The half-life of titanium-44 is 63 years. What is the constant k in the decay formula for the substance?

Jan 14, 2018

$k = {\text{0.011 years}}^{- 1}$

Explanation:

All you really need to use here is the fact that for a first-order reaction like radioactive decay, the integrated rate law takes the form

$\ln \left({\left[\text{A"]) - ln(["A}\right]}_{0}\right) = - k \cdot t$

Here

• $\left[\text{A}\right]$ is the concentration of the reactant after a given time $t$ passes
• ${\left[\text{A}\right]}_{0}$ is the initial concentration of the reactant
• $k$ is the rate constant

Now, the half-life of a radioactive nuclide, ${t}_{\text{1/2}}$, tells you the time needed for exactly half of an initial sample of the nuclide to undergo radioactive decay.

This means that after the passing of one half-life, you have

$t = {t}_{\text{1/2}}$

and

${\left[\text{A"] = 1/2 * [A}\right]}_{0}$

Plug this into the expression of the integrated rate law to get

ln(1/2 * ["A"]_0) - ln(["A"]_0) = - k * t_"1/2"

Rearrange to solve for $k$

ln( (1/2 * color(red)(cancel(color(black)(["A"]_0))))/color(red)(cancel(color(black)(["A"]_0)))) = - k * t_"1/2"

$\ln \left(\frac{1}{2}\right) = - k \cdot {t}_{\text{1/2}}$

You can rewrite this as

$\ln \left(1\right) - \ln \left(2\right) = - k \cdot {t}_{\text{1/2}}$

which will get you

$k = \ln \frac{2}{t} _ \text{1/2}$

Finally, plug in the value you have for the half-life of titanium-44 to find

k = ln(2)/"63 years" = color(darkgreen)(ul(color(black)("0.011 years"^(-1))))

The answer is rounded to two sig figs.