Structure of the Atom

What Is the Distance of Closest Approach?

In scattering experiments, a positively charged particle such as an alpha particle can be fired toward a positively charged atomic nucleus.

Because both the alpha particle and nucleus are positively charged, they repel each other electrically.

As the alpha particle approaches the nucleus:

• The electric repulsion becomes stronger.
• The alpha particle slows down.
• Its kinetic energy decreases.
• Its electric potential energy increases.

If the alpha particle is directed straight toward the centre of the nucleus, there is a point at which it momentarily stops before being repelled backward.

The distance between the centre of the alpha particle and the centre of the nucleus at this point is called the distance of closest approach.

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Why Doesn't the Alpha Particle Reach the Nucleus?

An alpha particle has a charge of:

+2e

A nucleus containing Z protons has a charge of:

+Ze

Since both charges are positive, there is an electrostatic repulsive force between them.

As the distance decreases, this force becomes stronger.

For a head-on approach, the alpha particle loses speed until its kinetic energy has been converted into electric potential energy.

At the closest point:

Speed = 0

and therefore:

Kinetic energy = 0

The alpha particle then accelerates away from the nucleus.

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Conservation of Energy

The calculation of closest approach relies on the principle of conservation of energy.

Energy cannot be created or destroyed. It can only be transferred or transformed.

Initially, far from the nucleus, the alpha particle has kinetic energy.

As it approaches the nucleus:

Kinetic energy → electric potential energy

At the distance of closest approach, for the idealized head-on case:

Initial kinetic energy = maximum electric potential energy

This provides the equation needed to calculate the closest distance.

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Electric Potential Energy

The electric potential energy of two point charges separated by distance r is:

Electric potential energy = kQ₁Q₂ / r

where:

• k = Coulomb constant
• Q₁ = charge of the alpha particle
• Q₂ = charge of the nucleus
• r = separation between their centres

Use:

k = 8.99 × 10⁹ N·m²/C²

The elementary charge is:

e = 1.60 × 10⁻¹⁹ C

Therefore, an alpha particle has charge:

Qα = +2e = +3.20 × 10⁻¹⁹ C

A nucleus with atomic number Z has charge:

Qnucleus = +Ze

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Closest-Approach Equation

At the distance of closest approach:

Initial kinetic energy = electric potential energy

Therefore:

KE = kQ₁Q₂ / r

Rearranging for r:

r = kQ₁Q₂ / KE

For an alpha particle approaching a nucleus:

Q₁ = 2e

and:

Q₂ = Ze

So we can also write:

r = 2kZe² / KE

This gives the closest centre-to-centre separation for an idealized head-on approach.

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Worked Example: Alpha Particle Approaching Gold

An alpha particle approaches a gold nucleus directly.

Gold has atomic number:

Z = 79

Suppose the alpha particle has kinetic energy:

KE = 8.0 × 10⁻¹³ J

Calculate the distance of closest approach.

The alpha particle charge is:

Q₁ = 2e = 3.20 × 10⁻¹⁹ C

The gold nucleus has charge:

Q₂ = 79e

So:

Q₂ = 79 × 1.60 × 10⁻¹⁹

Q₂ = 1.264 × 10⁻¹⁷ C

Now use:

r = kQ₁Q₂ / KE

Substitute:

r = (8.99 × 10⁹)(3.20 × 10⁻¹⁹)(1.264 × 10⁻¹⁷) / (8.0 × 10⁻¹³)

First multiply the charges:

(3.20 × 10⁻¹⁹)(1.264 × 10⁻¹⁷) = 4.0448 × 10⁻³⁶

Then multiply by k:

(8.99 × 10⁹)(4.0448 × 10⁻³⁶) ≈ 3.64 × 10⁻²⁶

Now divide by the kinetic energy:

r = (3.64 × 10⁻²⁶) / (8.0 × 10⁻¹³)

Therefore:

r ≈ 4.55 × 10⁻¹⁴ m

So the distance of closest approach is approximately:

4.6 × 10⁻¹⁴ m

or:

46 fm

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Interpreting the Result

The calculated value does not necessarily represent the radius of the gold nucleus.

