Nuclear radius (A-level only)

Welcome to the incredibly small world of the atomic nucleus! In earlier topics, you learned about the structure of the atom. Now, we are going to measure it.

Here is what you'll learn:

The typical size of a nucleus.
How to estimate nuclear radius using the closest approach of alpha particles.
How to determine nuclear radius more accurately using high-energy electron diffraction.
The mathematical link between a nucleus's radius and its nucleon number.
How to prove that all atomic nuclei share the same unbelievable density.

How big is a nucleus?

To describe the nucleus, we need a unit of measurement suited to the subatomic world. The standard SI unit is still the metre, but we use a very specific prefix.

Definition

Femtometre (fm)

A common unit of length in nuclear physics. $\text{ fm} = 10^{-15} \text{ m}$ .

For context, a typical atom has a radius of about $m10^{-10} \text{ m}$ . A typical nucleus has a radius of roughly $m10^{-15} \text{ m}$ (or $\text{ fm}$ ). This means the nucleus is about $100, 000$ times smaller than the atom itself!

To figure out exactly how large a specific nucleus is, physicists use two main experimental techniques. Let's look at both.

Method 1: Distance of closest approach

The oldest way to estimate the size of a nucleus is to fire an alpha particle directly at it. This is exactly what Ernest Rutherford did in his famous scattering experiment.

Because both the alpha particle and the target nucleus are positively charged, they repel each other via the electrostatic force. As the alpha particle gets closer, it slows down, losing kinetic energy and gaining electric potential energy.

Eventually, for a brief moment, the alpha particle comes to a complete stop before turning around and flying back. The distance between the centre of the nucleus and the point where the alpha particle stops is called the distance of closest approach.

Distance of closest approach

Key Idea

Conservation of Energy

At the point of closest approach, all of the alpha particle's initial kinetic energy ( $E_k$ ) has been converted into electric potential energy ( $E_p$ ).

Ek=Ep=Q1Q24πε0rE_k = E_p = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 r}

Where $r$ is the distance of closest approach, $Q_1$ is the charge of the alpha particle, $Q_2$ is the charge of the target nucleus, and $ε0\varepsilon_0$ is the permittivity of free space.

This distance $r$ provides an upper limit for the size of the nucleus. We know the nucleus must be smaller than $r$ , otherwise the alpha particle would have crashed into it!

Example

Calculating closest approach

An alpha particle with an initial kinetic energy of $\text{ MeV}$ is fired head-on at a stationary gold nucleus (atomic number $Z = 79$ ). Estimate the maximum radius of the gold nucleus.

Find the kinetic energy in Joules: Convert $\text{ MeV}$ into standard SI units using the conversion factor $\text{ eV} = 1.60 \times 10^{-19} \text{ J}$ . $JE_k = 4.0 \times 10^6 \times 1.60 \times 10^{-19} = 6.4 \times 10^{-13} \text{ J}$
Determine the charges involved: The charge of an alpha particle (2 protons) is $Q_1 = 2e$ . The charge of a gold nucleus (79 protons) is $Q_2 = 79e$ . Where the elementary charge $\times 10^{-19} \text{ C}$ .
Set up the Coulomb potential energy equation: Equate the initial kinetic energy to the electric potential energy at distance $r$ . $Ek=(2e)(79e)4πε0rE_k = \frac{(2e)(79e)}{4 \pi \varepsilon_0 r}$
Rearrange for the distance of closest approach ( $r$ ): $\frac{158 \times (1.60 \times 10^{-19})^2}{4 \pi \varepsilon_0 E_k}$
Substitute values and calculate: Using $m−1\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$ . $m\begin{aligned} r &= \frac{158 \times 2.56 \times 10^{-38}}{4 \pi (8.85 \times 10^{-12}) (6.4 \times 10^{-13})} \\ &= 5.68 \times 10^{-14} \text{ m} \end{aligned}$

The maximum radius of the gold nucleus is estimated to be $5.7 \times 10^{-14} \text{ m}$ .

Common Mistake

Atomic vs Mass Number

When finding the target nucleus charge $Q_2$ , students often mistakenly use the nucleon (mass) number $A$ instead of the atomic (proton) number $Z$ . Remember, neutrons have no charge! The charge of the nucleus is strictly $\times e$ .

Why is this only an estimate?

This method has a few flaws. First, it only gives an upper limit — it tells us where the alpha particle stopped, not where the nuclear surface actually begins. Second, alpha particles are hadrons, meaning they experience the strong nuclear force. If the alpha particle gets close enough to feel the strong force, the simple electrostatic equations we used above break down. Finally, the alpha particle has its own finite size, which blurs the measurement.

