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?

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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?

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Answer 1 · 2017-04-04T01:00:48

$A_{t} = 100 g \cdot {(\frac{1}{2})}^{\frac{t}{63 years}}$ $A_t = "100 g" * (1/2)^(t/"63 years")$

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, $t_{1/2}$ $t_"1/2"$ , 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 $63$ $63$ years, the mass of titanium-44 present in a sample is halved.

$t_{1/2} = 63 years$ $t_"1/2" = "63 years"$

Now, if you take $A_{0}$ $A_0$ to be the initial mass of titanium-44, you can say that you will have

$A_{0} \cdot \frac{1}{2} = A_{0} \cdot {(\frac{1}{2})}^{1} \to$ $A_0 * 1/2 = A_0 * (1/2)^color(red)(1)->$ after $1$ $color(red)(1)$ half-life

$\frac{A_{0}}{2} \cdot \frac{1}{2} = A_{0} \cdot \frac{1}{4} = A_{0} \cdot {(\frac{1}{2})}^{2} \to$ $A_0/2 * 1/2 = A_0 * 1/4 = A_0 * (1/2)^color(red)(2) ->$ after $2$ $color(red)(2)$ half-lives

$\frac{A_{0}}{4} \cdot \frac{1}{2} = A_{0} \cdot \frac{1}{8} = A_{0} \cdot {(\frac{1}{2})}^{3} \to$ $A_0/4 * 1/2 = A_0 * 1/8 = A_0 * (1/2)^color(red)(3) ->$ after $3$ $color(red)(3)$ half-lives

$\frac{A_{0}}{8} \cdot \frac{1}{2} = A_{0} \cdot \frac{1}{16} = A_{0} \cdot {(\frac{1}{2})}^{4} \to$ $A_0/8 * 1/2 = A_0 * 1/16 = A_0 * (1/2)^color(red)(4) ->$ after $4$ $color(red)(4)$ half-lives
$⋮$ $vdots$

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

$A_{t} = A_{0} \cdot {(\frac{1}{2})}^{n}$ $A_t = A_0 * (1/2)^color(red)(n)$

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

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

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

$n = \frac{t}{t_{1.2}}$ $color(red)(n) = t/t_"1.2"$

Plug this into the equation to get

$A_{t} = A_{0} \cdot {(\frac{1}{2})}^{\frac{t}{t_{1/2}}}$ $A_t = A_0 * (1/2)^(t/t_"1/2")$

Since you know that

$A_{0} = 100 g$ $A_0 = "100 g"$

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 $"12.5 g"$ .

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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?

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