Q. 5.67 - An Introduction to General, Organic, and Biological Chemistry

Question

A $120-mg$ sample of technetium- $99\mathrm{m}$ is used for a diagnostic test. If technetium- $99 \mathrm{m}$ has a half-life of $6.0 \mathrm{h}$. how many milligrams of the technetium-$99m$ sample remains active $24 \mathrm{h}$ after the test? $(5,4)$

Step-by-Step Solution

Answer

The value of the expression is $\frac{1}{16}$.

Step 1- Given information

We are given that

${\mathrm{N}}_0=120\mathrm{mg}, \mathrm{t}=24\mathrm{hr}, {\mathrm{t}}_{\frac{1}{2}}=6.0\mathrm{hr} \mathrm{N}=?$

The amount of radioactive sample (N) is known to remain after the $n$ half-life $=\frac{N_0}{2^n}$

 $N_{ e}=$ initial amount of radioactive material  

$\mathrm{N}=\text{Number of half periods}=\frac{\mathrm{t}}{{\mathrm{t}}_{\frac{1}{2}}}$

Step 2- Explanation

Replace the value of the expression

 $\frac{\mathrm{N}}{{\mathrm{N}}_0}={\left(\frac{1}{2}\right)}^{\frac{1}{t_1}}$

$\frac{\mathrm{N}}{120\mathrm{mg}}={\left(\frac{1}{2}\right)}^{\frac{24\mathrm{mr}}{6.04\mathrm{r}}}$

$\frac{\mathrm{N}}{120\mathrm{mg}}={\left(\frac{1}{2}\right)}^4$

or $\frac{\mathrm{N}}{120\mathrm{mg}}=\frac{1}{16}$