Rutherford Scattering

SPEC 3.8.1.1, 3.8.1.2 • NUCLEAR PHYSICS

Discovering the nuclear model of the atom and the properties of ionising radiation

Our Knowledge of the Atom

Over time, experiments have revealed more about the atom and nucleus:

ScientistModelKey Ideas

DemocritusAtomosMatter made of indivisible, identical particles

DaltonSolid spheresDifferent atoms for different elements; atoms combine in fixed ratios

ThomsonPlum puddingDiscovered electrons; positive "pudding" with embedded negative electrons

RutherfordNuclearSmall, dense, positive nucleus; electrons orbit at large distances

The Plum Pudding Model

Before Rutherford's experiment, J.J. Thomson proposed the plum pudding model (1904):

Thomson's Model

The atom was thought to be a uniform sphere of positive charge with negatively charged electrons embedded throughout, like plums in a pudding.

Prediction

If this model were correct, alpha particles would pass through with only small deflections.

Rutherford's Experiment

In 1909, Geiger and Marsden (under Rutherford's direction) fired alpha particles at thin gold foil:

Why This Setup?

Gold foil: Very malleable, can be hammered extremely thin (~10⁻⁶ m)

Lead block: Collimates the alpha beam (absorbs stray particles)

Evacuated chamber: Alpha particles only travel ~5 cm in air

Fluorescent screen: Shows where alpha particles hit

Key Observations

ObservationProportionConclusion

Most α-particles passed straight through~99%The atom is mostly empty space

Some deflected through small angles (<10°)Small fractionThere is a positive centre causing repulsion

Very few deflected through large angles (>90°)~1 in 8000The nucleus is small, dense , and contains most of the mass

Rutherford's Famous Quote

"It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

The Nuclear Model

Atoms consist of a small, dense, positively charged nucleus surrounded by negatively charged electrons at relatively large distances. This replaced the plum pudding model.

The Rutherford Model

Key Features

Tiny, dense, positive nucleus

Most of atom is empty space

Electrons orbit at large distances

Size Comparison

Nucleus: ~10⁻¹⁵ m

Atom: ~10⁻¹⁰ m

Ratio: 1:100,000

Worked Example01

Explain why most alpha particles pass straight through gold foil in the Rutherford scattering experiment.

1

The atom is mostly empty space.

2

The nucleus is extremely small compared to the size of the atom (about 10⁻¹⁵ m compared to 10⁻¹⁰ m).

3

Most alpha particles pass through regions where there is no significant positive charge to repel them.

4

The electrons have negligible mass compared to alpha particles (about 7300 times lighter), so they cannot significantly deflect the alpha particles.

AnswerThe atom is mostly empty space with a very small nucleus, so most alpha particles do not pass close enough to the nucleus to experience significant Coulomb repulsion.

Worked Example02

Explain why a small number of alpha particles are deflected through angles greater than 90°.

1

Alpha particles are positively charged (they are helium nuclei with charge +2e).

2

The nucleus is also positively charged.

3

When an alpha particle passes very close to a nucleus, it experiences a strong electrostatic (Coulomb) repulsion.

4

If the alpha particle is heading almost directly at the nucleus, the repulsive force can be strong enough to reverse its direction.

5

The nucleus must be very small and contain most of the atom's mass for this to occur without the alpha particle passing through it.

AnswerAlpha particles heading directly towards the small, dense, positive nucleus experience strong Coulomb repulsion that reverses their direction.

Alpha, Beta and Gamma Radiation

Unstable nuclei can emit three main types of ionising radiation:

PropertyAlpha (α)Beta-minus (β⁻)Gamma (γ)

NatureHelium nucleus (⁴₂He)Electron (⁰₋₁e)EM radiation (photon)

Charge+2e-1e0

Mass4u (6.64 × 10⁻²⁷ kg)~0.0005u (9.11 × 10⁻³¹ kg)0

Speed<1% of cUp to 90% of cc (3 × 10⁸ m/s)

Ionising powerHigh (10⁴ per cm)Medium (10² per cm)Low (1 per cm)

Penetrating powerLow (few mm air, paper)Medium (~5 mm Al, 30 cm air)High (cm of Pb, m of concrete)

Alpha Emission Rule

Alpha particles can only be emitted by nuclei with atomic number Z > 60. This is because the strong nuclear force must be overcome, which requires a large nucleus.

Key Relationship

Ionising power and penetrating power are inversely related. Alpha is highly ionising so loses energy quickly and is stopped easily. Gamma is weakly ionising so can travel far.

Nature of Radioactive Decay

Radioactive decay has two fundamental characteristics:

Spontaneous

Decay happens without any external input of energy. It cannot be started, stopped, or changed by temperature, pressure, or chemical reactions.

