Structure of the Atom

The Size of the Nucleus

The nucleus is an extremely small region at the centre of an atom. It contains the atom's protons and neutrons and therefore contains almost all of the atom's mass.

A typical atom has a radius of about:

10⁻¹⁰ m

A typical nucleus has a radius of about:

10⁻¹⁵ m

This means that the radius of an atom is roughly 100,000 times greater than the radius of its nucleus.

The nucleus therefore occupies only a tiny fraction of the atom's volume.

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What Is Nuclear Radius?

The nuclear radius is a measure of the size of an atomic nucleus.

Nuclei do not have perfectly sharp boundaries, so nuclear radius should be understood as an approximate measure of the extent of the nucleus.

Nuclear dimensions are often measured in femtometres (fm).

1 fm = 10⁻¹⁵ m

For example:

5 fm = 5 × 10⁻¹⁵ m

The femtometre is particularly useful because most nuclear radii are only a few femtometres.

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Measuring Nuclear Size

Nuclei are far too small to measure directly with an ordinary microscope.

Instead, scientists use experiments such as high-energy electron scattering.

High-energy electrons are directed toward nuclei. Scientists measure how the electrons scatter and use the scattering pattern to estimate the size and charge distribution of the nucleus.

This is one reason scattering experiments are so important in nuclear physics: they allow scientists to investigate structures far smaller than can be seen directly.

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Nuclear Radius and Nucleon Number

Experiments show that nuclear radius depends on the number of nucleons in the nucleus.

The relationship is approximately:

R = R₀A^(1/3)

where:

• R = nuclear radius
• R₀ = nuclear radius constant
• A = nucleon number
• A^(1/3) means the cube root of A

A commonly used approximate value is:

R₀ ≈ 1.2 × 10⁻¹⁵ m

or:

R₀ ≈ 1.2 fm

Therefore, a convenient version of the relationship is:

R ≈ 1.2A^(1/3) fm

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Understanding the Cube Root

The expression A^(1/3) means the cube root of A.

For example:

8^(1/3) = 2

because:

2 × 2 × 2 = 8

Similarly:

27^(1/3) = 3

because:

3 × 3 × 3 = 27

And:

64^(1/3) = 4

because:

4 × 4 × 4 = 64

Therefore, nuclear radius does increase as nucleon number increases, but it increases much more slowly than A itself.

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Worked Example: Nuclear Radius

Estimate the radius of a nucleus with nucleon number:

A = 64

Use:

R = R₀A^(1/3)

Since:

R₀ = 1.2 fm

and:

64^(1/3) = 4

then:

R = 1.2 × 4

Therefore:

R = 4.8 fm

In metres:

R = 4.8 × 10⁻¹⁵ m

So the estimated nuclear radius is 4.8 fm.

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Worked Example: A Larger Nucleus

Consider a nucleus with:

A = 125

Since:

125^(1/3) = 5

use:

R = 1.2 × 5

Therefore:

R = 6.0 fm

or:

R = 6.0 × 10⁻¹⁵ m

The nucleus with more nucleons is larger.

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Why Does Nuclear Radius Increase?

A larger nucleon number means the nucleus contains more:

• Protons
• Neutrons

Adding nucleons increases the size of the nucleus.

However, the radius does not increase directly in proportion to nucleon number.

For example, multiplying A by 8 only doubles the radius because:

8^(1/3) = 2

Therefore:

Nucleon number × 8 → radius × 2

This relationship tells us something important about the internal structure of nuclei.

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Nuclear Volume

To understand nuclear density, we first need to consider nuclear volume.

A nucleus can be approximated as a sphere.

The volume of a sphere is:

V = (4/3)πR³

where:

• V = volume
• R = radius

Because nuclear radius follows:

R = R₀A^(1/3)

we can cube both sides conceptually:

R³ = R₀³A

Therefore:

Nuclear volume is proportional to A.

In simpler terms:

More nucleons → proportionally larger nuclear volume

This relationship is extremely important when explaining nuclear density.

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Worked Example: Nuclear Volume

Suppose a nucleus has radius:

R = 4.0 × 10⁻¹⁵ m

Use:

V = (4/3)πR³

First cube the radius:

R³ = (4.0 × 10⁻¹⁵)³

R³ = 64 × 10⁻⁴⁵

which is:

R³ = 6.4 × 10⁻⁴⁴ m³

Now calculate the volume:

V = (4/3)π(6.4 × 10⁻⁴⁴)

This gives approximately:

V = 2.7 × 10⁻⁴³ m³

This is an extraordinarily small volume.

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Nuclear Mass

The mass of a nucleus depends mainly on the number of nucleons it contains.

A proton and neutron each have a mass of approximately:

1.67 × 10⁻²⁷ kg

For an estimate, we can therefore write:

Nuclear mass ≈ A × 1.67 × 10⁻²⁷ kg

where A is the nucleon number.