Instead, it tells us that the alpha particle could not approach any closer than about 46 fm under those conditions.

The actual radius of a gold nucleus is considerably smaller.

Therefore, the result gives an upper limit on nuclear size:

Nuclear radius < distance of closest approach

This was historically important because scattering experiments showed that the positive charge and mass of the atom must be concentrated within a region much smaller than the atom itself.

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Why Faster Alpha Particles Get Closer

From:

r = kQ₁Q₂ / KE

we can see that:

r is inversely proportional to kinetic energy.

Therefore:

Greater kinetic energy → smaller closest approach

A faster alpha particle can overcome more of the electrical repulsion and travel closer to the nucleus before stopping.

For example, if the kinetic energy doubles:

KE × 2 → r ÷ 2

If the kinetic energy triples:

KE × 3 → r ÷ 3

This makes high-energy particles particularly useful for investigating very small structures.

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Comparing Two Alpha Particles

Suppose Alpha Particle A has kinetic energy:

4.0 × 10⁻¹³ J

Alpha Particle B has:

8.0 × 10⁻¹³ J

Both approach the same nucleus directly.

Particle B has twice the kinetic energy.

Therefore:

rB = 1/2 rA

Particle B can approach twice as close to the nucleus.

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Effect of Nuclear Charge

The closest-approach equation also shows:

r ∝ Z

for alpha particles with the same kinetic energy.

A nucleus with more protons has a larger positive charge and therefore produces stronger electrostatic repulsion.

For the same incoming alpha-particle energy:

Larger Z → larger distance of closest approach

For example, an alpha particle approaching a nucleus with 80 protons will be stopped farther away than the same alpha particle approaching a nucleus with 20 protons.

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From Kinetic Energy to Closest Approach

Sometimes a problem gives the alpha particle's mass and speed instead of its kinetic energy.

In this case, first calculate:

KE = ½mv²

Then use:

r = kQ₁Q₂ / KE

For an alpha particle, the mass is approximately:

mα = 6.64 × 10⁻²⁷ kg

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Worked Example Using Speed

An alpha particle travels directly toward a nucleus at:

1.0 × 10⁷ m/s

Suppose the nucleus has:

Z = 50

First calculate the alpha particle's kinetic energy.

Use:

KE = ½mv²

Substitute:

KE = ½(6.64 × 10⁻²⁷)(1.0 × 10⁷)²

Square the speed:

(1.0 × 10⁷)² = 1.0 × 10¹⁴

Therefore:

KE = ½(6.64 × 10⁻²⁷)(1.0 × 10¹⁴)

KE = 3.32 × 10⁻¹³ J

The nucleus has charge:

Q = 50e

Q = 50(1.60 × 10⁻¹⁹)

Q = 8.00 × 10⁻¹⁸ C

Now use:

r = kQ₁Q₂ / KE

r = (8.99 × 10⁹)(3.20 × 10⁻¹⁹)(8.00 × 10⁻¹⁸) / (3.32 × 10⁻¹³)

This gives approximately:

r = 6.9 × 10⁻¹⁴ m

or:

r ≈ 69 fm

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Energy in Electronvolts

Nuclear energies are often given in electronvolts (eV) rather than joules.

One electronvolt is:

1 eV = 1.60 × 10⁻¹⁹ J

Common nuclear units include:

1 keV = 10³ eV

1 MeV = 10⁶ eV

Therefore:

1 MeV = 1.60 × 10⁻¹³ J

An alpha particle with energy 5.0 MeV therefore has:

KE = 5.0 × 1.60 × 10⁻¹³ J

KE = 8.0 × 10⁻¹³ J

This is the kinetic energy used in our earlier gold-nucleus example.

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Worked Example Using MeV

An alpha particle with kinetic energy 6.0 MeV approaches a nucleus with atomic number 60.

