Mathematical model for radioactive decay

The relationship between the number of undecayed radioactive nuclei (N) remaining from an original number (N₀) can be determined by:

${\text{N = N}}_0{\text{ e}}^{-\lambda \text{t}}$

Where:

• N = number of nuclei remaining
• N₀ = initial number of nuclei
• $\lambda$ = decay constant (s^-1)
• t = time (s)

As the nuclei decay over time, their activity (A) also changes. The activity at any given time t is:

${\text{A = A}}_0{\text{ e}}^{-\lambda \text{t}}$

Where:

• A = activity at time t (Bq)
• A₀ = initial activity at t = 0 (Bq)

Worked example - Calculating the number of remaining nuclei

Calculate the number of undecayed nuclei remaining after 10,000 years given the initial number of nuclei is 1.0 x 10⁶ and the decay constant for the substance is 1.209 x 10^-4 per year.

Step 1: Formula

$N=N_0{e}^{-\lambda \text{t}}$

Step 2: Substitution and correct evaluation

$N=1.0\times {10}^6{e}^{-1.209\times {10}^{-4}\times \mathrm{10,000}}=2.99\times {10}^3$

Radioactive dating using carbon-14

The radioactive isotope carbon-14 is extensively used in archaeological dating because:

• Living organisms absorb carbon-14 from the atmosphere.
• Upon death, the decay of carbon-14 in the organisms begins.
• The comparison of the remaining carbon-14 enables the dating of archaeological samples.

Key Points:

• Carbon-14 has a half-life of approximately 5,730 years.
• The measurement of the carbon-14 percentage facilitates the calculation of the sample's age.
• This method is applicable to materials that were once living, such as wood or bone.

Worked example - Calculating the age of an archaeological artefact

Estimate the age of an archaeological sample if the initial activity of carbon-14 was 0.25 Bq and has now decayed to 0.1 Bq, with the decay constant for carbon-14 being 1.21 x 10^-4 per year.

Step 1: Formula

$A=A_0\times e^{-\lambda \text{t}}$

Step 2: Rearranged formula

$\text{t = }\frac{\ln (\text{A}/{\text{A}}_0)}{-\lambda }$

Step 3: Substitution and correct evaluation

$t=\frac{\ln (0.1/0.25)}{-1.21\times {10}^{-4}}\approx \mathrm{7,573}\text{ years}$