Atomic Models and Rutherford’s Alpha-Particle Scattering Experiment: Distance of Closest Approach and Impact Parameter

Class 12/Modern Physics/NEET/JEE study materials

Introduction

J. J. Thomson proposed his atomic model in 1904 after discovering the electron. This model is commonly known as the plum pudding model.

Main features of Thomson’s atomic model

1. An atom is a sphere of uniformly distributed positive charge.
2. Negatively charged electrons are embedded within this positively charged sphere.
3. The total positive charge is equal in magnitude to the total negative charge of the electrons.
4. Therefore, the atom as a whole is electrically neutral.
5. The positive charge and mass were considered to be distributed throughout the volume of the atom.

Limitations of Thomson’s model

Although Thomson’s model explained the electrical neutrality of atoms, it could not explain the results of Rutherford’s alpha-particle scattering experiment.

In particular, it could not account for the large deflections of some alpha particles or the rare cases in which alpha particles were scattered backward.

These limitations led to the development of Rutherford’s nuclear model.

2. Rutherford’s Alpha-Particle Scattering Experiment

Ernest Rutherford, along with his collaborators Hans Geiger and Ernest Marsden, investigated the scattering of alpha particles by a thin metal foil, commonly gold foil.

An alpha particle is a positively charged particle consisting of two protons and two neutrons. Its charge is +2e, where e is the magnitude of the elementary charge.

Experimental arrangement

The experiment involved the following components:

• Radioactive source: Produced high-speed alpha particles.
• Lead block with a narrow opening: Formed a directed beam of alpha particles.
• Thin gold foil: Served as the target for alpha-particle scattering.
• Fluorescent zinc sulphide screen: Produced tiny flashes when struck by alpha particles.
• Microscope: Helped observe and count the flashes at different scattering angles.

Gold was chosen because it can be beaten into an extremely thin foil, allowing most alpha particles to pass through a small number of atomic layers.

Observations of Rutherford’s experiment

Rutherford and his collaborators observed the following:

1. Most alpha particles passed straight through the foil. This indicated that most of the volume of an atom is empty space.
2. Some alpha particles were deflected through small angles. This suggested that positive charge is concentrated rather than uniformly distributed.
3. A very small fraction were deflected through large angles. This indicated the presence of a strong repulsive electric field in a small region.
4. A very small number of alpha particles were scattered backward. This suggested that almost all the positive charge and a large fraction of the atomic mass are concentrated in a tiny central region.

Conclusions of Rutherford’s experiment

Rutherford proposed the nuclear model of the atom based on these observations.

• An atom contains a tiny, dense, positively charged nucleus at its centre.
• Nearly all the mass of the atom is concentrated in the nucleus.
• Electrons occupy the region surrounding the nucleus.
• Most of the volume of the atom is empty space.
• The nucleus is much smaller than the atom itself.

The nuclear model represented a major advance over Thomson’s model. However, Rutherford’s model could not explain the stability of atoms according to classical electrodynamics or the discrete line spectra emitted by atoms.

3. Why Are Alpha Particles Deflected?

Alpha particles carry a positive charge of +2e, and the nucleus is also positively charged. As an alpha particle approaches a nucleus, the electrostatic force between them is repulsive.

The closer the alpha particle approaches the nucleus, the stronger the repulsive force becomes.

The direction and magnitude of the deflection depend on factors such as:

• The initial speed and kinetic energy of the alpha particle.
• The positive charge of the nucleus.
• The impact parameter of the incoming alpha particle.
• The distance between the alpha particle and the nucleus during its motion.

Alpha particles passing far from the nucleus experience relatively small deflections, whereas particles approaching the nucleus more closely can experience much larger deflections.

4. Distance of Closest Approach

The distance of closest approach is the minimum distance between an incoming alpha particle and the nucleus during a head-on approach.

Consider an alpha particle moving directly towards a stationary nucleus. Because both are positively charged, the alpha particle slows down due to electrostatic repulsion.

At the point of closest approach, the alpha particle momentarily comes to rest in the radial direction before moving away from the nucleus. For an ideal head-on collision, its kinetic energy is converted into electrostatic potential energy.

Derivation of the distance of closest approach

Let:

• Z = atomic number of the target nucleus.
• e = elementary charge.
• E = initial kinetic energy of the alpha particle.
• r₀ = distance of closest approach.

The charge of the target nucleus is +Ze, while the charge of the alpha particle is +2e.

Initially, the alpha particle is sufficiently far from the nucleus that the electrostatic potential energy can be taken as zero. By conservation of energy:

Initial kinetic energy = Electrostatic potential energy at the closest approach

Therefore,

E = (1 / 4πε₀) × (2Ze² / r₀)

Rearranging,

Distance of closest approach:

r₀ = (1 / 4πε₀) × (2Ze² / E)

This expression applies to a head-on approach when the target nucleus can be treated as stationary and the electrostatic interaction is considered dominant.

