Nuclear Instability

SPEC 3.8.1.4–3.8.1.5 • NUCLEAR PHYSICS

Decay modes, nuclear radius, and nuclear density

The N-Z Stability Graph

Stable nuclei form a narrow band on a graph of neutron number (N) against proton number (Z). This is called the line of stability.

The N-Z stability graph: nuclei above the band undergo β⁻ decay, those below undergo β⁺ decay

Key Feature

For light nuclei (Z < 20), N ≈ Z. For heavier nuclei, N > Z because more neutrons are needed to dilute proton-proton electrostatic repulsion.

Alpha (α) Decay

Alpha decay occurs in nuclei with too many nucleons (typically Z > 82). The nucleus emits an α-particle (helium nucleus).

α Decay Equation

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

PropertyChange

Proton number (Z)Decreases by 2 (ΔZ = −2)

Neutron number (N)Decreases by 2 (ΔN = −2)

Mass number (A)Decreases by 4 (ΔA = −4)

Example

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

Uranium-238 decays to Thorium-234 by emitting an α-particle.

Worked Example01

Polonium-210 ( $_{84}^{210} Po$ ) undergoes α decay. Write the decay equation and identify the daughter nucleus.

1

Apply α decay rules: ΔZ = −2, ΔA = −4

2

Calculate daughter: Z = 84 − 2 = 82 (Lead), A = 210 − 4 = 206

3

Equation:

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

AnswerLead-206 (

_{82}^{206} Pb

)

Beta-minus (β⁻) Decay

β⁻ decay occurs in nuclei with too many neutrons (above the stability line). A neutron converts to a proton.

β⁻ Decay Equation

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

PropertyChange

Proton number (Z)Increases by 1 (ΔZ = +1)

Neutron number (N)Decreases by 1 (ΔN = −1)

Mass number (A)No change (ΔA = 0)

Example

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

Carbon-14 decays to Nitrogen-14. The antineutrino (

\overset{ν}{ˉ}_{e}

) carries away energy.

Remember

In β⁻ decay:

n \to p + e^{-} + \overset{ν}{ˉ}_{e}

— a neutron becomes a proton.

Worked Example02

Strontium-90 ( $_{38}^{90} Sr$ ) undergoes β⁻ decay. Write the complete decay equation.

1

In β⁻ decay: n → p + e⁻ + ν̄ₑ, so ΔZ = +1, ΔA = 0

2

Calculate daughter: Z = 38 + 1 = 39 (Yttrium), A = 90

3

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

AnswerStrontium-90 decays to Yttrium-90

Beta-plus (β⁺) Decay

β⁺ decay occurs in nuclei with too many protons (below the stability line). A proton converts to a neutron.

β⁺ Decay Equation

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

PropertyChange

Proton number (Z)Decreases by 1 (ΔZ = −1)

Neutron number (N)Increases by 1 (ΔN = +1)

Mass number (A)No change (ΔA = 0)

Example

_{11}^{22} Na \to_{10}^{22} Ne +_{+ 1}^{0} β^{+} + ν_{e}

Sodium-22 decays to Neon-22. A positron (β⁺) and neutrino are emitted.

Remember

In β⁺ decay:

p \to n + e^{+} + ν_{e}

— a proton becomes a neutron.

Electron Capture

Electron capture is an alternative to β⁺ decay for proton-rich nuclei. An inner orbital electron is captured by the nucleus and combined with a proton.

Electron Capture Equation

_{Z}^{A} X +_{- 1}^{0} e^{-} \to_{Z - 1}^{A} Y + ν_{e}

PropertyChange

Proton number (Z)Decreases by 1 (ΔZ = −1)

Neutron number (N)Increases by 1 (ΔN = +1)

Mass number (A)No change (ΔA = 0)

Example

_{4}^{7} Be +_{- 1}^{0} e^{-} \to_{3}^{7} Li + ν_{e}

Beryllium-7 captures an electron to form Lithium-7.

Key Difference

Electron capture has the same effect as β⁺ decay (Z decreases by 1) but involves capturing an electron rather than emitting a positron. No β⁺ particle is emitted.