Method 2: Electron diffraction

To get a much more accurate measurement of nuclear radius, physicists use high-energy electron diffraction.

Why electrons? Because electrons are leptons. They do not experience the strong nuclear force, so their paths aren't altered by the complicated nuclear forces that affect alpha particles.

From quantum physics, we know that particles can behave like waves. If we accelerate electrons to extremely high energies, their de Broglie wavelength ( $λ=hp\lambda = \frac{h}{p}$ ) becomes comparable to the size of a nucleus (around $m10^{-15} \text{ m}$ ). When a beam of these high-energy electrons is directed at a thin sample of atoms, they diffract around the nuclei.

A detector is moved around the sample to measure the number of electrons scattered at different angles.

$Electron diffraction graph$

Key Idea

Intensity-Angle Graph

When the scattered electron intensity is plotted against the diffraction angle $θ\theta$ , the graph shows a very tall central maximum, followed by a sharp drop, and then a series of smaller "ripples" called secondary maxima.

By finding the angle $θ\theta$ of the first minimum (the dip before the first ripple), physicists can use wave diffraction equations to calculate the exact diameter of the nucleus. This is currently our most precise way to measure nuclear dimensions!

The Nuclear Radius Equation

By repeating these diffraction experiments on many different elements, physicists discovered a clear mathematical relationship between the radius of a nucleus ( $R$ ) and its total number of nucleons ( $A$ ).

Definition

Nuclear Radius Formula

The radius of a nucleus is directly proportional to the cube root of its nucleon number:

R = R_0 A^{1/3}

Where:

$R$ is the nuclear radius (in $m\text{m}$ ).
$R_0$ is a constant, approximately $1.2 \times 10^{-15} \text{ m}$ (or $\text{ fm}$ ).
$A$ is the nucleon number (total number of protons and neutrons).

If you plot a graph of $R$ against $A^{1/3}$ , you will get a straight line passing through the origin. The gradient of this line is the constant $R_0$ .

Example

Using the radius formula

Calculate the radius of a Carbon-12 nucleus. Use $mR_0 = 1.2 \times 10^{-15} \text{ m}$ .

Identify the variables: For Carbon-12, the nucleon number $A = 12$ .
State the formula: $R = R_0 A^{1/3}$
Substitute and calculate: $m\begin{aligned} R &= (1.2 \times 10^{-15}) \times 12^{1/3} \\ &= (1.2 \times 10^{-15}) \times 2.289 \\ &= 2.7 \times 10^{-15} \text{ m} \end{aligned}$

Nuclear Density

One of the most fascinating consequences of the $R = R_0 A^{1/3}$ equation is what it tells us about the density of nuclear matter.

Let's calculate the density ( $ρ\rho$ ) of a generic nucleus. We will assume the nucleus is perfectly spherical, and we will use the atomic mass unit ( $\times 10^{-27} \text{ kg}$ ) as an approximation for the mass of a single nucleon.

Example

Proving nuclear density is constant

Show that the density of nuclear material is independent of the nucleon number $A$ .

Write the formula for density: Density is mass divided by volume. $ρ=mV\rho = \frac{m}{V}$
Express the mass ( $m$ ): The total mass of the nucleus is the number of nucleons ( $A$ ) multiplied by the mass of a single nucleon ( $u$ ). $\cdot u$
Express the volume ( $V$ ): Assuming a spherical nucleus, $\frac{4}{3} \pi R^3$ . Substitute the radius equation $R = R_0 A^{1/3}$ into the volume formula. $V=43π(R0A1/3)3=43πR03A\begin{aligned} V &= \frac{4}{3} \pi (R_0 A^{1/3})^3 \\ &= \frac{4}{3} \pi R_0^3 A \end{aligned}$
Substitute mass and volume back into the density formula: $ρ=A⋅u43πR03A\rho = \frac{A \cdot u}{\frac{4}{3} \pi R_0^3 A}$
Cancel the $A$ terms: $ρ=u43πR03\rho = \frac{u}{\frac{4}{3} \pi R_0^3}$

Because $u$ and $R_0$ are both constants, the density $ρ\rho$ is a constant too! The nucleon number $A$ completely cancels out.

This proves that every nucleus, from lightweight helium to super-heavy uranium, has exactly the same density. If you calculate the value, it comes out to roughly $1.4 \times 10^{17} \text{ kg m}^{-3}$ . A mere teaspoon of nuclear material would weigh over a billion tonnes!

