Stable and Unstable Nuclei

The strong nuclear force, alpha decay, and beta decay

SPEC 3.2.1.2 • PARTICLES AND RADIATION

The Nuclear Problem

Consider the nucleus: protons are packed incredibly close together (about 10⁻¹⁵ m apart). But protons are positively charged...

The Problem

Electrostatic repulsion between protons should blow the nucleus apart. The force between two protons at nuclear distances is enormous - about 230 N!

Key Question

What holds the nucleus together? There must be another force - one that is stronger than electrostatic repulsion at nuclear distances.

This force is called the strong nuclear force (or just "strong force").

The Strong Nuclear Force

Definition

The strong nuclear force is an attractive force that acts between all nucleons (protons and neutrons), holding the nucleus together.

Key properties of the strong force:

PropertyDescription

RangeVery short range - only effective up to about 3 fm (femtometres)

StrengthMuch stronger than electrostatic force at short range

Acts onAll nucleons equally: p-p, n-n, and p-n pairs

At very short rangeBecomes repulsive below about 0.5 fm

Remember

1 fm (femtometre) = 10⁻¹⁵ m - about the size of a proton!

Force vs Separation Graph

The strong nuclear force varies dramatically with nucleon separation:

Strong Nuclear Force vs Nucleon Separation

Three Regions

< 0.5 fm: Repulsive (prevents nuclear collapse) | 0.5-3 fm: Attractive (holds nucleus together) | > 3 fm: Negligible

Nuclear Stability

A nucleus is stable when the strong force attraction balances the electrostatic repulsion between protons.

Small Nuclei

Few protons means weak electrostatic repulsion. The strong force easily holds the nucleus together.

Large Nuclei

Many protons means strong electrostatic repulsion. The strong force struggles to hold it together.

Role of Neutrons

Neutrons act as "nuclear glue" - they add strong force attraction without adding electrostatic repulsion. This is why heavier stable nuclei need more neutrons than protons.

Key Insight

Beyond about Z = 82 (lead), no arrangement of neutrons can make a nucleus completely stable. All heavier nuclei are radioactive.

Unstable Nuclei

When a nucleus is unstable, it will undergo radioactive decay to become more stable.

Key Concept

Unstable nuclei emit radiation to reduce their energy and move towards a more stable configuration.

Two main types of decay at AS Level:

Decay TypeWhat is EmittedWhen it Occurs

Alpha (α)Helium nucleus (2p + 2n)Heavy nuclei (too many nucleons)

Beta-minus (β⁻)Electron + antineutrinoNeutron-rich nuclei

Exam Tip

Always think: "What makes this nucleus unstable?" - this tells you what type of decay will occur.

Alpha Decay

The emission of an alpha particle (α) from a nucleus. An alpha particle is a helium-4 nucleus: 2 protons + 2 neutrons.

Alpha particle notation:

_{2}^{4} α

or

_{2}^{4} He

General equation for alpha decay:

_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} α

What Changes?

Mass number (A): Decreases by 4 | Proton number (Z): Decreases by 2

The element changes (transmutation) - the daughter nucleus is a different element!

Worked Example01

Write the equation for the alpha decay of uranium-238 ( $_{92}^{238} U$ ).

1

Write the parent nucleus:

_{92}^{238} U

2

In alpha decay: A decreases by 4, Z decreases by 2

3

New mass number: 238 - 4 = 234

4

New proton number: 92 - 2 = 90 (this is Thorium, Th)

5

Conservation check:

A: 238 → 234 + 4 = 238 ✓

Z: 92 → 90 + 2 = 92 ✓

Answer

_{92}^{238} U \to_{90}^{234} Th +_{2}^{4} α

Beta-Minus Decay

A neutron in the nucleus converts into a proton, emitting an electron (β⁻) and an electron antineutrino (ν̄ₑ).

The process at nucleon level:

n \to p + e^{-} + \overset{ν}{ˉ}_{e}

General equation for beta-minus decay:

_{Z}^{A} X \to_{Z + 1}^{A} Y +_{- 1}^{0} β + \overset{ν}{ˉ}_{e}

What Changes?

Mass number (A): Stays the same | Proton number (Z): Increases by 1

The neutron number decreases by 1 (makes sense - a neutron became a proton!)