Gamma (γ) Emission

Gamma emission occurs when a nucleus has too much energy. This usually happens after α or β decay when the daughter nucleus is in an excited state.

γ Emission Equation

_{Z}^{A} X^{*} \to_{Z}^{A} X + γ

PropertyChange

Proton number (Z)No change (ΔZ = 0)

Neutron number (N)No change (ΔN = 0)

Mass number (A)No change (ΔA = 0)

Key Features

γ-rays are high-energy photons (electromagnetic radiation)

The * indicates an excited nuclear state

Energy of γ-ray equals the difference between nuclear energy levels

γ emission does not change the element — only releases excess energy

Energy Level Diagrams

You can represent nuclear decays using energy level diagrams which show the differences in energy of particles.

β⁻ Decay (Mo-99 → Tc-99m)

Mo-99 → Tc-99m

_{42}^{99} Mo \to_{43}^{99 m} Tc +_{- 1}^{0} β^{-} + \overset{ν}{ˉ}_{e}

The Tc-99m is formed in an excited state (*) and emits a γ-ray to reach ground state.

Nuclear Excited States

Nucleons occupy discrete energy levels. After α or β decay, the daughter nucleus is often in an excited state and releases energy as a γ-ray photon.

Isomeric Transition

When a nucleus transitions from an excited state to a lower energy state by emitting a γ-ray.

The energy of the γ-ray equals the energy difference between the two nuclear states:

E_{γ} = E_{e x c i t e d} - E_{g ro u n d}

Multiple γ-rays may be emitted as the nucleus cascades through several energy levels.

Key Point

Nuclei may decay through several types of emission before finally becoming stable. γ emission is evidence for discrete nuclear energy levels.

Technetium-99m

Tc-99m is the most widely used radioisotope in medical imaging. The "m" stands for metastable — a long-lived excited state.

Properties

Half-life: 6.0 hours

γ energy: 140 keV

Decay: Isomeric transition

Pure γ emitter (no α or β)

Why Ideal for Imaging

140 keV detected by gamma cameras

6 hr half-life: short enough to limit exposure, long enough for tests

γ-rays only weakly ionising

Easily prepared on site

Decay Equation

_{43}^{99 m} Tc \to_{43}^{99} Tc + γ

Nuclear Radius: Closest Approach

An α-particle fired at a positively charged nucleus is repelled by Coulomb repulsion. At closest approach, all kinetic energy converts to electrical potential energy.

Equations

V = \frac{1}{4 π ε _{0}} \frac{Q}{r} and E_{k} = \frac{1}{4 π ε _{0}} \frac{Q _{1} Q _{2}}{r}

Variables

E_{k}

= kinetic energy of α

Q_{1}

= nuclear charge = Ze

Q_{2}

= α charge = 2e

ε_{0}

= 8.85 × 10⁻¹² F m⁻¹

Limitation

This method gives an overestimate of nuclear radius — the α-particle may not reach the actual nuclear surface due to repulsion.

Important

Closest approach is not very accurate as it always overestimates. Electron diffraction gives more precise results.

Typical Mistake

Using MeV directly instead of converting to joules! Always convert: 1 MeV = 1.6 × 10⁻¹³ J

Worked Example03

An α-particle with kinetic energy 7.7 MeV is fired at a gold nucleus (Z = 79). Calculate the distance of closest approach.

e = 1.60 × 10⁻¹⁹ C, k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻², 1 MeV = 1.60 × 10⁻¹³ J

1

E_{k} = 7.7 \times 1.60 \times 1 0^{- 13} = 1.232 \times 1 0^{- 12}

J

2

Q₁ = 79 × 1.60 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C; Q₂ = 3.20 × 10⁻¹⁹ C

3

r = \frac{k Q _{1} Q _{2}}{E _{k}} = \frac{8.99 \times 1 0 ^{9} \times 1.264 \times 1 0 ^{- 17} \times 3.20 \times 1 0 ^{- 19}}{1.232 \times 1 0 ^{- 12}}

4

r = 2.95 \times 1 0^{- 14}

m

Answerr ≈ 3.0 × 10⁻¹⁴ m (this is an overestimate of nuclear radius)

Nuclear Radius: Electron Diffraction

Electrons are leptons — they do not interact via the strong force, only electromagnetically. This makes them ideal probes for measuring nuclear size.

