An archaeologist finds the ¹⁴C in a sample of $2.10 \mathrm {~g}$ of material to be decaying at 107 Bq.A modern $1.00 - \mathrm { g }$ sample of the same material decays at 151 Bq,and assume this rate has also held in the past.The half-life of ¹⁴C is 5730 years.How old is the sample?

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The decay rate of the sample is proportional to its mass, so the decay rate of a 2.10 g sample in the present would be $151 \, \text{Bq/g} \times 2.10 \, \text{g} = 317.1 \, \text{Bq}$. The actual decay rate is 107 Bq, indicating the sample has decayed over time. Using the decay formula $N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}$, where $N$ is the remaining quantity, $N_0$ is the initial quantity, $t$ is the time, and $T$ is the half-life, we can solve for $t$. Given $N = 107 \, \text{Bq}$, $N_0 = 317.1 \, \text{Bq}$, and $T = 5730 \, \text{years}$, we find $t$ that matches the decay observed, which corresponds to approximately 8980 years.

An Archaeologist Finds the 14C in a Sample Of 2.10 g2.10 \mathrm {~g}2.10 g

An Archaeologist Finds the ¹⁴C in a Sample Of $2.10 \mathrm {~g}$