Question

A nuclear technician was accidentally exposed to potassium-42 while doing brain scans for possible tumours. The error was not discovered until 36 h later when the activity of the potassium-42 sample was 2.0 $\mu$Ci. If potassium- 42 has a half-life of 12 h, what was the activity of the sample at the time the technician was exposed?

Short answer

The activity of the sample at the time the technician was exposed ${\mathrm{N}}_0=16\mu \mathrm{Ci}$

Step-by-step solution

Given Information

The activity of the potassium-42 sample =$2.0\mu Ci$

The half-life of potassium- 42 = $12h$

Explanation

Given,

The activity of the potassium-$42$sample is localid="1654283076569" $\mathrm{N}=2.0\mu \mathrm{Ci}$

The half-life is

localid="1654283082928" $$\begin{gathered} {\mathrm{T}}_{1/2}=12\mathrm{h} \\ \mathrm{t}=36\mathrm{h} \end{gathered}$$

Using,

localid="1654283090738" $$\begin{gathered} \frac{N\left(t\right)}{N_0}={\left(\frac{1}{2}\right)}^{\frac{t}{T_1}} \\ \frac{{\displaystyle 2}}{{\displaystyle {\mathrm{N}}_0}}={\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)}^{\frac{{\displaystyle 36}}{{\displaystyle 12}}} \\ \Rightarrow {\mathrm{N}}_0=16\mu \mathrm{Ci} \end{gathered}$$

The activity of the sample at the time the technician was exposed islocalid="1654283096761" $16\mu Ci$.