Key Points

Electrons accelerated through potential V

De Broglie wavelength: λ = h/p = hc/E

At ~400 MeV, λ ≈ 3 × 10⁻¹⁵ m (nuclear scale)

More accurate than closest approach

Central maximum with first minima at angle ±θ

First Minimum Equations

sin θ = \frac{1.22 λ}{d} or sin θ = \frac{0.61 λ}{R}

Where d is diameter and R is radius (d = 2R)

03b

Worked Example

A 380 MeV electron is diffracted through an angle of 40° by a nucleus. Calculate the diameter of the nucleus.

h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J

1

Calculate wavelength:

λ = \frac{h c}{E} = \frac{6.63 \times 1 0 ^{- 34} \times 3.0 \times 1 0 ^{8}}{380 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}}

2

λ = 3.11 \times 1 0^{- 15}

m

3

Apply formula:

sin 40° = 0.643 = \frac{1.22 λ}{d}

4

d = \frac{1.22 λ}{s i n 40°} = \frac{1.22 \times 3.11 \times 1 0 ^{- 15}}{0.643} = 5.90 \times 1 0^{- 15}

m

AnswerDiameter d = 5.9 fm (so radius R ≈ 3.0 fm)

The Nuclear Radius Equation

Experimental measurements show nuclear radius follows a simple relationship with nucleon number:

Nuclear Radius

R = R_{0} A^{1/3}

QuantityValue

R_{0}

1.2 × 10⁻¹⁵ m = 1.2 fm

ANucleon number

Curved graph — not a straight line

Physical Meaning

Since

R \propto A^{1/3}

, we get

R^{3} \propto A

, so Volume ∝ A. This means nucleons are tightly packed like hard spheres → constant nuclear density.

Finding the R-A Relationship

Plotting nuclear radius (R) against nucleon number (A) shows no clear linear trend. However, a log-log graph reveals the relationship.

The Log Method

Assume: $R = k A^{n}$

Take logs:

ln R = ln k + n ln A

This is

y = m x + c

form:

Gradient = n = ⅓

y-intercept = ln k

k ≈ 1.2–1.4 fm (renamed R₀)

ln(R) vs ln(A) Graph

Result

R = R_{0} A^{1/3}

Where R₀ ≈ 1.2 fm is the Fermi radius constant.

Worked Example04

Calculate the ratio of the nuclear radius of uranium-238 to carbon-12.

1

R_{U} = R_{0} \times 23 8^{1/3}

,

R_{C} = R_{0} \times 1 2^{1/3}

2

\frac{R _{U}}{R _{C}} = (\frac{238}{12})^{1/3} = (19.83)^{1/3}

3

\frac{R _{U}}{R _{C}} = 2.71

AnswerUranium is 2.7× larger in radius

Worked Example05

Calculate the radius of an iron-56 nucleus. Take $R_{0}$ = 1.2 fm.

1

R = R_{0} A^{1/3} = 1.2 \times 1 0^{- 15} \times 5 6^{1/3}

2

5 6^{1/3} = 3.826

3

R = 1.2 \times 1 0^{- 15} \times 3.826 = 4.59 \times 1 0^{- 15}

m

AnswerR = 4.6 fm

The Liquid Drop Model

We assume the nucleus is spherical and has constant density — like a drop of liquid.

Principle

Volume is proportional to nucleon number:

V = k A = \frac{4}{3} π r^{3}

Rearranging for radius:

r = 3 \frac{3 k A}{4 π} = r_{0} A^{1/3}

Where r₀ is the radius of a single nucleon (≈1 fm)

Validity of Model

Electron diffraction confirms

R \propto A^{1/3}

This proves the liquid drop model is correct

Nuclear material has constant density

Since

r^{3} = r_{0}^{3} A

, volume ∝ mass

Key Consequence

All nuclei have the same density — they are like drops of the same incompressible liquid. Nucleons are tightly packed with no gaps.

Nuclear Density

All nuclei have the same density. We can derive a single value for nuclear density.