Important conclusions

1. The distance of closest approach is directly proportional to the atomic number Z of the target nucleus.
2. It is inversely proportional to the initial kinetic energy E of the alpha particle.
3. Higher-energy alpha particles can approach the nucleus more closely.
4. For the same alpha-particle energy, a nucleus with a larger positive charge produces a greater distance of closest approach.

Numerical example

An alpha particle with kinetic energy 5 MeV approaches a gold nucleus with atomic number Z = 79 in a head-on collision. Calculate the distance of closest approach.

Use the approximate relation:

r₀ = 1.44 × 2Z / E MeV fm

Substituting Z = 79 and E = 5 MeV:

r₀ = (1.44 × 2 × 79) / 5 fm

r₀ = 45.5 fm

Therefore,

r₀ ≈ 4.55 × 10⁻¹⁴ m

This result shows that the alpha particle turns back at a distance of approximately 45.5 femtometres from the centre of the gold nucleus under the stated ideal conditions.

5. Impact Parameter

The impact parameter is the perpendicular distance between the initial direction of motion of an incoming alpha particle and the centre of the nucleus, assuming the particle would continue along its original straight-line path if there were no interaction.

It is usually represented by the symbol b.

Types of scattering based on impact parameter

• Zero impact parameter: The alpha particle approaches the nucleus head-on and undergoes the greatest possible deflection, ideally 180 degrees for repulsive Coulomb scattering.
• Small impact parameter: The particle passes relatively close to the nucleus and experiences a large deflection.
• Large impact parameter: The particle passes farther from the nucleus and experiences a smaller deflection.

The impact parameter is not the same as the distance of closest approach. The impact parameter describes the initial geometry of the incoming particle, whereas the distance of closest approach is the minimum actual separation during the interaction.

For a non-head-on trajectory, the alpha particle generally retains some transverse motion and does not come to rest completely at its closest approach.

6. Relationship Between Impact Parameter and Scattering Angle

For ideal Rutherford scattering under a repulsive Coulomb force, the impact parameter b is related to the scattering angle θ by:

b = (1 / 4πε₀) × (2Ze² / 2E) × cot(θ/2)

Equivalently,

b = (Ze² / 4πε₀E) × cot(θ/2)

Here:

• b = impact parameter.
• Z = atomic number of the target nucleus.
• E = initial kinetic energy of the alpha particle.
• θ = scattering angle.

This relationship assumes a point-like Coulomb interaction, a stationary target nucleus and non-relativistic motion.

Important results

1. As the impact parameter decreases, the scattering angle increases.
2. As the impact parameter increases, the scattering angle decreases.
3. For a head-on collision, b = 0 and θ = 180° in the ideal Rutherford model.
4. For a very large impact parameter, the scattering angle approaches zero.

These results explain why most alpha particles pass through the foil with little deflection, while only a small fraction undergo large-angle scattering.

7. Difference Between Distance of Closest Approach and Impact Parameter

Distance of closest approach	Impact parameter
Minimum actual separation between the alpha particle and nucleus.	Perpendicular distance from the nucleus to the initial straight-line trajectory.
Represented here by r₀ for a head-on collision.	Represented by b.
For a head-on collision, it depends on kinetic energy and nuclear charge.	It determines the scattering angle for a given energy and nuclear charge.
For a head-on collision, the alpha particle’s kinetic energy is converted into electrostatic potential energy at the turning point.	For non-head-on scattering, the particle retains transverse motion at the point of closest approach.

Exam tip: Do not use the head-on distance-of-closest-approach formula for every scattering trajectory. For non-head-on scattering, the impact parameter and angular momentum must also be considered.

8. Thomson’s Model vs Rutherford’s Model

Thomson’s atomic model	Rutherford’s nuclear model
Positive charge is distributed throughout the atom.	Positive charge is concentrated in a tiny nucleus.
Electrons are embedded in the positive sphere.	Electrons occupy the region surrounding the nucleus.
Does not explain large-angle alpha-particle scattering.	Explains the observed alpha-particle scattering results.
Does not contain a central nucleus.	Contains a small, dense, positively charged nucleus.

9. Important Formulas for CBSE, NEET and JEE

1. Charge of an alpha particle: +2e
2. Charge of a nucleus: +Ze
3. Electrostatic potential energy: U = (1 / 4πε₀) × (2Ze² / r)
4. Distance of closest approach for a head-on collision: r₀ = (1 / 4πε₀) × (2Ze² / E)
5. Impact parameter relation: b = (Ze² / 4πε₀E) × cot(θ/2)

Remember that E in these formulas is the initial kinetic energy of the alpha particle, and θ is the scattering angle.

10. Numerical Problems: Rutherford’s Alpha-Particle Scattering Experiment

Calculate the distance of closest approach

An alpha particle with kinetic energy 5 MeV approaches a gold nucleus (atomic number 79) in a head-on collision. Calculate the distance of closest approach.

