Question

The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 \({{\bf{d}}^{{\bf{ - 1}}}}\). What is the half-life for this decay?

Short answer

The half-life for this decay is 5.02 d.

Step-by-step solution

Rate of a Reaction 

The rate of reaction may be defined as the speed of the reactant reacting to obtain a product in a particular reaction at a particular time. The concentration of the reactant and product are represented into mole/L.

\({\bf{Rate = k}}{\left( {\bf{A}} \right)^{\bf{m}}}{\left( {\bf{B}} \right)^{\bf{n}}}^{}\)

Half-Life Period

The half-life of a reaction \(\left( {{{\bf{t}}_{{\bf{1/2}}}}} \right)\) is the time when the half of the reactant is consumed to produce product. Every half-life after that consumes half of the remaining reactant concentration.

\(\begin{aligned}{{}{}}{\,\,\,\,\,\,\,{\bf{ln }}{{\left( {\bf{A}} \right)}_{\bf{0}}}{\bf{/ }}\left( {\bf{A}} \right){\bf{ = kt}}}\\\begin{aligned}{l}{\bf{Hence, when t = }}\left( {{{\bf{t}}_{\frac{1}{2}}}} \right){\bf{, }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{A}} \right){\bf{ = }}\frac{1}{2}{\bf{ }}{\left( {\bf{A}} \right)_{\bf{0}}}{\bf{.}}\end{aligned}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\bf{t}}_{\frac{1}{2}}}{\bf{ = }}\frac{{{\bf{0}}{\bf{.693}}}}{{\bf{k}}}{\bf{ }}}\end{aligned}\)

Explanation

\(\begin{aligned}{{}{}}{{{\bf{t}}_{\frac{1}{2}}}{\bf{ = 0}}{\bf{.693 / k}}}\\{\,\,{{\bf{t}}_{\frac{1}{2}}}{\bf{ = 0}}{\bf{.693 /k }}}\\{\,\,{{\bf{t}}_{\frac{1}{2}}}{\bf{ = 0}}{\bf{.693 / 0}}{\bf{.138 }}{{\bf{d}}^{{\bf{ - 1}}}}}\\{\,\,{{\bf{t}}_{\frac{1}{2}}}{\bf{ = 5}}{\bf{.02 d}}}\end{aligned}\)