# 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

#### Answer:

#### 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, **half** of an initial sample of said nuclide to undergo radioactive decay.

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

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

Now, if you take

#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 **undecayed** after a period of time

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

Here **number of half-lives** that pass in the given period of time

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

you can express the number of half-lives that pass in the period of time

#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

Notice that if **multiple** of the half-life

#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