Question #b3120

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Question #b3120

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Answer 1 · 2017-06-08T00:30:10

Here's what I got.

Explanation:

The idea here is that a radioactive isotope's nuclear half-life, $t_{1/2}$ $t_"1/2"$ , tells you the time needed for half of the atoms present in a sample of said isotope to undergo radioactive decay.

This means that if you start with $A_{0}$ $A_0$ radioactive isotope, you can say that you will end up with

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

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

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

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

and so on. You can thus say that the number of radioactive isotopes that remain undecayed after a period of time $t$ $t$ is 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 time $t$ $t$ .

$n = \frac{t}{t_{1/2}}$ $color(red)(n) = t/t_"1/2"$

In your case, you know that you start with $100$ $100$ radioactive isotopes and that $88$ $88$ have already decayed to their daughter nuclides. This, of course, implies that you're left with $12$ $12$ radioactive isotopes after an unknown time $t$ $t$ passes.

You can thus say that

$12 = 100 \cdot {(\frac{1}{2})}^{n}$ $12 = 100 * (1/2)^color(red)(n)$

Rearrange to get

$\frac{12}{100} = {(\frac{1}{2})}^{n}$ $12/100 = (1/2)^color(red)(n)$

This will be equivalent to

$ln (\frac{12}{100}) = ln [{(\frac{1}{2})}^{n}]$ $ln(12/100) = ln[(1/2)^color(red)(n)]$

which gets you

$n = \frac{ln (\frac{12}{100})}{ln (\frac{1}{2})} = 3.06$ $color(red)(n) = ln(12/100)/ln(1/2) = 3.06$

So, you know that $3.06$ $3.06$ half-lives must pass in order for the number of radioactive isotopes to go from $100$ $100$ to $12$ $12$ .

You now have

$t = n \cdot t_{1/2}$ $t = color(red)(n) * t_"1/2"$

and since you know that

$t_{1/2} = 1000 years$ $t_"1/2" = "1000 years"$

you can say that the rock is

$"age of rock" = 3.06 * "1000 years" = color(darkgreen)(ul(color(black)("3060 years")))$

I'll leave the answer rounded to three sig figs, but keep in mind that your values do not justify this value.

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Question #b3120

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