Structure of the atom

**Proton** (+1) and **neutron** (0) in the nucleus; **electron** (−1) around it.

A particle found **in the nucleus** — i.e. a **proton or a neutron**.

A specific type of nucleus, fixed by its number of **protons and neutrons** (e.g. carbon-14).

**A** (top) = nucleon number = protons + neutrons. **Z** (bottom) = proton number = number of protons.

**N = A − Z** (nucleon number minus proton number).

**electrons = Z − q.** A 2+ ion has Z − 2 electrons; a 1− ion has Z + 1.

Only the **electron** count. Protons and neutrons (the nucleus) are unchanged.

Most passed **straight through**; a few deflected through **large angles**; a very few **bounced straight back**.

The atom is **mostly empty space**.

The positive charge and almost all the mass are in a **tiny, dense, positively charged nucleus**.

A small, fast, positive particle = **2 protons + 2 neutrons** (a helium nucleus).

An electron's mass is about **1/2000** of a nucleon's — negligible next to protons and neutrons.

Spread-out charge gives a **weak field**: it could never turn a fast alpha through a large angle. Large deflections need the charge **concentrated** in a tiny nucleus.

Only those heading **almost straight at** a nucleus; the rest pass far from any nucleus and are barely deflected.

Thin: each alpha meets **at most one** nucleus. Vacuum: alphas would be **stopped by air** within a few cm.

An atom can only have certain **fixed** allowed energies — never the values in between (like stairs, not a ramp).

A single tiny **packet of light energy**. Its energy is given by E = hf = hc/λ.

It **emits a photon** whose energy equals the **gap** between the two levels (an emission line).

An electron **jumps up** to a higher level — but only if the photon's energy exactly matches a level **gap**.

$E = hf$ — energy = Planck constant × frequency (given in the data booklet).

$E = \dfrac{hc}{\lambda}$ — bigger energy means shorter wavelength (given).

The one with the **smallest** energy drop — because E = hc/λ, a small energy means a large wavelength.

The **biggest** energy drop — more energy means a shorter wavelength (and higher frequency).

**n(n − 1) ÷ 2** distinct wavelengths. E.g. n = 3 → 3 lines; n = 4 → 6 lines.

Emission = **bright lines** on dark (electron falls, photon out). Absorption = **dark lines** in a rainbow (electron rises, photon in). Same atom → same line positions.

Each element has its **own** set of energy levels, so its own unique pattern of lines — you can match a spectrum to an element.

λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (3.0 × 10⁻¹⁹) ≈ 6.6 × 10⁻⁷ m.

The photons are re-emitted in **all directions**, so far fewer travel on in the original beam towards you.

They are at the **same wavelengths**: both come from the same energy gaps.

Measure the wavelengths of its absorption lines and **match** them with lab spectra of elements; every line of an element must be present.

The **energy gained by one electron** when it moves through a potential difference of **one volt**. It is a unit of energy.

**1 eV = 1.60 × 10⁻¹⁹ J** — given in the data booklet.

Energy = charge × voltage. The electron's charge e = 1.60 × 10⁻¹⁹ C, so crossing 1 V gives it 1.60 × 10⁻¹⁹ J.

**Multiply** the number of eV by 1.60 × 10⁻¹⁹.

**Divide** the energy in joules by 1.60 × 10⁻¹⁹.

**1 keV = 10³ eV** (a kilo-electronvolt).

**1 MeV = 10⁶ eV** (a mega-electronvolt). Nuclear energies are usually quoted in MeV.

Atomic and nuclear energies are tiny fractions of a joule; the eV gives convenient, easy-to-read numbers.

A **few eV** (about 2 eV) — that is why atomic transitions emit visible light.

A **few MeV** — about a million times bigger than an atomic-transition energy.

**Joules (J).** Convert to eV at the end (÷ 1.60 × 10⁻¹⁹) only if the question asks for eV.

5.0 × 1.60 × 10⁻¹⁹ = **8.0 × 10⁻¹⁹ J** (eV → J, so multiply).

**1000 eV (1 keV)**: energy in eV = charge in e × pd in V = 2 × 500.

2.0 × 10³ eV × 1.60 × 10⁻¹⁹ = **3.2 × 10⁻¹⁶ J**.

Work done = qV = (ne)V, and e × 1 V is one eV, so the energy is **nV electronvolts**.

Charge only comes in **whole-number multiples** of the elementary charge e — never a fraction of e. It changes in fixed steps.

