# 8.3.5 Closest Approach Estimate

### Closest Approach Estimate

• The Coulomb equation can be used to give an estimate for the radius of a nucleus
• When the alpha particle reaches its closest approach (to the nucleus), all of its kinetic energy has been converted to electric potential energy
• Initially, the alpha particles have kinetic energy equal to: • The electric potential energy between the two charges is equal to: • At this point, the kinetic energy Ek lost by the α particle approaching the nucleus is equal to the potential energy gain Ep, so: • Where:
• m = mass of an α particle (kg)
• v = initial speed of the α particles (m s–1)
• q = charge of an α particle (C)
• Q = charge of the nucleus being investigated (C)
• r = the radius of closest approach (m)
• ε0 = permittivity of free space

#### Worked Example

The first artificially produced isotope, phosphorus-30 (15P) was formed by bombarding an aluminium-27 isotope (13Al) with an α particle.

For the reaction to take place, the α particle must come within a distance, r, from the centre of the aluminium nucleus.

Calculate the distance, r, if the nuclear reaction occurs when the α particle is accelerated to a speed of at least 2.55 × 107 m s–1.

Step 1: List the known quantities

• Mass of an α particle, m = 4u = 4 × (1.66 × 10–27) kg
• Speed of the α particle, v = 2.55 × 107 m s–1
• Charge of an α particle, q = 2e = 2 × (1.6 × 10–19) C
• Charge of an aluminium nucleus, Q = 13e = 13 × (1.6 × 10–19) C
• Permittivity of free space, ε0 = 8.85 × 10–12 F m–1

Step 2: Write down the equations for kinetic energy and electric potential energy Step 3: Rearrange for distance, r Step 4: Calculate the distance, r r = 2.77 × 10–15 m

#### Exam Tip

Make sure you’re comfortable with the calculations involved with the alpha particle closest approach method, as this is a common exam question.

You will be expected to remember that the charge of an α is the charge of 2 protons (2 × the charge of an electron)

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