Random

It is impossible to predict which nucleus will decay next or when. We can only give probabilities for large numbers of nuclei.

Why Beta Has Continuous Spectrum

In beta decay, an antineutrino is also emitted. The available energy is shared randomly between the beta particle and antineutrino, giving a range of beta energies.

Gamma Emission

Gamma radiation is emitted when a nucleus remains in an excited state after alpha or beta decay. The nucleus releases excess energy as a gamma photon to reach its ground state.

Absorption of Radiation

Experimental Identification

To identify radiation type using a GM tube:

Measure count rate with source present (no absorber)

Add paper: significant drop → alpha present

Add aluminium (few mm): further drop → beta present

Add thick lead: remaining drop → gamma present

Beta-plus (β⁺)

Beta-plus particles are positrons (anti-electrons). When they meet an electron, they annihilate, producing two gamma photons. This is used in PET scanning.

Deflection in Fields

The different charges of radiation types cause different deflections in electric and magnetic fields:

RadiationIn Electric FieldIn Magnetic Field

Alpha (α)Deflects towards negative plateDeflects (small radius)

Beta (β⁻)Deflects towards positive plateDeflects opposite to α (large radius)

Gamma (γ)No deflectionNo deflection

Inverse-Square Law for Gamma

Gamma radiation is part of the electromagnetic spectrum (highest frequency/shortest wavelength). It spreads out uniformly in all directions from a point source. The intensity follows the inverse-square law:

Formula

I = \frac{k}{x ^{2}}

where I = intensity, k = constant, x = distance from source

Why Inverse-Square?

The same amount of radiation spreads over the surface area of a sphere (4πr²). Double the distance → 4× the area → ¼ the intensity.

Important

This law applies to gamma radiation in air/vacuum. It does NOT apply to alpha or beta due to their absorption in air.

Worked Example03

A gamma source gives a corrected count rate of 400 counts per second at a distance of 20 cm. Calculate the corrected count rate at a distance of 50 cm.

1

Using the inverse-square law:

I \propto \frac{1}{x ^{2}}

2

Therefore:

\frac{I _{2}}{I _{1}} = \frac{x _{1}^{2}}{x _{2}^{2}}

3

Substitute values:

\frac{I _{2}}{400} = \frac{2 0 ^{2}}{5 0 ^{2}} = \frac{400}{2500}

4

Calculate:

I_{2} = 400 \times \frac{400}{2500} = 400 \times 0.16 = 64

counts per second

Answer: 64 counts per second

Worked Example04

The intensity of gamma radiation at 2.0 m from a source is 45 W m⁻². At what distance is the intensity 5.0 W m⁻²?

1

Using:

\frac{I _{2}}{I _{1}} = \frac{x _{1}^{2}}{x _{2}^{2}}

2

Rearrange for x₂:

x_{2}^{2} = x_{1}^{2} \times \frac{I _{1}}{I _{2}}

3

Substitute:

x_{2}^{2} = 2. 0^{2} \times \frac{45}{5.0} = 4.0 \times 9 = 36

4

x_{2} = 36 = 6.0

m

Answer: 6.0 m

Activity vs Count Rate

Activity (A)

The total number of decays per second in all directions from a source. Measured in becquerels (Bq).

1 Bq = 1 decay per second

Count Rate

The number of decays detected per second by a detector (e.g., GM tube). Always less than the activity.

Key Point

Count rate is always much less than activity because the detector only captures radiation emitted in its direction. A GM tube typically detects only a small fraction of the total emissions.

In Calculations

Count rate is proportional to activity, so we can still use it in inverse-square law calculations: C ∝ 1/x²

Background Radiation

Background radiation is always present in our environment. When measuring radiation, we must account for it:

Correction Formula

Corrected count = Measured count − Background count

Natural Sources (~85%)

Radon gas (~50%) - from decay of uranium in rocks, builds up in buildings

Cosmic rays - from space

Rocks and soil - uranium, thorium

Food and drink - K-40, C-14

Internal - K-40 in body

Artificial Sources (~15%)

Medical - X-rays, scans, radiotherapy

Nuclear weapons testing - fallout

Nuclear power - small contribution

Industrial uses

Typical Background

Background count rate is typically 0.3-0.5 counts per second (or about 20-30 counts per minute) depending on location.

Worked Example05

In an experiment, the background count over 5 minutes was 75 counts. With a gamma source present at 15 cm, the count over 1 minute was 340 counts. Calculate the corrected count rate due to the source only.

1

Background count rate = 75 ÷ 5 = 15 counts per minute

2

Measured count rate with source = 340 ÷ 1 = 340 counts per minute

3

Corrected count rate = Measured − Background

4

Corrected count rate = 340 − 15 = 325 counts per minute

Answer: 325 counts per minute (or 5.42 counts per second)

Applications of Radiation

ApplicationRadiation TypeWhy This Type?