**e = 1.60 × 10⁻¹⁹ C** — the charge on one proton (+e) or one electron (−e). The smallest 'lump' of charge. Given in the data booklet.

$Q = N e$ — total charge = whole number N of elementary charges. Rearranged: $N = \dfrac{Q}{e}$.

Use **N = Q ÷ e**. The answer must be a **whole number**.

Because you can only add or remove **whole** electrons — charge changes in steps of e, so N is always a whole number.

It has **gained extra electrons**. (A positively charged object has **lost** electrons.) Each electron carries −e.

Every measured drop charge was a **whole-number multiple of the same smallest step**, e — the experimental proof that charge is **quantised**.

**No.** N = Q ÷ e = 2.4 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 1.5, not a whole number — so it is not allowed.

N = Q ÷ e = 6.4 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = **4** electrons.

**No** — it is the definition of charge quantisation, so memorise it. But the constant **e = 1.60 × 10⁻¹⁹ C** IS given.

Each half gets **4e** (8e ÷ 2). Still a whole multiple of e, so allowed.

**1e and 2e**: charge comes in whole e, so it cannot split as 1.5e + 1.5e; the total stays 3e.

The electric force balances the weight: **qE = mg**, so q = mg/E.

Every measured charge was a **whole-number multiple** of the same smallest value, e = 1.60 × 10⁻¹⁹ C.

$R = R_0 A^{1/3}$, with $R_0 = 1.2\times10^{-15}$ m (a constant the same for every nucleus).

The **nucleon number** — the total number of **protons + neutrons** in the nucleus.

R ∝ **A^(1/3)** — it grows as the **cube root** of the nucleon number.

$R_0 = 1.2\times10^{-15}$ m (≈ 1.2 fm).

$R = 1.2\times10^{-15} \times \sqrt[3]{64} = 1.2\times10^{-15}\times 4.0 = 4.8\times10^{-15}$ m.

Volume ∝ R³ ∝ A and mass ∝ A, so density = mass/volume is the **same for every nucleus**.

Radius × **2** (∛8 = 2); volume × **8** — so density is unchanged.

Fire a beam of **high-energy electrons** at it and analyse the **diffraction / scattering** pattern (the first ring gives the diameter).

To resolve a 10⁻¹⁵ m nucleus you need λ ≈ 10⁻¹⁵ m; $\lambda = h/p$ so a small λ needs a **large momentum** (high energy).

The de Broglie relation $\lambda = h/p$ — bigger momentum gives a shorter wavelength.

Electrons are **point-like** and feel only the EM force (not the strong force), so the scattering pattern is clean and easy to interpret.

About **10¹⁷ kg m⁻³** — and it is the same for essentially every nucleus.

All its kinetic energy becomes electric potential energy: **E_k = k(2e)(Ze)/d**, so d = k(2e)(Ze)/E_k.

The alpha gets close enough (d ≈ R) to feel the **strong nuclear force**; Rutherford's model assumes the electric force only.

**No**: d depends on the charge Ze, and isotopes have the same Z.

$E_n = -\dfrac{13.6}{n^2}$ eV, with n = 1, 2, 3, … The levels are **discrete** (quantized).

The zero is set at the just-free electron (n → ∞), so any **bound** electron has **less** energy — hence negative.

n = 1, with $E_1 = -13.6$ eV — the **lowest** (most tightly bound) level.

$E_2 = -13.6/4 = -3.40$ eV.

$hf = E_i - E_f$ — the photon energy equals the size of the energy drop.

**Emission**: electron jumps **down**, atom gives out a photon. **Absorption**: electron jumps **up**, atom takes a photon in.

Its levels are **discrete**, so only certain energy drops — and hence certain photon energies — are possible.

**13.6 eV** — the energy to lift the electron from the ground state (−13.6 eV) up to 0 eV (free).

$(-1.51) - (-3.40) = 1.89$ eV $= 3.0\times10^{-19}$ J.

Multiply by $1.60\times10^{-19}$ (1 eV = 1.60×10⁻¹⁹ J).

The one with the **largest** energy drop — falling down to n = 1.

The gaps **shrink** (1/n²), so the levels **crowd together** approaching 0 eV.

The electron's **angular momentum**: mvr = nh/2π, n = 1, 2, 3 … That fixes the allowed orbits and so the energy levels.

Atoms and ions with **one electron**: H, He⁺, Li²⁺. It fails when there are two or more electrons.

Find **k** from one given transition energy, then use it for the transition asked about.