Question

Technetium$-99$ is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of $6.0\mathrm{h}$. Calculate the rate constant for the decay of ${}_{43}^{99}\mathrm{T}\mathrm{c}$.

Short answer

The rate constant for the decay of ${}_{43}^{99}\mathrm{T}\mathrm{c}$is $3.2\times 10^{-5}{\mathrm{s}}^{-1}$.

Step-by-step solution

Concept Introduction

The formula to calculate the decay constant is:

$\lambda =\frac{\mathrm{l}\mathrm{n}(2)}{{\mathrm{t}}_{1/2}}$

Where$\mathrm{t}$is time, and$\lambda$is the decay constant.

Calculating the decay constant

The value of the half-life is given as$6$hours:

$\begin{array}{c}1\mathrm{h}=60\mathrm{m}\mathrm{i}\mathrm{n}=3600\mathrm{s}\\ 6\mathrm{h}=21600\mathrm{s}\end{array}$

Substituting the values into the equation and solving:

$\begin{array}{c}\lambda =\frac{\mathrm{l}\mathrm{n}(2)}{21600\mathrm{s}}\\ \lambda =3.2\times 10^{-5}{\mathrm{s}}^{-1}\end{array}$

Therefore, the value of the rate constant is obtained as $3.2\times 10^{-5}{\mathrm{s}}^{-1}$.