First convert the energy to joules:

KE = 6.0 × 1.60 × 10⁻¹³

KE = 9.60 × 10⁻¹³ J

The alpha particle has charge:

Q₁ = 3.20 × 10⁻¹⁹ C

The nucleus has charge:

Q₂ = 60 × 1.60 × 10⁻¹⁹

Q₂ = 9.60 × 10⁻¹⁸ C

Now:

r = kQ₁Q₂ / KE

r = (8.99 × 10⁹)(3.20 × 10⁻¹⁹)(9.60 × 10⁻¹⁸) / (9.60 × 10⁻¹³)

Therefore:

r ≈ 2.88 × 10⁻¹⁴ m

or:

r ≈ 28.8 fm

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Closest Approach and Nuclear Size

Closest-approach experiments can help establish limits on nuclear size.

Suppose an alpha particle turns around at:

r = 3.0 × 10⁻¹⁴ m

If the interaction can be explained entirely by electrostatic repulsion at this distance, then the nucleus must fit inside that separation.

Therefore:

Nuclear radius must be less than about 3.0 × 10⁻¹⁴ m.

Experiments using particles with greater kinetic energies can probe even smaller distances.

This provides progressively better information about nuclear dimensions.

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Connection to Rutherford Scattering

Rutherford scattering provided powerful evidence for the nuclear model of the atom.

When alpha particles were fired at thin metal foil:

• Most passed through almost undeflected.
• Some were deflected through small angles.
• A very small number were deflected through large angles.
• A tiny fraction were scattered backward.

The large-angle scattering could only be explained if the atom's positive charge and most of its mass were concentrated in a very small region.

That region is the nucleus.

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Closest Approach and the Nuclear Model

The distance-of-closest-approach calculation strengthens the nuclear interpretation.

If positive charge were spread throughout the entire atom, as suggested by the older plum pudding model, an alpha particle would not experience the intense concentrated repulsion required to reverse its motion over such a small distance.

Instead, the results indicate that:

• Positive charge is concentrated.
• The concentration occupies a tiny region.
• The region is much smaller than the atom.
• Strong electrostatic forces occur when alpha particles approach this region.

These observations support the existence of a small, positively charged nucleus.

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Head-On and Non-Head-On Approaches

The simple closest-approach calculation assumes a head-on collision.

The alpha particle travels directly toward the centre of the nucleus.

In this special case, the alpha particle can momentarily stop before reversing direction.

Most particles in a scattering experiment do not approach the nucleus exactly head-on.

A particle passing to one side of the nucleus follows a curved path and is deflected through some angle rather than simply reversing direction.

The closer its path takes it to the nucleus, the greater the deflection generally becomes.

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Assumptions in the Simple Model

The basic closest-approach calculation makes several simplifying assumptions.

It assumes:

• The nucleus and alpha particle can be treated approximately as point charges until very small separations.
• The interaction is dominated by electrostatic repulsion.
• The nucleus remains approximately stationary.
• The approach is directly head-on.
• Energy is conserved.
• Other interactions can be neglected at the calculated separation.

These assumptions make the model useful, but they also establish its limitations.

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When the Simple Model Breaks Down

At sufficiently small distances, the simple electrostatic model becomes incomplete.

When particles approach nuclear dimensions, the strong nuclear interaction can become important.

The nucleus also has a finite size rather than being a perfect point charge.

Therefore, closest-approach calculations based only on Coulomb repulsion work best when the particles remain sufficiently far apart that electrostatic interaction dominates.

Deviations from the simple model can themselves provide information about nuclear structure.

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Connecting Closest Approach to Nuclear Radius

It is useful to distinguish these two quantities:

Nuclear radius

An estimate of the physical size of the nucleus.

Distance of closest approach

The minimum centre-to-centre separation reached by an incoming charged particle under specified conditions.

They are not automatically equal.

If:

rclosest > Rnucleus

then the alpha particle turns around before reaching the nuclear surface.

Increasing the alpha particle's energy reduces the distance of closest approach and allows scientists to probe smaller distances.

