Dual Nature of Matter, Atoms and Nuclei

1The photoelectric effect

Shine light on a metal and electrons come out — but only if the frequency clears a threshold, however bright the light is. Einstein's account is one line of accounting: a single photon of energy hν arrives, spends φ (the work function) getting the electron out of the surface, and the rest leaves as kinetic energy. So K_{max} = h\nu - \phi, and below \nu_0 = \phi/h nothing is emitted at all, however long you wait.

Two consequences follow that no wave picture can produce. Brightness is photons per second, so it sets how many electrons come out and never how fast. Frequency is energy per photon, so it alone sets K_max. And because the transaction is one photon to one electron, emission starts within about 10⁻⁹ s of switching the lamp on, instead of after the slow build-up a wave would need.

Figure. Both metals give the same tilt. The slope is h and nothing else — not the metal, not the intensity — which is why this graph is how Planck's constant is measured. Raising the work function slides the line to the right without turning it: the threshold moves, the slope cannot. Extended back to ν = 0 either line would cut the axis at −φ.

How it works

1. One photon, one electronThe photon is absorbed whole. Two photons cannot club together to pay one work function.
2. Pay the work functionφ is the least it can cost to escape the surface; electrons bound deeper pay more.
3. Keep the changeK_max = hν − φ, and only the least tightly bound electrons come out with all of it.

2Stopping potential and the photocell characteristic

To measure K_max you make the electrons climb a hill. Reverse the collector's polarity and only electrons energetic enough to reach it get counted; wind the retarding voltage up until the current just vanishes and that is the stopping potential, with eV_0 = K_{max}. Written out, V_0 = (h/e)\nu - \phi/e: a straight line against frequency with slope h/e = 4.14 × 10⁻¹⁵ V s for every metal ever tested.

The full current–voltage characteristic carries the rest of the story. Push the anode positive and the current climbs to a saturation value fixed by how many photoelectrons are produced per second — a photon-counting number, so it follows intensity. Pull it negative and the current dies at −V₀ — an energy number, so it follows frequency. One graph, two independent controls.

Figure. Reading this graph is two separate questions. How high is the plateau? — a photon-counting question, so intensity: doubling it lifts the plateau and leaves the foot exactly where it was. How far to the left does the curve leave the axis? — an energy question, so frequency: raising ν drags the foot further into the retarding region while the plateau stays put, because the same number of photons still arrive each second.

3De Broglie matter waves

De Broglie's move was to run the photon relation backwards. If light of wavelength λ carries momentum p = h/λ, then anything with momentum p has a wavelength λ = h/p — an electron, a neutron, a cricket ball. For a particle of kinetic energy K, non-relativistically, \lambda = h/\sqrt{2mK}.

Accelerate an electron from rest through V volts and K = eV, which collapses the whole thing to λ = 12.27/√V Å. At 100 V that is 1.23 Å, the spacing of atoms in a crystal — which is why Davisson and Germer saw electrons diffract off nickel, and why an electron microscope resolves what light cannot. The same formula gives a 100 g ball thrown at 10 m s⁻¹ a wavelength of 6.6 × 10⁻³⁴ m: real, and forever unmeasurable.

Figure. The curve is 1/√V, so it flattens fast: getting from 1.23 Å down to 0.61 Å costs four times the voltage, not twice. A few tens to a few hundred volts is exactly the window where the wavelength matches the spacing of atoms in a crystal, which is the whole of electron diffraction. A proton at the same 100 V would sit at 0.029 Å — forty-three times lower, and indistinguishable from the axis at this scale.

How it works

1. The same relation, both waysp = h/λ describes a photon; λ = h/p describes everything that is not one.
2. Energy, not speedλ = h/√(2mK), so at equal kinetic energy the heavier particle has the shorter wave.
3. Accelerated by a voltageK = qV, and for an electron that is λ = 12.27/√V Å with V in volts.

4Rutherford's nuclear atom

Geiger and Marsden fired α particles at a thin gold foil and watched where they went. Almost all passed nearly straight through, but about one in eight thousand turned through more than 90°, and a rare few came straight back. A diffuse 'plum-pudding' atom, with its charge spread out, could deflect nothing sharply; only a tiny, massive, positively charged core could throw an α straight back. So Rutherford placed essentially all the mass and all the positive charge in a nucleus about 10⁻¹⁴ m across — ten thousand times smaller than the atom — with the electrons far outside.

For a head-on approach the α stops where all its kinetic energy has turned into electrostatic potential energy: the distance of closest approach r₀ = (1/4πε₀)(2Ze²)/K, so r₀ shrinks as the α is made faster and grows with the target's charge. The model's fatal flaw was left for Bohr — a classical electron orbiting the nucleus must radiate and spiral in within nanoseconds, and matter does not.

Figure. The impact parameter — how far off-centre the α is aimed — decides everything. Aimed well to one side (top) it barely deviates; aimed closer it is thrown aside; aimed almost dead-on it climbs the Coulomb hill, stops at the distance of closest approach and returns the way it came. Because the nucleus is so small, the near-head-on case is rare, about one α in eight thousand, and that rarity is what made the experiment decisive.

