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

1 Answer
Jan 14, 2018

Answer:

#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.