Smoke detectorsAlphaHighly ionising, creates current in air; smoke absorbs alpha, reducing current

Paper thickness controlBetaPartially absorbed by paper; too thick = lower count = rollers move closer

Steel thickness controlGammaPenetrates metal; intensity indicates thickness

Medical tracersGamma (Tc-99m)Penetrates body, detected externally; short half-life (6 hrs)

Cancer treatmentGammaPenetrates to tumour, ionising damages cancer cells

PET scanningBeta-plusPositron annihilation produces detectable gamma pairs

Food irradiationGammaKills bacteria to preserve food; does not make food radioactive

Sterilise equipmentGammaKills bacteria on surgical instruments; penetrates packaging

Safe Handling of Sources

Ionising radiation can damage cells and DNA. Safety measures reduce exposure:

The Three Principles

1. TIME: Minimise time spent near sources

2. DISTANCE: Maximise distance from sources (inverse-square law helps)

3. SHIELDING: Use appropriate shielding between you and the source

Practical Precautions

Handle with long tongs (never hands)

Store in lead-lined containers

Point source away from people

Use minimum activity needed

Wear monitoring badges

Never eat/drink near sources

Risk vs Benefit

In medicine, the benefit of diagnosis or treatment must outweigh the radiation risk. Doses are kept "as low as reasonably achievable" (ALARA principle).

Worked Example06

A worker must not exceed a dose rate of 2.0 μSv h⁻¹. At 0.50 m from a gamma source, the dose rate is 32 μSv h⁻¹. Calculate the minimum safe working distance from the source.

1

Using inverse-square law:

\frac{I _{2}}{I _{1}} = \frac{x _{1}^{2}}{x _{2}^{2}}

2

Rearrange:

x_{2}^{2} = x_{1}^{2} \times \frac{I _{1}}{I _{2}}

3

Substitute:

x_{2}^{2} = 0.5 0^{2} \times \frac{32}{2.0} = 0.25 \times 16 = 4.0

4

x_{2} = 4.0 = 2.0

m

Answer: Minimum safe distance = 2.0 m

Practice

Question 1: Explain why alpha particles, despite being the most ionising, are considered the least dangerous when the source is outside the body, but the most dangerous when the source is inside the body.

Outside the body:

•

Alpha particles have very low penetrating power

•

They are stopped by the dead outer layer of skin (epidermis)

•

They cannot reach living cells

Inside the body (inhaled or ingested):

•

Alpha particles are highly ionising (cause many ionisations per cm)

•

All their energy is deposited in a small volume of tissue

•

This causes significant cell damage and DNA mutations

•

They cannot escape so all radiation is absorbed

Practice

Question 2: The count rate from a gamma source is 1800 counts per minute at a distance of 10 cm. The background count rate is 24 counts per minute. What is the corrected count rate at 30 cm from the source?

1

Corrected count rate at 10 cm = 1800 − 24 = 1776 counts per minute

2

Using inverse-square law:

\frac{I _{2}}{I _{1}} = \frac{x _{1}^{2}}{x _{2}^{2}} = \frac{1 0 ^{2}}{3 0 ^{2}} = \frac{100}{900} = \frac{1}{9}

3

Corrected count rate at 30 cm = 1776 × (1/9) = 197.3 counts per minute

Answer: 197 counts per minute (or approximately 200 counts per minute)

Practice

Question 3: A student plots a graph of count rate (I) against 1/x² for a gamma source. The graph is a straight line but does not pass through the origin. Suggest two reasons why.

Possible reasons for a non-zero intercept:

1

Background radiation not corrected: If background count was not subtracted, there would be a positive y-intercept (count rate ≠ 0 even as x → ∞).

2

Distance measurement error: If the distance x was measured from the wrong point (e.g., from the case of the GM tube rather than the detector, or not from the centre of the source), all distances would be systematically wrong.

3

Source not a point source: At very small distances, the extended nature of the source means the inverse-square law doesn't hold exactly.

Summary

Key Concepts

ConceptKey Points

Rutherford ScatteringMost α pass through (empty space); few deflected back (small dense nucleus)

Alpha radiationHe nucleus, +2e, highly ionising, stopped by paper

Beta radiationElectron/positron, ±1e, medium ionising, stopped by aluminium

Gamma radiationEM wave, no charge, weakly ionising, reduced by lead

Inverse-square lawI = k/x² for gamma in air; double distance = quarter intensity

Background radiationAlways present; must subtract from measurements

Key Exam Skills

Explain Rutherford's observations in terms of atomic structure

Compare properties of α, β, γ radiation

Apply the inverse-square law to calculate intensities or distances

Describe RP12 and analyse graphical results

Evaluate risk vs benefit in medical applications