Atoms & Nuclei

Rutherford & Bohr · hydrogen spectrum · the nucleus · binding energy · fission, fusion & radioactivity · Class 12 Physics

1The nuclear atom (Rutherford)

Rutherford fired fast, positive α-particles at a wafer-thin gold foil and watched where they went. Almost all sailed straight through, but a tiny fraction — about one in eight thousand — bounced back through more than 90°. In his own words, it was “as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”

A big deflection needs a big, concentrated force. So the positive charge and nearly all the mass of an atom cannot be spread out — they must be packed into a minute nucleus at the centre, with the electrons occupying the vast empty space around it. That is the whole result: an atom is mostly empty, with a tiny, dense, positive nucleus. The nucleus turned out to be about 10⁻¹⁵ m across, roughly ten-thousand times smaller than the atom itself.

But Rutherford's planetary picture had a fatal flaw. An electron circling the nucleus is accelerating, and classical electromagnetism says an accelerating charge must radiate energy continuously. It would spiral into the nucleus in about 10⁻⁸ s, and while doing so emit a smooth rainbow of all frequencies. Atoms plainly do not collapse, and their light comes in sharp separate lines, not a smear. Something new was needed.

2Bohr's model, built up from two postulates

Bohr kept Rutherford's nucleus but bolted two bold quantum rules onto it. Everything about the hydrogen atom then falls out by ordinary mechanics.

Bohr's two postulates

(1) Stationary states. The electron may occupy only certain special orbits in which it does not radiate, even though it is accelerating. (2) Quantised angular momentum. Those orbits are exactly the ones where the angular momentum is a whole-number multiple of h/2π:

Quantisation condition m v r = n h2π n = 1, 2, 3, …

A third rule tells us where the light comes from: an atom emits (or absorbs) a photon only when the electron jumps between two allowed orbits, and the photon carries the exact energy difference, h ν = E_i − E_f. Sharp lines, at last.

Deriving the orbit radius

The electron of charge −e is held in its circle by the Coulomb pull of the nuclear charge +Z e. That electric force is the centripetal force:

14πε₀ Z e²r² = m v²r

Use the quantisation rule to replace v = n h / 2π m r, substitute, and solve for r. The messy constants collect into one number, and the clean result is:

Radius of the n-th orbit r_n = n² h² ε₀π m Z e² = 0.529 n²Z Å

The key physics is r_n ∝ n² — orbits balloon outward as the square of n. For hydrogen (Z = 1) the smallest orbit, n = 1, has radius 0.529 Å, the famous Bohr radius.

Deriving the energy

The electron's total energy is kinetic plus potential. From the force balance, ½ m v² = 18πε₀ Z e²r, while the Coulomb potential energy is −14πε₀ Z e²r. Adding them, the total is exactly minus the kinetic energy:

E = 12 m v² − 14πε₀ Z e²r = −18πε₀ Z e²r_n

Put in r_n from above and, again, the constants pack into one value:

Energy of the n-th level E_n = −13.6 Z²n² eV

The energy is negative — the electron is bound; you must supply energy to tear it free. For hydrogen the ground state is E₁ = −13.6 eV, so the ionisation energy of hydrogen (the work to lift the electron from n = 1 to n = ∞) is exactly 13.6 eV. The levels crowd together as 1/n² and pile up towards zero at n = ∞, where the electron is just free.

3The hydrogen spectrum

Each spectral line is one downward jump. A photon of energy E_i − E_f comes out, and since E = h c / λ, its wavelength follows the Rydberg formula:

Rydberg formula (hydrogen) 1λ = R (1n₁² − 1n₂²) n₂ > n₁

Here R = 1.097 × 10⁷ m⁻¹ is the Rydberg constant — it is nothing but the collection of Bohr constants m e⁴ / 8 ε₀² h³ c, and its value came out right, which is what convinced everyone the model was on to something. Lines are grouped into series by the level the electron lands on (n₁):

Downward jumps of the electron. Landing on n = 1 gives the Lyman series (ultraviolet), on n = 2 the Balmer series (the visible lines), on n = 3 the Paschen series (infrared).

