Answer:
λ = 1.3252 x 10⁻⁴
Step-by-step explanation:
Since we are already given the half-life, the decay expression can be simplified as:
[tex]N(t) = N_0*e^{-\lambda t}\\\frac{N(half-life)}{N_0}=0.5[/tex]
For a half-life of t =5230 years:
[tex]0.5 = e^{(-\lambda t)} \\ln(0.5) = -\lambda * 5230\\\lambda = 1.3252*10^{-4}[/tex]
The decay-rate parameter λ for C-14 is 1.3252 x 10⁻⁴