What is an Antineutrino?

The electron antineutrino (ν̄ₑ) is the antiparticle of the electron neutrino. It has no charge, nearly zero mass, and barely interacts with matter - making it extremely difficult to detect.

Important

The electron in beta decay is NOT from the electron shells - it is created during the decay process.

Why the Neutrino?

When scientists first studied beta decay, they found a serious problem:

The Problem

The emitted electrons had a continuous range of energies - not a single fixed energy as expected from energy conservation.

Pauli's Solution (1930)

Wolfgang Pauli proposed that an undetected particle (later named the neutrino) carries away the "missing" energy and momentum. The electron and neutrino share the available energy.

Worked Example02

Write the equation for the beta-minus decay of carbon-14 ( $_{6}^{14} C$ ).

1

Write the parent nucleus:

_{6}^{14} C

2

In beta-minus decay: A stays same, Z increases by 1

3

New mass number: 14 (unchanged)

4

New proton number: 6 + 1 = 7 (this is Nitrogen, N)

5

Conservation check:

A: 14 → 14 + 0 = 14 ✓

Z: 6 → 7 + (-1) = 6 ✓

Answer

_{6}^{14} C \to_{7}^{14} N +_{- 1}^{0} β + \overset{ν}{ˉ}_{e}

Worked Example03

Tritium ( $_{1}^{3} H$ ) undergoes beta-minus decay. Write the decay equation and identify the daughter nucleus.

1

Parent:

_{1}^{3} H

(tritium - hydrogen with 2 neutrons)

2

Beta-minus: A unchanged, Z increases by 1

3

New A = 3, New Z = 1 + 1 = 2 (Helium)

4

Conservation check:

A: 3 → 3 + 0 = 3 ✓

Z: 1 → 2 + (-1) = 1 ✓

The daughter nucleus is helium-3.

Answer

_{1}^{3} H \to_{2}^{3} He +_{- 1}^{0} β + \overset{ν}{ˉ}_{e}

Worked Example04

Polonium-210 ( $_{84}^{210} Po$ ) undergoes alpha decay. Identify the daughter nucleus.

1

Parent:

_{84}^{210} Po

2

Alpha decay: A decreases by 4, Z decreases by 2

3

New A = 210 - 4 = 206, New Z = 84 - 2 = 82 (Lead)

4

Conservation check:

A: 210 → 206 + 4 = 210 ✓

Z: 84 → 82 + 2 = 84 ✓

The daughter nucleus is lead-206.

Answer

_{84}^{210} Po \to_{82}^{206} Pb +_{2}^{4} α

Worked Example05

Cobalt-60 ( $_{27}^{60} Co$ ) is neutron-rich. Explain why it undergoes beta-minus decay rather than alpha decay.

1

Co-60 has: 27 protons, 60 - 27 = 33 neutrons

2

Neutron-to-proton ratio: 33/27 = 1.22 (too many neutrons)

3

Beta-minus decay converts a neutron into a proton, reducing the neutron excess

4

Alpha decay would remove 2p + 2n, not addressing the neutron imbalance

5

Decay equation:

_{27}^{60} Co \to_{28}^{60} Ni +_{- 1}^{0} β + \overset{ν}{ˉ}_{e}

6

Conservation check:

A: 60 → 60 + 0 = 60 ✓

Z: 27 → 28 + (-1) = 27 ✓

AnswerBeta-minus decay specifically addresses the instability caused by excess neutrons by converting n → p, whereas alpha decay would not correct the neutron-proton imbalance.

Worked Example06

A nucleus undergoes two alpha decays followed by one beta-minus decay. If it starts as $_{92}^{235} U$ , find the final nucleus.

