Question

If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 0.4 mg to 0.1 mg in 26.4 h, what is the half-life of iodine-123?

Short answer

The half-life of iodine-$123$is $13.2\text{h}$

Step-by-step solution

Given Information

The amount of radioactive iodine-$123$, used to treat thyroid cancer, in a sample decreases from $0.4$mg to $0.1$mg in $26.4$h.

Explanation

We know,

The initial concentration of I-$123$$N_o$= $0.4$mg and the concentration after $26.4$h N = $0.1$mg

Time t = $26.4$h

Using, $\mathrm{In}N-\mathrm{In}{N}_0=-\lambda t$

$\mathrm{In}\frac{0.1}{0.4}=-\lambda \times 26.4\mathrm{h}$

role="math" localid="1653899395922" $\Rightarrow \mathrm{\lambda }=0.05265/\mathrm{h}$

The half-life is

role="math" localid="1653899475118" $$\begin{gathered} t_{\frac{1}{2}}=\frac{0.693}{\lambda }==\frac{0.693}{0.05265} \\ =13.2\text{h} \end{gathered}$$

Hence the half-life of iodine-$123$is$13.2$h.