QB_PH_SEC9_001: Nuclear Properties & Nuclear Forces

Using the nuclear radius empirical formula $R = R_0 A^{1/3}$ with $R_0 = 1.2\text{ fm}$, what is the radius (in fm) of the $_{27}^{64}\text{Cu}$ nucleus? (Give your answer as an exact decimal)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Nuclear Radius Relation**: From electron-nucleus scattering measurements, charge distribution follows: $R = R_0 A^{1/3}$ 2. **Parameter Substitution**: - Mass number $A = 64$ for $_{27}^{64}\text{Cu}$ - $R_0 = 1.2\text{ fm}$ 3. **Evaluation**: $A^{1/3} = 64^{1/3} = 4$ $R = 1.2\text{ fm} \times 4 = 4.8\text{ fm}$ 4. **Official Exact Answer**: 4.8 fm.