Analogy

Grapes in a jar

Why is nuclear density constant? Think of the nucleus like a jar filled with identically sized grapes (the nucleons). If you add more grapes, the jar needs to be bigger to hold them (volume increases). However, the individual grapes don't get squished to become smaller, nor do they spread out. Because the mass increases at the exact same rate as the required volume, the overall density of the "grape cluster" stays perfectly constant.

Exam technique

In the exam

Always convert energy values from $MeV\text{MeV}$ or $eV\text{eV}$ into Joules before plugging them into the Coulomb potential energy formula.
In electron diffraction questions, remember to explicitly state that electrons are used because they are leptons and do not feel the strong nuclear force.
If asked to plot a graph to prove $\propto A^{1/3}$ , explicitly state that plotting $R$ against $A^{1/3}$ yields a straight line through the origin, and the gradient equals $R_0$ .
Be ready to reproduce the algebraic proof showing that nuclear density is constant. It is a highly common multi-mark question.

Self review

Check yourself

Can you describe the energy transfer that takes place as an alpha particle approaches a nucleus?
Why does the closest approach method usually overestimate the true size of the nucleus?
What does the graph of intensity against angle look like for electron diffraction by a nucleus?
Why does the formula $R = R_0 A^{1/3}$ imply that all nuclei have the same density?

Nuclear radius (A-level only)

Welcome to the incredibly small world of the atomic nucleus! In earlier topics, you learned about the structure of the atom. Now, we are going to measure it.

Here is what you'll learn:

The typical size of a nucleus.
How to estimate nuclear radius using the closest approach of alpha particles.
How to determine nuclear radius more accurately using high-energy electron diffraction.
The mathematical link between a nucleus's radius and its nucleon number.
How to prove that all atomic nuclei share the same unbelievable density.

How big is a nucleus?

To describe the nucleus, we need a unit of measurement suited to the subatomic world. The standard SI unit is still the metre, but we use a very specific prefix.

Definition

Femtometre (fm)

A common unit of length in nuclear physics. $\text{ fm} = 10^{-15} \text{ m}$ .

To figure out exactly how large a specific nucleus is, physicists use two main experimental techniques. Let's look at both.

Method 1: Distance of closest approach

The oldest way to estimate the size of a nucleus is to fire an alpha particle directly at it. This is exactly what Ernest Rutherford did in his famous scattering experiment.

Distance of closest approach

Key Idea

Conservation of Energy

At the point of closest approach, all of the alpha particle's initial kinetic energy ( $E_k$ ) has been converted into electric potential energy ( $E_p$ ).

Ek=Ep=Q1Q24πε0rE_k = E_p = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 r}

Where $r$ is the distance of closest approach, $Q_1$ is the charge of the alpha particle, $Q_2$ is the charge of the target nucleus, and $ε0\varepsilon_0$ is the permittivity of free space.

This distance $r$ provides an upper limit for the size of the nucleus. We know the nucleus must be smaller than $r$ , otherwise the alpha particle would have crashed into it!

Example

Calculating closest approach

An alpha particle with an initial kinetic energy of $\text{ MeV}$ is fired head-on at a stationary gold nucleus (atomic number $Z = 79$ ). Estimate the maximum radius of the gold nucleus.

Find the kinetic energy in Joules: Convert $\text{ MeV}$ into standard SI units using the conversion factor $\text{ eV} = 1.60 \times 10^{-19} \text{ J}$ . $JE_k = 4.0 \times 10^6 \times 1.60 \times 10^{-19} = 6.4 \times 10^{-13} \text{ J}$
Determine the charges involved: The charge of an alpha particle (2 protons) is $Q_1 = 2e$ . The charge of a gold nucleus (79 protons) is $Q_2 = 79e$ . Where the elementary charge $\times 10^{-19} \text{ C}$ .
Set up the Coulomb potential energy equation: Equate the initial kinetic energy to the electric potential energy at distance $r$ . $Ek=(2e)(79e)4πε0rE_k = \frac{(2e)(79e)}{4 \pi \varepsilon_0 r}$
Rearrange for the distance of closest approach ( $r$ ): $\frac{158 \times (1.60 \times 10^{-19})^2}{4 \pi \varepsilon_0 E_k}$
Substitute values and calculate: Using $m−1\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$ . $m\begin{aligned} r &= \frac{158 \times 2.56 \times 10^{-38}}{4 \pi (8.85 \times 10^{-12}) (6.4 \times 10^{-13})} \\ &= 5.68 \times 10^{-14} \text{ m} \end{aligned}$

The maximum radius of the gold nucleus is estimated to be $5.7 \times 10^{-14} \text{ m}$ .

