Q. 5.76 - An Introduction to General, Organic, and Biological Chemistry

Question

If the amount of radioactive iodine-$123$, used to treat thyroid cancer, in a sample decreases from $0.4$ my to$0.1\mathrm{mg}$ in $26.4 \mathrm{h}$, what is the half-life of iodinc-$123$? (5.4)

Step-by-Step Solution

Answer

The iodin half-life-$123$is$13.2hr$

Step 1:Introduction

This is due to the radioactive iodine in the drink or capsule being absorbed by your body and picked up by thyroid cancer cells, even if they have spread to other parts of the body. The cancer cells are then destroyed by the radiation.

Step 2:Given data

Given data is amount of radioactive iodine-$123$, used to treat thyroid cancer, in a sample decreases

To find half-life of  iodine-  $123$

Step 3:Explanation

$a=0,4mg$

$a-x=0,1\text{ mg }$

$\begin{array}{r}t=26.4\mathrm{h}\\ t=26.4\mathrm{hr}\end{array}$

$k=\frac{2\cdot 303}{t}\log \left(\frac{a}{a-x}\right)$

$k=\frac{0.693}{t\frac{1}{2}}$

$$\begin{gathered} \frac{0.693}{t+2} \\ =\frac{2.303}{t}\log \left(\frac{a}{a-x}\right) \end{gathered}$$

$$\begin{gathered} \frac{0.693}{1} \\ =2.303\log \left(\frac{0.4}{1.1}\right)=0.0525 \end{gathered}$$

$$\begin{gathered} t_2=(0,693(0,0525) \\ =13.2\mathrm{hr} \end{gathered}$$

$t1/2=13.2 hr$