Derivation

Mass:

m = A \times m_{n u c l eo n}

where

m_{n u c l eo n} = 1.66 \times 1 0^{- 27}

kg

Volume:

V = \frac{4}{3} π R^{3} = \frac{4}{3} π (R_{0} A^{1/3})^{3} = \frac{4}{3} π R_{0}^{3} A

Density:

ρ = \frac{m}{V} = \frac{A m _{n u c l eo n}}{\frac{4}{3} π R _{0}^{3} A} = \frac{m _{n u c l eo n}}{\frac{4}{3} π R _{0}^{3}}

Nuclear Density

ρ = \frac{3 m _{n u c l eo n}}{4 π R _{0}^{3}} \approx 1.45 \times 1 0^{17} kg m^{- 3}

Note: A cancels — density is the same for ALL nuclei!

Comparison

Nuclear density (10¹⁷ kg m⁻³) is much larger than atomic density. This shows atoms are mostly empty space with most mass concentrated in the nucleus.

Worked Example06

Calculate nuclear density using $R_{0}$ = 1.2 × 10⁻¹⁵ m and $m_{n u c l eo n}$ = 1.66 × 10⁻²⁷ kg.

1

ρ = \frac{3 m _{n u c l eo n}}{4 π R _{0}^{3}}

2

R_{0}^{3} = (1.2 \times 1 0^{- 15})^{3} = 1.728 \times 1 0^{- 45}

m³

3

ρ = \frac{3 \times 1.66 \times 1 0 ^{- 27}}{4 π \times 1.728 \times 1 0 ^{- 45}} = \frac{4.98 \times 1 0 ^{- 27}}{2.17 \times 1 0 ^{- 44}}

4

ρ = 2.29 \times 1 0^{17}

kg m⁻³

Answerρ ≈ 2.3 × 10¹⁷ kg m⁻³

Practice Question 1

Cobalt-60 ( $_{27}^{60} Co$ ) undergoes β⁻ decay to form nickel-60 in an excited state. The nickel then emits two γ-rays.

(a) Write the equation for the β⁻ decay.

(b) Explain why γ emission does not change A or Z.

(a)

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

(b)

γ-rays are photons with zero mass and zero charge. They carry energy but don't change proton or neutron numbers.

Practice Question 2

The radius of carbon-12 is 2.7 fm. Calculate the radius of lead-208.

1

\frac{R _{P b}}{R _{C}} = (\frac{208}{12})^{1/3} = (17.33)^{1/3} = 2.59

2

R_{P b} = 2.59 \times 2.7 = 7.0

fm

AnswerR = 7.0 fm

Practice Question 3

A nucleus with A = 64 has radius 4.8 fm. Estimate $R_{0}$ and calculate nuclear density.

1

R_{0} = \frac{R}{A ^{1/3}} = \frac{4.8}{6 4 ^{1/3}} = \frac{4.8}{4} = 1.2

fm

2

ρ = \frac{3 \times 1.66 \times 1 0 ^{- 27}}{4 π \times ( 1.2 \times 1 0 ^{- 15} ) ^{3}}

3

ρ = 2.3 \times 1 0^{17}

kg m⁻³

Answer

R_{0}

= 1.2 fm, ρ ≈ 2.3 × 10¹⁷ kg m⁻³

Summary

Decay Modes

α: ΔZ = −2, ΔA = −4 (too many nucleons)

β⁻: ΔZ = +1, ΔA = 0 (too many neutrons)

β⁺: ΔZ = −1, ΔA = 0 (too many protons)

EC: ΔZ = −1, ΔA = 0 (too many protons)

γ: ΔZ = 0, ΔA = 0 (too much energy)

Nuclear Radius

Closest approach: overestimate

Electron diffraction:

sin θ = \frac{1.22 λ}{d}

R = R_{0} A^{1/3}

,

R_{0}

≈ 1.2 fm

Liquid drop model: V ∝ A

Key Results

R = R_{0} A^{1/3} ∣ λ = \frac{h c}{E} ∣ ρ \approx 1.45 \times 1 0^{17} kg m^{- 3}

Constant density → nucleons are incompressible (liquid drop model)

Tc-99m

Half-life 6.0 hours, γ energy 140 keV, pure γ emitter, easily prepared on site — ideal for medical imaging