Common Mistake

Atomic vs Mass Number

Why is this only an estimate?

Method 2: Electron diffraction

To get a much more accurate measurement of nuclear radius, physicists use high-energy electron diffraction.

Why electrons? Because electrons are leptons. They do not experience the strong nuclear force, so their paths aren't altered by the complicated nuclear forces that affect alpha particles.

A detector is moved around the sample to measure the number of electrons scattered at different angles.

$Electron diffraction graph$

Key Idea

Intensity-Angle Graph

The Nuclear Radius Equation

Definition

Nuclear Radius Formula

The radius of a nucleus is directly proportional to the cube root of its nucleon number:

R = R_0 A^{1/3}

Where:

$R$ is the nuclear radius (in $m\text{m}$ ).
$R_0$ is a constant, approximately $1.2 \times 10^{-15} \text{ m}$ (or $\text{ fm}$ ).
$A$ is the nucleon number (total number of protons and neutrons).

If you plot a graph of $R$ against $A^{1/3}$ , you will get a straight line passing through the origin. The gradient of this line is the constant $R_0$ .

Example

Using the radius formula

Calculate the radius of a Carbon-12 nucleus. Use $mR_0 = 1.2 \times 10^{-15} \text{ m}$ .

Identify the variables: For Carbon-12, the nucleon number $A = 12$ .
State the formula: $R = R_0 A^{1/3}$
Substitute and calculate: $m\begin{aligned} R &= (1.2 \times 10^{-15}) \times 12^{1/3} \\ &= (1.2 \times 10^{-15}) \times 2.289 \\ &= 2.7 \times 10^{-15} \text{ m} \end{aligned}$

Nuclear Density

One of the most fascinating consequences of the $R = R_0 A^{1/3}$ equation is what it tells us about the density of nuclear matter.

Example

Proving nuclear density is constant

Show that the density of nuclear material is independent of the nucleon number $A$ .

Write the formula for density: Density is mass divided by volume. $ρ=mV\rho = \frac{m}{V}$
Express the mass ( $m$ ): The total mass of the nucleus is the number of nucleons ( $A$ ) multiplied by the mass of a single nucleon ( $u$ ). $\cdot u$
Express the volume ( $V$ ): Assuming a spherical nucleus, $\frac{4}{3} \pi R^3$ . Substitute the radius equation $R = R_0 A^{1/3}$ into the volume formula. $V=43π(R0A1/3)3=43πR03A\begin{aligned} V &= \frac{4}{3} \pi (R_0 A^{1/3})^3 \\ &= \frac{4}{3} \pi R_0^3 A \end{aligned}$
Substitute mass and volume back into the density formula: $ρ=A⋅u43πR03A\rho = \frac{A \cdot u}{\frac{4}{3} \pi R_0^3 A}$
Cancel the $A$ terms: $ρ=u43πR03\rho = \frac{u}{\frac{4}{3} \pi R_0^3}$

Because $u$ and $R_0$ are both constants, the density $ρ\rho$ is a constant too! The nucleon number $A$ completely cancels out.

Analogy

Grapes in a jar

Exam technique

In the exam

Always convert energy values from $MeV\text{MeV}$ or $eV\text{eV}$ into Joules before plugging them into the Coulomb potential energy formula.
In electron diffraction questions, remember to explicitly state that electrons are used because they are leptons and do not feel the strong nuclear force.
If asked to plot a graph to prove $\propto A^{1/3}$ , explicitly state that plotting $R$ against $A^{1/3}$ yields a straight line through the origin, and the gradient equals $R_0$ .
Be ready to reproduce the algebraic proof showing that nuclear density is constant. It is a highly common multi-mark question.

Self review

Check yourself

Can you describe the energy transfer that takes place as an alpha particle approaches a nucleus?
Why does the closest approach method usually overestimate the true size of the nucleus?
What does the graph of intensity against angle look like for electron diffraction by a nucleus?
Why does the formula $R = R_0 A^{1/3}$ imply that all nuclei have the same density?

Nuclear radius (A-level only)

How big is a nucleus?

Method 1: Distance of closest approach

Why is this only an estimate?

Method 2: Electron diffraction

The Nuclear Radius Equation

Nuclear Density

Nuclear radius (A-level only) Revision Guide

Nuclear radius (A-level only)

How big is a nucleus?

Method 1: Distance of closest approach

Why is this only an estimate?

Method 2: Electron diffraction

The Nuclear Radius Equation

Nuclear Density

Nuclear radius (A-level only) Revision Guide