Titanium-44 is radioactive isotope such that every 63 years, its mass decreases by half. For a sample of titanium-44 with an initial mass of 100 grams, what is a function that will give the mass of the sample remaining after any amount of time?

1 Answer
Apr 4, 2017

Answer:

#A_t = "100 g" * (1/2)^(t/"63 years")#

Explanation:

The first thing to notice here is that the problem is giving you the half-life of titanium-44.

As you know, the half-life of a radioactive nuclide, #t_"1/2"#, tells you the amount of time needed for half of an initial sample of said nuclide to undergo radioactive decay.

In this case, you know that every #63# years, the mass of titanium-44 present in a sample is halved.

#t_"1/2" = "63 years"#

Now, if you take #A_0# to be the initial mass of titanium-44, you can say that you will have

  • #A_0 * 1/2 = A_0 * (1/2)^color(red)(1)-># after #color(red)(1)# half-life
  • #A_0/2 * 1/2 = A_0 * 1/4 = A_0 * (1/2)^color(red)(2) -># after #color(red)(2)# half-lives
  • #A_0/4 * 1/2 = A_0 * 1/8 = A_0 * (1/2)^color(red)(3) -># after #color(red)(3)# half-lives
  • #A_0/8 * 1/2 = A_0 * 1/16 = A_0 * (1/2)^color(red)(4) -># after #color(red)(4)# half-lives
    #vdots#

and so on. You can thus say that #A_t#, which will be the amount of titanium-44 that remains undecayed after a period of time #t#, will be equal to

#A_t = A_0 * (1/2)^color(red)(n)#

Here #color(red)(n)# represents the number of half-lives that pass in the given period of time #t#. Since you know that

#"number of half-lives" = "total time"/"one half-life"#

you can express the number of half-lives that pass in the period of time #t# by using the half-life of the nuclide

#color(red)(n) = t/t_"1.2"#

Plug this into the equation to get

#A_t = A_0 * (1/2)^(t/t_"1/2")#

Since you know that

#A_0 = "100 g"#

you can rewrite the equation as

#color(darkgreen)(ul(color(black)(A_t = "100 g" * (1/2)^(t/"63 years"))))#

This equation will allow you to find the mass of titanium-44 that remains undecayed after a given period of time #t# passes.

Notice that if #t# is a multiple of the half-life #t_"1/2"#, let's say

#t = 3 xx t_"1/2"#

you have

#color(red)(n) = (3 * color(red)(cancel(color(black)(t_"1/2"))))/color(red)(cancel(color(black)(t_"1/2"))) = color(red)(3)#

and

#A_t = "100 g" * (1/2)^color(red)(3)#

#A_t = "100 g" * 1/8 = "12.5 g"#

This means that after

#3 * "63 years" = "189 years"#

pass, the initial #"100-g"# sample of titanium-44 will be reduced to #"12.5 g"#.