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Nuclear Decay

The decay constant and activity

  • 3 things to know
  • A2 Unit 3

Overview

Recall questions on The decay constant and activity.

What you need to know

Answer each one before you reveal it.

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Numbers

Cobalt-60 has a half-life of 5.27 years. Calculate its decay constant in s⁻¹, taking one year as 3.16 × 10⁷ s.

Key term

State the unit of the decay constant.

Key term

State the equation linking activity, decay constant and number of undecayed nuclei, defining each symbol with its unit.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the unit of the decay constant.

    Model answer

    The decay constant has the unit s⁻¹, although h⁻¹ or y⁻¹ are often more convenient for long-lived nuclides.

  2. 2

    State the equation linking activity, decay constant and number of undecayed nuclei, defining each symbol with its unit.

    Model answer

    A = λN, where A is the activity in Bq, λ the decay constant in s⁻¹ and N the number of undecayed nuclei present.

  3. 3

    Define the becquerel.

    Model answer

    One becquerel is an activity of one decay per second.

  4. 4

    Why is there a minus sign in the relation dN/dt = −λN?

    Model answer

    The number of undecayed nuclei is falling as time passes, so the rate of change of N is negative while the activity λN is a positive quantity.

  5. 5

    Derive the relationship between the decay constant and the half-life.

    Model answer

    After one half-life the number left is N₀/2, so substituting into N = N₀e^(−λt) gives ½ = e^(−λt½). Taking natural logarithms gives ln2 = λt½, so λ = ln2/t½.

  6. 6

    Cobalt-60 has a half-life of 5.27 years. Calculate its decay constant in s⁻¹, taking one year as 3.16 × 10⁷ s.

    Model answer

    The half-life is 5.27 × 3.16 × 10⁷ = 1.66 × 10⁸ s, so λ = 0.693 ÷ 1.66 × 10⁸ = 4.17 × 10⁻⁹ s⁻¹. Substituting back, 0.693 ÷ 4.17 × 10⁻⁹ returns 1.66 × 10⁸ s, which is 5.27 years.

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