# 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?

Apr 4, 2017

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, ${t}_{\text{1/2}}$, 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.

${t}_{\text{1/2" = "63 years}}$

Now, if you take ${A}_{0}$ to be the initial mass of titanium-44, you can say that you will have

• ${A}_{0} \cdot \frac{1}{2} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{1}} \to$ after $\textcolor{red}{1}$ half-life
• ${A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} \cdot \frac{1}{4} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{2}} \to$ after $\textcolor{red}{2}$ half-lives
• ${A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} \cdot \frac{1}{8} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{3}} \to$ after $\textcolor{red}{3}$ half-lives
• ${A}_{0} / 8 \cdot \frac{1}{2} = {A}_{0} \cdot \frac{1}{16} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{4}} \to$ after $\textcolor{red}{4}$ 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

${A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

Here $\textcolor{red}{n}$ represents the number of half-lives that pass in the given period of time $t$. Since you know that

$\text{number of half-lives" = "total time"/"one half-life}$

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

$\textcolor{red}{n} = \frac{t}{t} _ \text{1.2}$

Plug this into the equation to get

${A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \text{1/2}}$

Since you know that

${A}_{0} = \text{100 g}$

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 ${t}_{\text{1/2}}$, let's say

$t = 3 \times {t}_{\text{1/2}}$

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

${A}_{t} = \text{100 g} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{3}}$

${A}_{t} = \text{100 g" * 1/8 = "12.5 g}$

This means that after

$3 \cdot \text{63 years" = "189 years}$

pass, the initial $\text{100-g}$ sample of titanium-44 will be reduced to $\text{12.5 g}$.