Solution

Given: Z = 79, E = 5 MeV.

r₀ = 2.88 Z / E

r₀ = (2.88 × 79) / 5

r₀ = 45.50 fm

Answer: 4.55 × 10⁻¹⁴ m

Effect of increasing kinetic energy

An alpha particle approaches a gold nucleus in a head-on collision. Calculate the distance of closest approach when its kinetic energy is increased from 5 MeV to 10 MeV.

Solution

r₀ = 2.88 × 79 / 10

r₀ = 22.75 fm

Answer: 2.275 × 10⁻¹⁴ m

The kinetic energy doubles, so the distance of closest approach becomes half its original value.

Determine the atomic number of a nucleus

An alpha particle with kinetic energy 4 MeV has a distance of closest approach of 36 fm during a head-on collision. Find the atomic number of the target nucleus.

Solution

r₀ = 2.88 Z / E

Rearranging:

Z = r₀ E / 2.88

Z = (36 × 4) / 2.88

Z = 50

Answer: Atomic number = 50

Calculate the impact parameter

An alpha particle of kinetic energy 5 MeV is scattered by a gold nucleus (Z = 79) through an angle of 90°. Calculate the impact parameter.

Solution

b = (1.44 Z / E) cot(θ/2) fm

For θ = 90°, θ/2 = 45° and cot 45° = 1.

b = (1.44 × 79) / 5

b = 22.75 fm

Answer: 2.275 × 10⁻¹⁴ m

Find the scattering angle

An alpha particle of kinetic energy 4 MeV approaches a gold nucleus (Z = 79) with an impact parameter of 28.44 fm. Calculate its scattering angle.

Solution

b = (1.44 Z / E) cot(θ/2)

Rearranging:

cot(θ/2) = bE / (1.44 Z)

cot(θ/2) = (28.44 × 4) / (1.44 × 79)

cot(θ/2) ≈ 1

Therefore, θ/2 ≈ 45°.

Answer: θ ≈ 90°

Compare the scattering of two alpha particles

Two alpha particles of equal kinetic energy approach the same nucleus. The first has an impact parameter of 10 fm, while the second has an impact parameter of 20 fm. Which particle undergoes the larger scattering angle?

Solution

For the same nucleus and kinetic energy:

b is proportional to cot(θ/2).

A smaller impact parameter corresponds to a larger scattering angle.

Since 10 fm is smaller than 20 fm, the first alpha particle undergoes the larger deflection.

Answer: The first alpha particle.

Calculate electrostatic potential energy

An alpha particle is at a distance of 10 fm from a gold nucleus (Z = 79). Calculate its electrostatic potential energy.

Solution

Electrostatic potential energy, U = 2kZe² / r

Using k e² = 1.44 MeV·fm:

U = (2 × 1.44 × 79) / 10 MeV

U = 22.75 MeV

Answer: 22.75 MeV

Note: This is the electrostatic potential energy relative to zero potential energy at infinite separation.

11. Frequently Asked Questions

Q1. Why did Rutherford use gold foil?

Gold can be beaten into an extremely thin sheet, allowing alpha particles to pass through it while interacting with individual atomic nuclei.

Q2. Why did most alpha particles pass through the foil without deflection?

Most of the volume of an atom is empty space, so most alpha particles do not pass sufficiently close to a nucleus to experience a large deflection.

Q3. What does the distance of closest approach indicate?

For a head-on collision, it gives the minimum separation between the alpha particle and nucleus, where the alpha particle’s initial kinetic energy has been converted into electrostatic potential energy.

Q4. What is the physical significance of the impact parameter?

The impact parameter specifies how closely the initial path of an alpha particle would pass to the nucleus and helps determine the scattering angle.

Q5. What happens when the impact parameter is zero?

The alpha particle follows a head-on trajectory. In ideal Rutherford scattering, it approaches the nucleus, comes to a turning point and returns along the same line, giving a scattering angle of 180 degrees.

Q6. How does increasing alpha-particle energy affect the distance of closest approach?

For the same nucleus in a head-on collision, increasing the alpha particle’s kinetic energy decreases the distance of closest approach.

Conclusion

Thomson’s atomic model was an important early attempt to describe atomic structure, but Rutherford’s alpha-particle scattering experiment demonstrated that positive charge and most atomic mass are concentrated in a tiny nucleus.

The distance of closest approach explains how far an alpha particle can approach a nucleus in a head-on collision, while the impact parameter helps explain the angular deflection of alpha particles in non-head-on encounters.

A clear understanding of these concepts, their physical meaning and their formulas is essential for mastering atomic structure and solving numerical problems in CBSE, NEET and JEE Physics.

Explore more Physics notes, practical manuals, formula sheets, MCQs and previous-year questions at edu-physics.com.