How it works

1. The rare event is the resultThe many tiny deflections say little; the one-in-eight-thousand backscatter demands charge concentrated in a hard core.
2. Stop where the energy runs outHead-on, K = (1/4πε₀)(2Ze²)/r₀, so r₀ ∝ Z/K — faster α, closer approach.
3. Hand the flaw to BohrA classical orbit radiates and collapses in nanoseconds; quantised orbits are the repair.

5The Bohr model

Classical physics says an orbiting electron radiates and spirals into the nucleus within nanoseconds. Bohr's repair was to declare that it does not: certain orbits are stationary, singled out by angular momentum coming in whole multiples of h/2π, so that mvr = nh/2\pi. Radiation happens only in the jump between two of them, and then hν = E_i − E_f.

Impose that on a Coulomb orbit and everything else is forced: r_n = 0.529\,n^2/Z Å, v_n = 2.19\times10^{6}(Z/n) m s⁻¹ and E_n = -13.6\,Z^2/n^2 eV. Notice what the model covers. It is exact for one electron around a nucleus of charge Ze — H, He⁺, Li²⁺ — and it fails outright for neutral helium, which has two electrons that push on each other.

Figure. Stationary energies E_n = −13.6/n² eV. The gaps shrink toward the continuum at n = ∞; photons match level differences, not the orbit picture the vocabulary cannot draw.

How it works

1. Quantise the orbitOnly radii with mvr = nh/2π survive; everything in between is simply not allowed.
2. Balance the forcesCoulomb attraction supplies the centripetal force, which then fixes r_n and v_n.
3. Read the energyE_n = −13.6Z²/n² eV — negative, equal to minus the kinetic energy and to half the potential energy.

6Energy levels and the hydrogen spectrum

Hydrogen emits a set of sharp lines, and the ladder of Bohr levels accounts for every one. A jump from n_i down to n_f emits a photon of exactly E_i − E_f; group the jumps by where they land and the named series appear. Everything ending on n = 1 is the Lyman series, on n = 2 the Balmer, on n = 3 the Paschen.

In wavelength form, 1/\lambda = RZ^2(1/n_f^2 - 1/n_i^2) with R = 1.097 × 10⁷ m⁻¹. Two habits pay for themselves. Within a series the longest wavelength comes from the smallest jump, always n_f + 1 → n_f, and the shortest — the series limit — from n = ∞. And because the levels crowd towards zero, only Balmer has lines in the visible: Lyman is entirely ultraviolet and everything from Paschen onwards is infrared.

Figure. The rungs are not evenly spaced: they pile up towards E = 0 as 1/n², which is why every series has a limit — a shortest wavelength it can never pass. The two unlabelled rungs above n = 3 are n = 4 and n = 5, and infinitely many more are packed into the sliver below the top line. The two downward arrows emit a photon each; the long upward one is ionisation, the 13.6 eV that takes a ground-state electron out to n = ∞ and frees it.

7Nuclear size and density

Scattering experiments give the nucleus a radius R = R_0A^{1/3} with R₀ = 1.2 fm. The cube root is the whole content: nucleons pack against one another like touching marbles rather than spreading out, so eight times the mass number buys only twice the radius.

Volume then goes as A while mass goes as A, and the ratio is free of A altogether — every nucleus from helium to uranium has the same density, about 2.3 × 10¹⁷ kg m⁻³. That is 10¹⁴ times the density of water, and it is the most quoted single fact about nuclear matter.

Figure. The curve flattens because the radius follows the cube root: from A = 27 to A = 216 the nucleus gains eight times the nucleons and doubles its radius, 3.6 fm to 7.2 fm. Since the volume has risen eightfold as well, the density has not moved at all — which is what makes the nucleus a liquid drop rather than a gas.

How it works

1. Cube root, not linearR = 1.2 A^(1/3) fm, so two radii are in the ratio of the cube roots of their mass numbers.
2. Volume tracks AV = (4/3)πR³ ∝ A, meaning each nucleon occupies its own fixed volume.
3. So density cannot varyρ = mass/volume is the same 2.3 × 10¹⁷ kg m⁻³ for every nucleus there is.

8Mass defect and binding energy

Weigh a nucleus and it comes out lighter than its parts. Two hydrogen atoms and two neutrons total 4.03298 u; the ⁴He atom is 4.002603 u — atomic masses on both sides, so the two electrons each side cancel and never enter the answer. The missing 0.030377 u has not vanished — it is the energy given out when the nucleons bound together, and it is exactly what you must pay to take them apart again: B = \Delta m\,c^2.

Work in atomic mass units and the conversion is a single number, 1 u = 931.5 MeV; going round by kilograms and joules is four times the work for the same answer. Then divide by A before comparing two nuclei. Total binding energy mostly says how big a nucleus is; binding energy per nucleon says how tightly it is held.