1

Start:

_{92}^{235} U

2

1st alpha decay: A = 235 - 4 = 231, Z = 92 - 2 = 90

_{92}^{235} U \to_{90}^{231} Th +_{2}^{4} α

3

2nd alpha decay: A = 231 - 4 = 227, Z = 90 - 2 = 88

_{90}^{231} Th \to_{88}^{227} Ra +_{2}^{4} α

4

Beta-minus decay: A = 227 (unchanged), Z = 88 + 1 = 89

_{88}^{227} Ra \to_{89}^{227} Ac +_{- 1}^{0} β + \overset{ν}{ˉ}_{e}

5

Conservation check (overall):

A: 235 → 227 + 4 + 4 + 0 = 235 ✓

Z: 92 → 89 + 2 + 2 + (-1) = 92 ✓

AnswerThe final nucleus is

_{89}^{227} Ac

(actinium-227)

Worked Example07

Explain why electrons emitted in beta decay have a continuous energy spectrum, and how this led to the hypothesis of the neutrino.

1

In beta decay, a fixed amount of energy is released (from mass difference)

2

If only the electron were emitted, it should have a single, fixed kinetic energy

3

However, experiments showed electrons had a continuous range of energies, from zero up to a maximum

4

This appeared to violate conservation of energy

5

Pauli proposed that another particle (the neutrino) is also emitted:

_{Z}^{A} X \to_{Z + 1}^{A} Y +_{- 1}^{0} β + \overset{ν}{ˉ}_{e}

6

The electron and antineutrino share the available energy in variable proportions

AnswerWhen the electron has maximum energy, the neutrino has minimum (nearly zero), and vice versa. This explains the continuous spectrum while conserving energy and momentum.

Conservation Laws in Decay

Every nuclear decay equation must obey these conservation laws:

QuantityRuleCheck

Mass Number (A)Total A before = Total A afterAdd up A values on both sides

Proton Number (Z)Total Z before = Total Z afterAdd up Z values on both sides

ChargeTotal charge before = Total charge afterSame as Z for nuclei; β⁻ has charge -1

EnergyTotal energy conservedIncludes mass-energy and kinetic energy

MomentumTotal momentum conservedDaughter + emitted particles share momentum

Exam Tip

Always check your decay equations by adding up A and Z on both sides. If they don't balance, something is wrong!

Practice

Question 1: Radon-222 (

_{86}^{222} Rn

) undergoes alpha decay. Write the complete decay equation.

Alpha decay: A decreases by 4, Z decreases by 2

New A = 222 - 4 = 218

New Z = 86 - 2 = 84 (Polonium, Po)

Conservation check:

A: 222 → 218 + 4 = 222 ✓

Z: 86 → 84 + 2 = 86 ✓

Answer

_{86}^{222} Rn \to_{84}^{218} Po +_{2}^{4} α

Practice

Question 2: Strontium-90 (

_{38}^{90} Sr

) undergoes beta-minus decay. Write the decay equation and identify the daughter nucleus.

Beta-minus decay: A unchanged, Z increases by 1

New A = 90 (same)

New Z = 38 + 1 = 39 (Yttrium, Y)

Conservation check:

A: 90 → 90 + 0 = 90 ✓

Z: 38 → 39 + (-1) = 38 ✓

The daughter nucleus is yttrium-90.

Answer

_{38}^{90} Sr \to_{39}^{90} Y +_{- 1}^{0} β + \overset{ν}{ˉ}_{e}

Practice

Question 3: A nucleus X undergoes alpha decay to form radium-228 (

_{88}^{228} Ra

). Identify the original nucleus X.

Work backwards! If the daughter is

_{88}^{228} Ra

after alpha decay...

Parent A = 228 + 4 = 232

Parent Z = 88 + 2 = 90 (Thorium, Th)

Conservation check:

A: 232 → 228 + 4 = 232 ✓

Z: 90 → 88 + 2 = 90 ✓

AnswerThe original nucleus is

_{90}^{232} Th

(thorium-232)

Summary

Key Concepts

TopicKey Points

Strong ForceAttractive 0.5-3 fm, repulsive <0.5 fm, acts on all nucleons equally

Alpha DecayEmits ⁴₂α, A decreases by 4, Z decreases by 2

Beta-Minus Decayn → p + e⁻ + ν̄ₑ, A unchanged, Z increases by 1

NeutrinoExplains continuous beta energy spectrum; conserves energy/momentum

Exam Essentials

Know the three regions of the strong force graph (repulsive, attractive, negligible)

Write decay equations with correct notation including the antineutrino for β⁻

Check conservation of A and Z in all equations

Explain why the neutrino was hypothesised