# Nuclear Radius & Density ## Nuclear Radius - Nuclear radii are typically of the order $10^{-15}$ m or 1 fm (femtometre) - The radii of different nuclei vary depending on their size | Element | Nuclear radius $R / 10^{-15}$ m | Mass number $A$ | |---------|----------------------------------|-----------------| | Carbon | 2.66 | 12 | | Silicon | 3.43 | 28 | | Iron | 4.35 | 56 | | Tin | 5.49 | 120 | | Lead | 6.66 | 208 |

$1 \text{ fm} = 1 \times 10^{-15} \text{ m}$

### Relationship Between Nuclear Radius and Nucleon Number - The nuclear radius, $R$, varies with nucleon number, $A$, following a specific relationship - Experimental data reveals a clear pattern: as nuclei get larger (more nucleons), their radius increases - The relationship is **not** linear - larger nuclei don't grow proportionally to the number of nucleons added **Key observations from the $R$ vs $A$ graph:** - The graph starts with a steep gradient at the origin - The gradient gradually decreases, becoming almost horizontal for very large nuclei - This indicates that as more nucleons are added to a nucleus, the nucleus gets bigger - However, the number of nucleons $A$ is **not proportional** to the nuclear radius $R$ The relationship between nuclear radius and mass number can be visualized from the experimental data:

- Notice how the rate of increase in radius slows down as mass number increases - This non-linear relationship suggests a power law, not a simple direct proportion ### Calculating the Nuclear Radius - The radius of a nucleus depends on its nucleon number, $A$ - This makes physical sense: more nucleons occupy more space, resulting in a larger radius - Analysis of experimental data led physicists to deduce the following relationship: $$R = r_0 A^{1/3}$$ **Where:** - $R$ = nuclear radius (m) - $A$ = nucleon / mass number - $r_0$ = constant of proportionality = 1.2 fm = $1.2 \times 10^{-15}$ m (the radius of a proton)

The nuclear radius formula: $R = r_0 A^{1/3}$ where $r_0 = 1.2 \times 10^{-15}$ m

## Mean Densities of Atoms and Nuclei ### Equation for Nuclear Density - Assuming the nucleus is spherical, its volume is: $$V = \frac{4}{3}\pi R^3$$ - Where $R$ is the nuclear radius, related to mass number $A$ by: $$R = r_0 A^{1/3}$$ - Where $r_0$ is a constant of proportionality - Combining these equations gives: $$V = \frac{4}{3}\pi (r_0 A^{1/3})^3 = \frac{4}{3}\pi r_0^3 A$$ - This shows that nuclear volume, $V$, is proportional to the mass of the nucleus, $A$: $$V \propto A$$ - Mass ($m$), volume ($V$), and density ($\rho$) are related by: $$\rho = \frac{m}{V}$$ - The mass, $m$, of a nucleus equals: $$m = Au$$ **Where:** - $A$ = mass number - $u$ = atomic mass unit ($1.661 \times 10^{-27}$ kg) - Using the equations for mass and volume, nuclear density equals: $$\rho = \frac{Au}{\frac{4}{3}\pi r_0^3 A} = \frac{3u}{4\pi r_0^3}$$ ### Key Property of Nuclear Density - Since the mass number $A$ cancels out, the remaining quantities are all constant - This shows the density of the nucleus is: - **Constant** - **Independent of the radius** - Nuclear density is approximately $2.3 \times 10^{17}$ kg m$^{-3}$ for all nuclei - The fact that nuclear density is constant shows that **nucleons are evenly separated throughout the nucleus regardless of their size** - Think of it like a perfectly uniform material: whether you have a small piece or a large chunk, the density stays exactly the same

Nuclear density is constant for all nuclei - it doesn't depend on the size of the nucleus or the number of nucleons

## Worked Example ### Question Calculate the approximate density of an aluminium nucleus. Assume the atomic mass of aluminium to be 27u. ### Answer **Step 1: Write down the equations** - Density: $\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{Au}{\frac{4}{3}\pi R^3}$ - Volume of a sphere: $V = \frac{4}{3}\pi R^3$ - Nuclear radius: $R = r_0 A^{1/3}$ **Where:** - $r_0$ = constant = $1.2 \times 10^{-15}$ m - $A$ = mass number = 27 for aluminium **Step 2: Combine equations** $$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (r_0 A^{1/3})^3 = \frac{4}{3}\pi r_0^3 A$$ **Step 3: Calculate the volume of the nucleus** $$V = \frac{4}{3}\pi r_0^3 A = \frac{4}{3}\pi \times (1.2 \times 10^{-15})^3 \times 27 = 1.95 \times 10^{-43} \text{ m}^3$$ **Step 4: Calculate mass of aluminium nucleus** $$\text{Mass} = Au = 27 \times (1.661 \times 10^{-27}) = 4.48 \times 10^{-26} \text{ kg}$$ **Step 5: Calculate the density** $$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{4.48 \times 10^{-26}}{1.95 \times 10^{-43}} = 2.30 \times 10^{17} \text{ kg m}^{-3}$$ **Step 6: Finalise your answer** The density of an aluminium nucleus is $2.3 \times 10^{17}$ kg m$^{-3}$ (2 s.f.)

Don't let the powers confuse you. Work through step by step: it's just mass/volume with substitution

--- **Key Points Summary:** - Nuclear radii are of order $10^{-15}$ m (1 fm) - Nuclear radius formula: $R = r_0 A^{1/3}$ where $r_0 = 1.2 \times 10^{-15}$ m - Nuclear volume is proportional to mass number: $V \propto A$ - Nuclear density is constant at approximately $2.3 \times 10^{17}$ kg m$^{-3}$ for all nuclei - Nucleons are evenly separated throughout all nuclei regardless of size