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Common Mistakes

Assuming Closest Approach Equals Nuclear Radius

The closest approach usually gives an upper limit on nuclear size, not necessarily the actual nuclear radius.

Forgetting That Both Charges Are Positive

An alpha particle and a normal atomic nucleus are both positively charged.

Their interaction is therefore repulsive.

Forgetting the Alpha Particle Has Charge +2e

Use:

Qα = +2e

not +e.

Using Atomic Mass Instead of Atomic Number

The nuclear electric charge depends on the number of protons.

Therefore:

Qnucleus = +Ze

where Z is the atomic number, not the nucleon number.

Forgetting to Convert MeV to Joules

If using SI values for k and charge, energy should be converted to joules.

Remember:

1 MeV = 1.60 × 10⁻¹³ J

Assuming the Alpha Particle Has Zero Energy at Closest Approach

Its kinetic energy is momentarily zero in the idealized head-on model.

Its energy has not disappeared. It has been converted into electric potential energy.

Applying the Simple Equation to Every Scattering Path

The straightforward energy calculation applies to a head-on approach.

Non-head-on scattering requires consideration of the particle's trajectory and angular momentum.

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Check Your Understanding

1. Define the distance of closest approach.

2. Why does an alpha particle slow down as it approaches a positively charged nucleus?

3. What happens to the alpha particle's kinetic energy during its approach?

4. What happens to its electric potential energy?

5. What is the alpha particle's kinetic energy at the closest point in an ideal head-on approach?

6. State the energy relationship used to calculate the distance of closest approach.

7. What is the charge of an alpha particle in terms of e?

8. What is the charge of a nucleus with atomic number Z?

9. Why can a higher-energy alpha particle approach a nucleus more closely?

10. If an alpha particle's kinetic energy doubles, what happens to its distance of closest approach to the same nucleus?

11. Why does a nucleus with a larger atomic number produce a larger closest-approach distance for alpha particles of equal energy?

12. Convert 4.0 MeV into joules.

13. Explain why a calculated closest approach can provide an upper limit for nuclear radius.

14. Why is the simple calculation most appropriate for a head-on approach?

15. Explain how closest-approach and scattering experiments provide evidence for a small, positively charged nucleus.

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Key Terms

• Distance of closest approach – the minimum separation between an incoming charged particle and a nucleus during an interaction.
• Alpha particle – a particle containing two protons and two neutrons with charge +2e.
• Electrostatic repulsion – the repulsive force between charges of the same sign.
• Kinetic energy – energy associated with motion.
• Electric potential energy – energy associated with the positions of electric charges.
• Conservation of energy – the principle that total energy remains constant in an isolated system.
• Electronvolt (eV) – a unit of energy commonly used in atomic and nuclear physics.
• Mega-electronvolt (MeV) – one million electronvolts.
• Head-on approach – a trajectory directed toward the centre of the target nucleus.
• Nuclear radius – an approximate measure of the size of a nucleus.
• Scattering – a change in the direction of a particle caused by an interaction.
• Coulomb interaction – the electrical attraction or repulsion between charged particles.

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Key Takeaways

• The distance of closest approach is the minimum separation reached by an incoming charged particle and a nucleus.
• An alpha particle and an atomic nucleus are both positively charged, so they repel.
• As an alpha particle approaches a nucleus, kinetic energy is converted into electric potential energy.
• For an ideal head-on approach, the alpha particle momentarily stops at its closest point.
• Conservation of energy gives: KE = kQ₁Q₂ / r.
• The closest distance can therefore be calculated using: r = kQ₁Q₂ / KE.
• An alpha particle has charge +2e, while a nucleus has charge +Ze.
• Increasing the alpha particle's kinetic energy decreases its distance of closest approach.
• A larger nuclear charge produces stronger electrostatic repulsion.
• Closest-approach calculations can provide an upper limit on nuclear size.
• Distance of closest approach and nuclear radius are related concepts, but they are not necessarily equal.
• Closest-approach and scattering experiments support the model of the atom containing a tiny, concentrated, positively charged nucleus.