For example, a nucleus containing 100 nucleons has an approximate mass of:

m ≈ 100 × 1.67 × 10⁻²⁷

Therefore:

m ≈ 1.67 × 10⁻²⁵ kg

This is an approximation because actual nuclear masses are affected slightly by nuclear binding energy.

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Nuclear Density

Density describes how much mass is contained within a particular volume.

The density equation is:

Density = mass / volume

Using symbols:

ρ = m / V

where:

• ρ = density in kg/m³
• m = mass in kg
• V = volume in m³

Because nuclei contain considerable mass in extremely small volumes, nuclear density is enormous.

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Worked Example: Calculating Nuclear Density

Suppose a nucleus has:

Mass = 1.0 × 10⁻²⁵ kg

and:

Radius = 5.0 × 10⁻¹⁵ m

First calculate its volume.

V = (4/3)πR³

Substitute the radius:

V = (4/3)π(5.0 × 10⁻¹⁵)³

Cube the radius:

(5.0 × 10⁻¹⁵)³ = 1.25 × 10⁻⁴³ m³

Therefore:

V ≈ 5.24 × 10⁻⁴³ m³

Now calculate density:

ρ = m / V

ρ = (1.0 × 10⁻²⁵) / (5.24 × 10⁻⁴³)

Therefore:

ρ ≈ 1.9 × 10¹⁷ kg/m³

This is an extraordinarily high density.

Typical nuclear densities are on the order of:

10¹⁷ kg/m³

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How Dense Is Nuclear Matter?

Nuclear density is difficult to imagine because it is so much greater than densities encountered in everyday life.

For comparison:

• Water has a density of about 1.0 × 10³ kg/m³.
• Many metals have densities around 10³ to 10⁴ kg/m³.
• Nuclear matter has a density around 10¹⁷ kg/m³.

This means nuclear matter is roughly 100 trillion times denser than water.

The enormous density results from packing protons and neutrons into an extremely small volume.

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Why Is Nuclear Density Approximately Constant?

One of the most interesting properties of nuclei is that their densities are approximately the same, regardless of their nucleon number.

We can understand why from the relationships for mass and volume.

Nuclear mass is approximately proportional to nucleon number:

m ∝ A

Nuclear radius follows:

R ∝ A^(1/3)

Because volume depends on radius cubed:

V ∝ R³

Substitute the radius relationship:

V ∝ [A^(1/3)]³

Therefore:

V ∝ A

So both mass and volume increase approximately in proportion to A.

Density is:

ρ = m / V

Since both the numerator and denominator increase approximately in the same proportion, their ratio stays approximately constant.

Therefore:

Nuclear density is approximately independent of nucleon number.

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A Simple Comparison

Imagine one nucleus has 8 times as many nucleons as another.

Its mass is approximately:

8 times greater

Its radius is:

8^(1/3) = 2 times greater

If radius doubles, volume increases by:

2³ = 8

So:

• Mass increases × 8
• Volume increases × 8

Density therefore remains approximately the same.

This is a powerful consequence of the relationship:

R ∝ A^(1/3)

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What Constant Nuclear Density Tells Us

The approximately constant density of nuclei suggests that nucleons are packed together in a broadly similar way in different nuclei.

A larger nucleus is not simply a much more tightly packed version of a smaller nucleus.

Instead, adding nucleons generally increases the volume available to contain them.

This behaviour is sometimes compared with adding more closely packed objects to a cluster: the cluster grows while the average packing density remains similar.

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Radius Does Not Increase Linearly with Nucleon Number

A common mistake is to assume:

Twice as many nucleons → twice the radius

This is incorrect.

Because:

R ∝ A^(1/3)

doubling A changes the radius by a factor of:

2^(1/3) ≈ 1.26

So if the nucleon number doubles, the nuclear radius increases by only about 26%.

The volume, however, approximately doubles.

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Worked Example: Comparing Nuclear Radii

Nucleus A has nucleon number:

A = 27

Nucleus B has nucleon number:

A = 216

Calculate their approximate radii.

For Nucleus A:

27^(1/3) = 3

Therefore:

R = 1.2 × 3 = 3.6 fm

For Nucleus B:

216^(1/3) = 6

Therefore:

R = 1.2 × 6 = 7.2 fm

Nucleus B contains:

216 / 27 = 8 times as many nucleons

but its radius is only:

7.2 / 3.6 = 2 times larger

Its volume, however, is approximately 8 times larger.

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Estimating Nuclear Density from the Radius Relationship

We can combine the nuclear radius relationship with the density equation.

Use:

R = R₀A^(1/3)

Nuclear volume is:

V = (4/3)πR³

Substituting the radius relationship gives:

V = (4/3)πR₀³A

Approximate nuclear mass is:

m = Amₙ

where mₙ represents the approximate mass of one nucleon.