Lyman series — electron falls to n₁ = 1. Biggest energy drops, so shortest wavelengths: ultraviolet.
Balmer series — falls to n₁ = 2. These are the four lines you can actually see (Hα is the red 656 nm line).
Paschen series — falls to n₁ = 3. Small drops, long wavelengths: infrared. (Brackett to n = 4 and Pfund to n = 5 follow the same pattern, further into the IR.)

4Inside the nucleus — size & density

The nucleus holds Z protons and N = A − Z neutrons, together called nucleons; A is the mass number and Z the atomic number. Scattering experiments show the nuclear radius grows with the cube root of the number of nucleons:

Nuclear radius R = R₀ A^1/3 R₀ ≈ 1.2 × 10⁻¹⁵ m

Look at what that little cube root is telling you. Volume ∝ R³ ∝ A, so the volume is simply proportional to the number of nucleons. That means every nucleus has essentially the same density — pack in more nucleons and the ball just grows to keep the packing constant, like marbles in a bag. That density is colossal, about 2.3 × 10¹⁷ kg m⁻³: a matchbox of pure nuclear matter would weigh billions of tonnes.

5Mass defect & binding energy

Weigh a nucleus carefully and you find a surprise: it weighs less than the protons and neutrons that make it up, added separately. The shortfall is the mass defect:

Δm = [ Z m_p + (A − Z) m_n ] − M_nucleus

Where did the missing mass go? It was released as energy when the nucleons bound together, through Einstein's E = Δm c². To pull the nucleus apart again you would have to pay that energy back — which is why it is called the binding energy. A tightly bound nucleus has a large mass defect. The convenient conversion for nuclear work is:

Mass–energy conversion 1 u = 931.5 MeV

So a mass defect of just one atomic mass unit corresponds to a whopping 931.5 million electron-volts — millions of times more than the few eV of a chemical bond. That factor of a million is the entire reason nuclear energy dwarfs chemical energy.

6The binding-energy curve · fission & fusion

The number that really matters is the binding energy per nucleon, E_b/A — how tightly each nucleon is held, on average. Plot it against mass number and you get one of the most important curves in physics:

Binding energy per nucleon rises steeply for light nuclei, peaks at about 8.8 MeV near iron (A ≈ 56), then falls gently for the heaviest nuclei. Iron sits at the bottom of the energy valley — the most stable matter there is.

The curve is the key to nuclear energy. Energy is released whenever a reaction moves nucleons towards the iron peak, because more tightly bound means lower energy, and the surplus escapes.

Fission. A very heavy nucleus (uranium-235) is at the low, right-hand side. Split it into two medium nuclei nearer the peak and each nucleon becomes more tightly bound; the difference — about 200 MeV per fission — comes out. A stray neutron triggers the split, and the extra neutrons released can trigger more: a chain reaction, the principle of reactors and bombs.
Fusion. Very light nuclei (hydrogen isotopes) are at the low, left-hand side. Join them and, again, you climb towards the peak and release energy — this is what powers the Sun, where hydrogen fuses to helium. Fusion needs enormous temperatures to overcome the protons' mutual repulsion, which is why it is so hard to harness on Earth.

7Radioactivity — α, β, γ

Unstable nuclei shed energy by spitting out radiation. There are three kinds, first sorted by how far they penetrate and how a magnetic field bends them:

Alpha (α) — a helium-4 nucleus, ₂⁴He (2 protons + 2 neutrons). Heavy, slow, and stopped by a sheet of paper. Emitting one lowers Z by 2 and A by 4: _Z^AX → _Z−2^A−4Y + ₂⁴He.
Beta (β⁻) — a fast electron created when a neutron turns into a proton inside the nucleus. So Z rises by 1 while A is unchanged: _Z^AX → _Z+1^AY + ₋₁⁰e + ν̄. More penetrating than alpha; stopped by a few mm of aluminium.
Gamma (γ) — a high-energy photon, emitted when a nucleus left in an excited state (often just after an α or β decay) drops to a lower state. It carries away energy but changes neither Z nor A. The most penetrating of the three; needs thick lead or concrete.

