E.4 The nuclear atom

IB Physics HL· 6 min read

1. Rutherford Alpha Scattering Experiment★★☆☆☆⏱ 15 min

Before 1911, the accepted model of the atom was J.J. Thomson's 'plum pudding model', which described the atom as a uniform sphere of positive charge with electrons embedded evenly throughout. Geiger and Marsden, working under Rutherford's direction, tested this model by firing fast alpha particles at a thin gold foil.

📘 Definition

Rutherford Alpha Scattering

An experiment that measured the deflection of alpha particles passing through thin metal foil to probe internal atomic structure

Example:

Most alpha particles passed undeflected, while ~1 in 8000 were deflected by more than 90°

📐 Worked Example

State three key observations from Rutherford's experiment and one conclusion from each.

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Observation 1: Most alpha particles pass straight through the foil undeflected
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Conclusion 1: The atom is mostly empty space
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Observation 2: A small number of alpha particles are deflected through angles > 90°
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Conclusion 2: All positive charge and almost all mass of the atom are concentrated in a very small central nucleus
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Observation 3: Deflection rate decreases with increasing deflection angle
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Conclusion 3: The nucleus follows Coulomb's inverse-square law of electrostatic repulsion

2. Nuclear Size and Density★★★☆☆⏱ 20 min

Alpha scattering can be used to estimate the radius of a nucleus. The closest approach of an alpha particle to the nucleus gives an upper bound for the nuclear radius. Precise scattering experiments have confirmed an empirical relationship between nuclear radius and mass number:

r = r_0 A^{1/3}

Where $r_{0} \approx 1.2 \times 1 0^{- 15} m$ and $A$ is the mass number. Since the volume of a spherical nucleus is $V = \frac{4}{3} π r^{3}$ , substituting gives $V \propto A$ , meaning volume is proportional to the number of nucleons. All nuclei therefore have approximately the same constant density.

📐 Worked Example

Calculate the radius of an oxygen-16 nucleus, given $r_{0} = 1.2 \times 1 0^{- 15} m$ .

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1. Identify the mass number: $A = 16$
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1. Substitute into the nuclear radius formula:
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$r = (1.2 \times 10^{-15}) \times (16)^{1/3}$
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1. Calculate $1 6^{1/3} \approx 2.52$
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1. Final result:
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$r \approx 3.0 \times 10^{-15} \text{ m}$

3. Nuclear Composition and Isotopes★★☆☆☆⏱ 15 min

📘 Definition

Nuclear Composition

$_{Z}^{A} X$ (standard nuclide notation, X = chemical symbol)

Atomic number $Z$ = number of protons, Mass number $A$ = total number of nucleons ( $Z + N$ ), Neutron number $N = A - Z$ = number of neutrons

Example:

Helium-4: $_{2}^{4} He$ , 2 protons, 2 neutrons, $N = 2$

Isotopes are nuclides of the same element that share the same atomic number $Z$ but have different mass numbers $A$ and different neutron numbers $N$ . They have identical chemical properties because chemical reactions depend on electron configuration, which is determined by $Z$ , but different nuclear properties.

📐 Worked Example

State the number of protons, neutrons and nucleons for uranium-235 and uranium-238, and write their standard notation.

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1. Uranium has atomic number $Z = 92$ , so both isotopes have 92 protons and 92 nucleons from protons.
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1. For uranium-235: $A = 235$ , $N = 235 - 92 = 143$ neutrons. Notation:
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$^{235}_{92}\text{U}$
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1. For uranium-238: $A = 238$ , $N = 238 - 92 = 146$ neutrons. Notation:
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$^{238}_{92}\text{U}$

4. The Strong Nuclear Force★★★☆☆⏱ 10 min

Positively charged protons in the nucleus repel each other via the long-range electrostatic Coulomb force. There must be an attractive force stronger than this repulsion at short distances to hold the nucleus together: the strong nuclear force. Its key properties are:

It is attractive between nucleons at separations of ~ $0.5 \times 1 0^{- 15} m$ to $3 \times 1 0^{- 15} m$
It becomes repulsive at separations less than ~ $0.5 \times 1 0^{- 15} m$ , preventing nuclear collapse
It is very short range, and drops to zero at separations greater than ~ $3 \times 1 0^{- 15} m$
It acts equally between all nucleons, regardless of charge (proton-proton, proton-neutron, neutron-neutron)

5. Common Pitfalls

Wrong move:

Confusing mass number with actual atomic mass in u

Why:

Mass number is a count of nucleons, not a measured mass. Binding energy reduces actual nuclear mass below $A$ u

Correct move:

State that mass number is the total number of nucleons, and actual mass is approximately $A$ u for rough calculations

Wrong move:

Claiming large alpha deflection only proves the nucleus is positively charged

Why:

Deflection by positive charge was expected; the surprising result was deflection greater than 90°

Correct move:

Link large angle deflection to the conclusion that most mass is concentrated in a tiny nucleus

Wrong move:

Writing the nuclear radius formula as $r = r_{0} A^{3}$

Why:

Volume is proportional to $A$ , not radius. Radius scales with the cube root of volume

Correct move:

Remember $V \propto r^{3} \propto A ⟹ r \propto A^{1/3}$

Wrong move:

Claiming isotopes have different chemical properties

Why:

Chemical properties depend on electron configuration, which is determined by atomic number $Z$ , not mass number

Correct move:

State that isotopes have identical chemical properties but different nuclear properties

Wrong move:

Stating the strong nuclear force is attractive at all distances

Why:

The strong force becomes repulsive at very short distances, which is required for nuclear stability

Correct move:

Describe the strong force as repulsive < 0.5 fm, attractive 0.5-3 fm, and zero beyond 3 fm

6. Quick Reference Cheatsheet

Quantity/Property	Symbol/Rule	Value/Definition
Nuclear radius	$r = r_{0} A^{1/3}$	$r_{0} = 1.2 \times 1 0^{- 15}$ m
Atomic number	$Z$	Number of protons
Mass number	$A$	Total number of nucleons
Neutron number	$N$	$N = A - Z$
Nuclear density	$ρ$	$\approx 2 \times 1 0^{17}$ kg m⁻³ (constant)
Strong force range		Repulsive < 0.5 fm, attractive 0.5-3 fm, zero > 3 fm
Rutherford key conclusion		Atom is mostly empty space, mass/charge in small nucleus
Isotopes		Same Z, different A, same chemical properties

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Rutherford scattering conclusions
2024 · 2
Nuclear radius calculation
2023 · 1
Strong nuclear force properties

Going deeper

simulationRutherford Scattering PhET SimulationInteractive exploration of alpha deflection

What's Next

Understanding the structure of the nuclear atom is the foundation for all further topics in nuclear physics, from radioactive decay to nuclear fission and fusion. The properties of the nucleus you have learned here, including nuclear size, composition, and the strong force, underpin explanations of nuclear instability and energy release in nuclear reactions. You will now build on this knowledge to explore how unstable nuclei decay, and how binding energy explains the energy released in fission and fusion reactions.

→E.6 Fission and fusion (AHL)