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

##### 1 Answer

#### Answer:

#### 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 theconcentrationof the reactant after a given time#t# passes#["A"]_0# is theinitial concentrationof the reactant#k# is therate constant

Now, the **half-life** of a radioactive nuclide, **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

#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**.