Two bookkeeping rules never fail: the total charge (Z) and the total nucleon number (A) are conserved on both sides of every decay equation. Balance those and you can always name the daughter.

8The decay law & half-life

You can never predict when a particular nucleus will decay — it is genuinely random. But with huge numbers, the statistics are rock-solid: the number decaying in a moment is proportional to how many are still present. That single sentence is the whole law:

dNdt = − λ N

The constant λ is the decay constant — the probability per second that any one nucleus decays. Integrating (this is the same maths as compound interest, run backwards) gives the exponential decay law:

Radioactive decay law N = N₀ e^−λt

The natural yardstick is the half-life T_½: the time for half the nuclei to decay. Set N = N₀/2 and solve ½ = e^{−λ T_½}. Take logs: λ T_½ = ln 2, so

Half-life T_½ = ln 2λ = 0.693λ

After one half-life, half remain; after two, a quarter; after three, an eighth — the count halves every T_½ no matter where you start. After n half-lives the surviving fraction is simply (½)ⁿ. (The related activity, the counts-per-second you actually measure, is A = λN and decays in exactly the same exponential way.)

Remember this

Half-life doesn't depend on how much you start with. A gram or a tonne of the same isotope both drop to half in one T_½. Decay is a fixed fraction per unit time, never a fixed amount — which is exactly why the curve is an exponential and not a straight line.

◆ Activity — radioactive decay

A sample starts with N₀ = 1000 nuclei. Drag the slider to move time forward (measured in half-lives), or press the button to jump one whole half-life. Watch the curve N = N₀ e^−λt fall, and notice the survivors halve every half-life — 1000 → 500 → 250 → 125 … — never reaching a fixed “zero” amount.

Worked example

A radioactive sample has a half-life of 6 hours and an initial activity of 8000 counts per second. (a) What fraction is left after 18 hours? (b) What is its decay constant? (c) After how long does the activity fall to 1000 counts per second?

(a) 18 hours is exactly 18 / 6 = 3 half-lives. Each halves the sample, so the fraction remaining is (½)³ = 1/8.

(b) λ = 0.693T_½ = 0.6936 h = 0.1155 h⁻¹ = 3.2 × 10⁻⁵ s⁻¹.

(c) Going from 8000 to 1000 is a factor of 8 = 2³, i.e. 3 half-lives, so it takes 3 × 6 = 18 hours. Because activity A = λN, it falls off with the very same half-life as the number of nuclei — no separate calculation needed.

9Common confusions to clear up

Bohr energies are negative for a reason. The minus sign means bound. A “higher” level like n = 2 (−3.4 eV) is higher because it is less negative than n = 1 (−13.6 eV), i.e. closer to freedom.
The nucleus loses mass, and that mass is the binding energy. The nucleus is lighter than its parts precisely because energy was given up in binding them. Nothing is missing — mass became energy.
Both fission and fusion release energy, from opposite ends of the curve. The rule is not “heavy” or “light” but “move towards iron.” Splitting the heavy and joining the light both climb the binding-energy peak.
The beta electron is not an orbital electron. It is created on the spot when a neutron converts to a proton; it does not come from the electron cloud.
Half-life is not “half the lifetime,” and a sample never truly vanishes. Each half-life removes half of what's left, so the amount only ever approaches zero — it never gets there in finite time.
Gamma emission changes neither Z nor A. It is the nucleus shedding surplus energy after a decay, not a change of element.

10Check yourself

Class 12 Physics · Atoms & Nuclei · aligned to NCERT Chapters 12–13 · SmartStudy.School