Density is:

ρ = m / V

Therefore:

ρ = Amₙ / [(4/3)πR₀³A]

The A appears in both the numerator and denominator, so it cancels.

This leaves:

ρ = mₙ / [(4/3)πR₀³]

The important point is that A is no longer present.

This mathematical result explains why nuclear density is approximately constant for nuclei of different sizes.

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Nuclear Density and Neutron Stars

Nuclear density is not normally encountered in everyday matter because atoms contain enormous amounts of empty space compared with the size of their nuclei.

However, extremely dense astronomical objects called neutron stars contain matter compressed to densities comparable to those found in atomic nuclei.

Neutron stars form from the collapsed cores of certain massive stars.

They provide an extraordinary example of matter existing under conditions approaching nuclear density on an astronomical scale.

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Evidence and Limitations

The relationship:

R = R₀A^(1/3)

is an empirical approximation based on experimental evidence.

Real nuclei are more complicated than perfectly uniform spheres.

For example:

• Some nuclei are not perfectly spherical.
• Nuclear charge and matter do not have perfectly sharp boundaries.
• Proton and neutron distributions can differ slightly.
• Very light or unusual nuclei may show departures from the simple model.

Nevertheless, the relationship is extremely useful for estimating nuclear dimensions and explaining the approximately constant density of nuclear matter.

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Common Mistakes

Confusing Atomic Radius with Nuclear Radius

Atomic radii are typically around:

10⁻¹⁰ m

Nuclear radii are typically around:

10⁻¹⁵ m

These differ by roughly five orders of magnitude.

Forgetting the Cube Root

The relationship is:

R = R₀A^(1/3)

not:

R = R₀A

Assuming Twice the Nucleon Number Means Twice the Radius

Because radius depends on the cube root of A, doubling A increases the radius by only about 26%.

Forgetting to Cube the Radius When Finding Volume

For a spherical nucleus:

V = (4/3)πR³

The radius must be cubed.

Forgetting to Cube the Power of Ten

For example:

(10⁻¹⁵)³ = 10⁻⁴⁵

not 10⁻¹⁵.

Confusing Nuclear Mass with Nucleon Number

Nucleon number A is a count and has no unit.

Mass is measured in kilograms or atomic mass units.

Assuming Larger Nuclei Are Much Denser

Larger nuclei contain more nucleons, but they also have proportionally greater volume.

Their average nuclear densities therefore remain approximately constant.

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Check Your Understanding

1. What is meant by nuclear radius?

2. What is a femtometre?

3. State the approximate relationship between nuclear radius and nucleon number.

4. What does A represent in the nuclear radius relationship?

5. Estimate the radius of a nucleus with A = 8 using R₀ = 1.2 fm.

6. Estimate the radius of a nucleus with A = 27.

7. A nucleus has 8 times as many nucleons as another nucleus. How many times larger is its radius?

8. Why does nuclear radius increase as nucleon number increases?

9. State the equation for the volume of a spherical nucleus.

10. State the equation for density.

11. Why is nuclear matter extremely dense?

12. Explain why nuclear volume is approximately proportional to nucleon number.

13. Explain why nuclear mass is approximately proportional to nucleon number.

14. Use these relationships to explain why nuclear density is approximately constant.

15. Why should the simple nuclear radius relationship be regarded as an approximation rather than an exact description of every nucleus?

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Key Terms

• Nuclear radius – an approximate measure of the size of an atomic nucleus.
• Femtometre (fm) – a unit of length equal to 10⁻¹⁵ m.
• Nucleon – a proton or neutron within a nucleus.
• Nucleon number (A) – the total number of protons and neutrons in a nucleus.
• Nuclear radius constant (R₀) – the constant used in the approximate nuclear radius relationship.
• Nuclear volume – the volume occupied by a nucleus, often estimated by treating it as a sphere.
• Nuclear density – the mass of a nucleus divided by its volume.
• Density – mass per unit volume.
• Nuclear matter – matter consisting of closely packed nucleons.
• Empirical relationship – a relationship established from experimental observations.

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Key Takeaways

• Atomic nuclei have radii on the order of 10⁻¹⁵ m.
• Nuclear dimensions are often expressed in femtometres (fm).
• Nuclear radius can be estimated using R = R₀A^(1/3).
• A commonly used value is R₀ ≈ 1.2 fm.
• Nuclear radius increases as nucleon number increases, but it increases according to the cube root of A.
• Nuclear volume is approximately proportional to nucleon number.
• Nuclear mass is also approximately proportional to nucleon number.
• Nuclear density can be calculated using ρ = m / V.
• Typical nuclear densities are on the order of 10¹⁷ kg/m³.
• Because nuclear mass and nuclear volume both increase approximately in proportion to A, nuclear density remains approximately constant.
• The approximately constant density suggests that nucleons are packed with broadly similar average density in nuclei of different sizes.