A brief note on Radioactive Decay and Half-Life . Read it before a test. Try each question at the end before you open its answer.
The three decays Alpha: Z A X → Z − 2 A − 4 Y + 2 4 He {}^{A}_{Z}\text{X} \to {}^{A-4}_{Z-2}\text{Y} + {}^{4}_{2}\text{He} Z A X → Z − 2 A − 4 Y + 2 4 He . The alpha particle takes the larger share of the kinetic energy, K α = Q A − 4 A K_\alpha = Q\,\dfrac{A - 4}{A} K α = Q A A − 4 . Beta-minus: n → p + e − + ν ˉ n \to p + e^- + \bar{\nu} n → p + e − + ν ˉ . The mass number stays and Z Z Z rises by 1 1 1 . Beta-plus: p → n + e + + ν p \to n + e^+ + \nu p → n + e + + ν . The mass number stays and Z Z Z falls by 1 1 1 . Gamma: an excited nucleus sends out a photon. A A A and Z Z Z stay. Conserved: the total mass number, the total charge, the energy and the momentum. Energy: Q = Δ m × 931.494 MeV/u Q = \Delta m \times 931.494\ \text{MeV/u} Q = Δ m × 931.494 MeV/u with atomic masses. For beta-plus, subtract 2 m e = 1.022 2m_e = 1.022 2 m e = 1.022 MeV. The decay law Law: d N d t = − λ N \dfrac{dN}{dt} = -\lambda N d t d N = − λ N , so N = N 0 e − λ t = N 0 2 − t / T N = N_0e^{-\lambda t} = N_0\,2^{-t/T} N = N 0 e − λ t = N 0 2 − t / T . Half-life: T = ln 2 λ = 0.693 λ T = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda} T = λ ln 2 = λ 0.693 . Mean life: τ = 1 λ = 1.44 T \tau = \dfrac{1}{\lambda} = 1.44\,T τ = λ 1 = 1.44 T . At t = τ t = \tau t = τ , N N 0 = 1 e \dfrac{N}{N_0} = \dfrac{1}{e} N 0 N = e 1 . Two ways to decay: λ = λ 1 + λ 2 \lambda = \lambda_1 + \lambda_2 λ = λ 1 + λ 2 , and 1 T = 1 T 1 + 1 T 2 \dfrac{1}{T} = \dfrac{1}{T_1} + \dfrac{1}{T_2} T 1 = T 1 1 + T 2 1 . Activity: R = λ N = R 0 2 − t / T R = \lambda N = R_0\,2^{-t/T} R = λ N = R 0 2 − t / T . 1 Bq = 1 s − 1 1\ \text{Bq} = 1\ \text{s}^{-1} 1 Bq = 1 s − 1 and 1 Ci = 3.7 × 10 10 Bq 1\ \text{Ci} = 3.7 \times 10^{10}\ \text{Bq} 1 Ci = 3.7 × 1 0 10 Bq . Series and dating Series: an alpha decay lowers A A A by 4 4 4 , a beta decay does not change it. So A A A changes by multiples of 4 4 4 along a series. Radiocarbon age: t = T log 2 R 0 R t = T\log_2\dfrac{R_0}{R} t = T log 2 R R 0 , with T = 5730 T = 5730 T = 5730 y. Uranium-lead age: t = T ln 2 ln ( 1 + N P b N U ) t = \dfrac{T}{\ln 2}\ln\left(1 + \dfrac{N_{Pb}}{N_U}\right) t = ln 2 T ln ( 1 + N U N P b ) , with numbers of atoms. A worked example A source has an activity of 1200 1200 1200 Bq and a half-life of 15 15 15 min. Find its activity after 1.0 1.0 1.0 h, and the time in which it falls to 75 75 75 Bq.
After 1.0 h. The time is 60 60 60 min, which is 4 4 4 half-lives. R = 1200 Bq × 1 16 = 75 R = 1200\ \text{Bq} \times \dfrac{1}{16} = 75 R = 1200 Bq × 16 1 = 75 Bq. Down to 75 Bq. The ratio is 1200 75 = 16 = 2 4 \dfrac{1200}{75} = 16 = 2^4 75 1200 = 16 = 2 4 . So 4 4 4 half-lives are needed, which is 60 60 60 min. The two parts agree. Practise with new numbers
Check yourself Complete: 92 238 U → Z A Th + 2 4 He {}^{238}_{92}\text{U} \to {}^{A}_{Z}\text{Th} + {}^{4}_{2}\text{He} 92 238 U → Z A Th + 2 4 He . What are A A A and Z Z Z ? Stuck? Show a hint Show the answerA nucleus has a decay constant of 0.0231 min − 1 0.0231\ \text{min}^{-1} 0.0231 min − 1 . What is its half-life? Stuck? Show a hint Show the answerThe carbon-14 activity of a bone is a quarter of that of a living bone. How old is it? Stuck? Show a hint Show the answerSources NCERT Physics Part II (Class 12, rationalised edition), Chapter 13 Nuclei. OpenStax University Physics Volume 3, Chapter 10 Nuclear Physics. OpenStax Chemistry 2e, Chapter 21 Nuclear Chemistry.