Chapter 30, Problem 40P

To determine

Activity of the sample of carbon

Answer to Problem 40P

Solution:

$A\text{=2}{\text{.5×10}}^{\text{9}}\text{decays/sec}$

Explanation of Solution

Given Info:

The radioactive carbon sample contains $\text{6}{\text{.5×10}}^{\text{20}}\text{nuclei}$.

Formula Used:

Activity can be described as

$A=\lambda N$

$A=\frac{\ln 2}{T_{1/2}}N$

Where N is number of particles at time t, λ is decay constant, A is activity and T_1/2 is half-life.

Calculation:

The activity of the carbon sample can be determined by the above equation as

$\begin{array}{l}A=\frac{\ln 2}{T_{1/2}}N\\ \text{A=}\frac{\text{ln2}}{\text{(5730yr)(3}{\text{.16×10}}^{\text{7}}\text{s/yr)}}\text{(6}{\text{.5×10}}^{\text{20}}\text{nuclei)}\\ \text{A=2}{\text{.5×10}}^{\text{9}}\text{decays/sec}\end{array}$