Revision NotesExam QuestionsPast Papers
A LevelPhysicsAQARevision NotesNuclear PhysicsNuclear Instability & RadiusClosest Approach Estimate

Closest Approach Estimate (AQA A Level Physics)

Revision Note

Katie M

Author

Katie M

Last updated

Closest Approach Estimate

  • The Coulomb equation for electric potential energy can be used to estimate the radii of nuclei other than gold

  • Initially, the alpha particles have kinetic energy equal to:

E subscript k space equals space e V space equals space 1 half m v squared

  • The electric potential energy between the two charges is equal to:

E subscript p space equals space fraction numerator Q q over denominator 4 straight pi epsilon subscript 0 r end fraction

  • This can be expressed as the potential energy at the point of repulsion:

E subscript p space equals space fraction numerator open parentheses 2 e close parentheses open parentheses Z e close parentheses over denominator 4 straight pi epsilon subscript 0 r end fraction space equals space fraction numerator 2 Z e squared over denominator 4 straight pi epsilon subscript 0 r end fraction

  • Where:

    • Charge of an alpha particle, Q = 2e

    • Charge of the target nucleus, q = Ze

    • Z = proton number of the target nucleus

    • e = elementary charge (C)

    • r = the distance of closest approach (m)

    • ε0 = permittivity of free space

  • When the alpha particle reaches the distance of closest approach (to the target nucleus), all of its kinetic energy E subscript k has been transformed into electric potential energy E subscript p

E subscript k space equals space E subscript p space equals space 1 half m v squared space equals space fraction numerator 2 Z e squared over denominator 4 straight pi epsilon subscript 0 r end fraction

  • Rearranging for the distance of closest approach r:

r space equals space fraction numerator 2 Z e squared over denominator 4 straight pi epsilon subscript 0 E subscript k end fraction space equals space fraction numerator Z e squared over denominator straight pi epsilon subscript 0 m v squared end fraction

  • This gives an upper limit for the radius of the nucleus, assuming the alpha particle is fired at a high energy

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.

Answer:

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

  • Proton number of aluminium, Z = 13

  • 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

E subscript k space equals space E subscript p space equals space 1 half m v squared space equals space fraction numerator Q q over denominator 4 straight pi epsilon subscript 0 r end fraction

1 half m v squared space equals space fraction numerator 2 Z e squared over denominator 4 straight pi epsilon subscript 0 r end fraction

Step 3: Rearrange for distance, r

r space equals space fraction numerator Z e squared over denominator straight pi epsilon subscript 0 m v squared end fraction

Step 4: Calculate the distance, r

r space equals space fraction numerator 13 cross times open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses squared over denominator straight pi cross times open parentheses 8.85 cross times 10 to the power of negative 12 end exponent close parentheses cross times 4 cross times open parentheses 1.66 cross times 10 to the power of negative 27 end exponent close parentheses cross times open parentheses 2.55 cross times 10 to the power of 7 close parentheses squared end fraction

r = 2.77 × 10–15 m

Examiner Tips and Tricks

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)

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:Nuclear RadiusNext:Nuclear Radius Equation
    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.