Properties of nuclei and high-energy scattering (HL)

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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).

Answer

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

Answer

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

Answer

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