Question #e3336

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Question #e3336

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Answer 1 · 2017-12-27T12:36:49

$15.5 mg$ $"15.5 mg"$

Explanation:

The half-life of a radioactive nuclide, $t_{1/2}$ $t_"1/2"$ , is simply a measure of the time needed for half of a given sample to undergo radioactive decay.

So if you take $A_{0}$ $A_0$ to be the initial mass of carbon-14, you can say that this sample will be reduced to

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

and so on.

Now, let's say that $A_{t}$ $A_t$ represents the mass of carbon-14 that remains undecayed after a period of time $t$ $t$ . You can say that $A_{t}$ $A_t$ will be equal to

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

with

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

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

In your case, the sample has an initial mass of $52 mg$ $"52 mg"$ and the half-life of carbon-14 is equal to $5730$ $5730$ years, so you can say that after a period of time $t$ $t$ passes, the sample will be reduced to

$A_{t} = 52 mg \cdot {(\frac{1}{2})}^{\frac{t}{5730 years}}$ $A_t = "52 mg" * (1/2)^(t/"5730 years")$

To find the amount of carbon-14 that remains undecayed after $10,000$ $"10,000"$ years, simply plug in this value into the above equation.

You will have

$color(red)(n) = ("10,000" color(red)(cancel(color(black)("years"))))/(5730color(red)(cancel(color(black)("years"))))$

and

$A_t = "52 mg" * (1/2)^("10,000"/5730) = color(darkgreen)(ul(color(black)("15.5 mg")))$

The answer is rounded to the nearest tenth.

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Question #e3336

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