Question

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

Answer 1

`k = "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(["A"]) - ln(["A"]_0) = - k * t`

Here

• `["A"]` is the concentration of the reactant after a given time `t` passes
• `["A"]_0` is the initial concentration of the reactant
• `k` is the rate constant

Now, the half-life of a radioactive nuclide, `t_"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_"1/2"`

and

`["A"] = 1/2 * [A"]_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(1/2) = - k * t_"1/2"`

You can rewrite this as

`ln(1) - ln(2) = - k * t_"1/2"`

which will get you

`k = ln(2)/t_"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.