickel crystal produced a diffraction pattern (constructive/destructive interference) matching the wavelength de Broglie predicted — diffraction is a wave-only phenomenon, so this confirmed matter has wave properties.

Looking back: two theories for the same thing

By the late 1920s, physics had accepted a paradox: light behaves as a wave in diffraction and interference, and as a particle in the photoelectric effect and Compton scattering. As Einstein himself put it, we need "two theories... [which] together describe the phenomena of light." This same wave–particle duality applies to matter — and it is one of the central, still slightly unsettling, ideas at the heart of modern quantum mechanics.

Timeline: from plum pudding to wave–particle duality

Year	Discovery	What it showed
1897	J. J. Thomson discovers the electron	Atoms are divisible; leads to the "plum pudding" model
1909	Geiger–Marsden–Rutherford scattering experiment	Atoms have a tiny, dense, positively charged nucleus
1900–1905	Planck's quantum hypothesis; Einstein's photon model	Light delivers energy in discrete quanta, E = hf
1913	Bohr model of hydrogen	Electron energy levels and angular momentum are quantized
1923	Compton scattering	Photons carry momentum and scatter like particles
1924	de Broglie's hypothesis	Matter should have wave-like properties, λ = h/p
1927	Davisson–Germer electron diffraction	Electrons diffract — confirming wave–particle duality

Glossary

Nucleon: A proton or a neutron; the particles that make up an atomic nucleus.
Nuclide: A specific type of nucleus, defined by its proton number Z and nucleon number A, written as ^A_ZX.
Isotope: One of two or more atoms of the same element (same Z) with different numbers of neutrons (different A).
Strong nuclear force: The short-range force that binds nucleons together in the nucleus, strong enough to overcome electrostatic repulsion between protons but negligible beyond about 3 × 10⁻¹⁵ m.
Distance of closest approach: The minimum separation reached by a charged particle fired directly at a nucleus, found by equating its initial kinetic energy to the electric potential energy at that separation.
Photoelectric effect: The emission of electrons from a metal surface when light of sufficiently high frequency is incident on it.
Work function, Φ: The minimum energy needed to remove the most loosely bound electron from the surface of a particular metal.
Threshold frequency, f₀: The minimum frequency of incident light that can just cause photoelectric emission from a given metal, f₀ = Φ/h.
Compton scattering: The scattering of a photon off a free (or loosely bound) electron, in which the photon loses energy and momentum and its wavelength increases.
De Broglie wavelength: The wavelength λ = h/p associated with any moving particle of momentum p, responsible for wave-like behaviour such as diffraction.
Energy level: One of the discrete, allowed values of energy that an electron bound in an atom may have.
Ground state: The lowest-energy, most stable energy level available to an electron in an atom.