Figure. Binding energy is a depth, not a possession. The bound nucleus sits 28.3 MeV below the four free nucleons: that gap is both the energy released when it formed and the energy needed to pull it apart. On a balance the identical gap shows up as 0.0304 u of missing mass — the same fact, weighed instead of counted.

How it works

1. Add up the free partsZ hydrogen atoms plus N neutrons — atomic masses throughout, so the electrons cancel.
2. Subtract the measured massThe difference Δm is the mass defect, positive for every bound nucleus.
3. Convert once, then divideB = Δm × 931.5 MeV, and B/A is the number worth comparing.

9The binding energy curve, fission and fusion

Plot binding energy per nucleon against mass number and you get the most useful curve in nuclear physics. It climbs steeply through the light nuclei, flattens into a broad maximum near A = 56 at about 8.8 MeV, then sags slowly to 7.6 MeV at uranium. Iron sits at the bottom of the energy valley — there is nothing to be gained by taking it anywhere.

Everything else can move towards that peak, and both directions release energy. Light nuclei gain by fusing; heavy nuclei gain by splitting. The release is roughly the number of nucleons involved times the gain in binding energy per nucleon, which is why one fission gives about 200 MeV while burning one carbon atom gives about 4 eV — fifty million times less.

Figure. A nucleus releases energy only by moving up this curve, and both ends can do it: the light ones by joining, the heavy ones by splitting, both heading for iron. A star that has fused its way to iron has nothing left to burn. The steep left edge hides real structure — ⁴He is anomalously tightly bound and stands well above ⁶Li next door — too fine to resolve at this scale.

Reading the curve

1. Find the peakThe maximum near A = 56 at ~8.8 MeV per nucleon is the most tightly bound arrangement of matter.
2. Left of it, fuseLight nuclei gain by joining — this is what powers stars, and hydrogen bombs.
3. Right of it, splitHeavy nuclei gain by breaking in two — this is what powers reactors.

10Radioactivity and half-life

A radioactive nucleus has no memory and no schedule. In any short interval every surviving nucleus has the same probability λ dt of decaying, whatever its age, and that one statement integrates to N = N_0e^{-\lambda t}. The activity you actually measure on a counter obeys the same law, since A = λN.

Three constants describe the same λ and nothing else does: the half-life t_{1/2} = \ln 2/\lambda = 0.693/\lambda, and the mean life \tau = 1/\lambda = 1.44\,t_{1/2}. When the elapsed time is a whole number of half-lives, skip the exponential altogether — after n half-lives the surviving fraction is (1/2)ⁿ.

Figure. Every half-life removes half of what is left, so the curve approaches the axis without ever reaching it — it simply runs out of nuclei to count. The three marked points are the whole of the arithmetic: N₀/2 after one half-life, N₀/4 after two, N₀/8 after three, and 6.25% after four. Nothing about the sample's past changes the next step.

How it works

1. Count half-lives firstn = t/t½ — turn the elapsed time into a count before doing anything else.
2. Halve it n timesThe remaining fraction is (1/2)ⁿ: three half-lives leaves an eighth, not a third.
3. Only then reach for eFor a fractional number of half-lives use N = N₀e^(−λt) with λ = 0.693/t½.

11Alpha, beta and gamma decay

A nucleus sheds excess energy by one of three routes, and each moves it differently across the chart of nuclides. Alpha decay ejects a ⁴₂He nucleus, dropping Z by 2 and A by 4 — the route heavy nuclei take. Beta-minus turns a neutron into a proton (n → p + e⁻ + ν̄), raising Z by 1 at constant A, the route for neutron-rich nuclei; beta-plus does the reverse. Gamma emission changes neither Z nor A — the nucleus merely drops from an excited state, usually just after an α or β has left it there.

Their penetrating powers run in the opposite order to their charges: the heavy, doubly charged α is stopped by a sheet of paper, β by a few millimetres of aluminium, and the uncharged γ needs centimetres of lead. Every decay's energy release is its Q-value, the mass lost times c² (using 1 u = 931.5 MeV), shared as kinetic energy among the products — and by momentum conservation the lighter product carries away almost all of it.

Figure. The chart of nuclides, proton number Z across and neutron number N up, with a parent nucleus near the middle. Each decay is an arrow: α drops both Z and N by two (down-left), β⁻ trades a neutron for a proton so Z rises and N falls (down-right), and β⁺ does the reverse (up-left). Gamma emission leaves the nucleus exactly where it is. The arrows are schematic direction indicators; the exact steps in Z and N are the numbers on the labels.

How it works

1. Name the route by the chartToo heavy → α (Z−2, A−4); too many neutrons → β⁻ (Z+1, A same); excited → γ (no change).
2. Balance Z and AThe superscripts (A) and subscripts (Z) must each balance across the arrow — that is the displacement law.
3. Take Q from the mass lossQ = Δm × 931.5 MeV; the light product gets the larger share of the kinetic energy.