Question #d4494

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Question #d4494

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Answer 1 · 2018-01-03T12:31:21

Here's what I got.

Explanation:

The thing to remember about an isotope's nuclear half-life, $t_{1/2}$ $t_"1/2"$ , is that it represents the time needed for half of an initial sample of that isotope to undergo radioactive decay.

In other words, the half-life tells you how much must pass in order for your sample to be halved.

If you take $A_{t}$ $A_t$ to be the amount of a radioactive isotope that remains undecayed after a time $t$ $t$ passes and $A_{0}$ $A_0$ to be the initial amount of that isotope, you can say that you will have

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

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

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

and so on. So with every passing half-life, you get to divide the initial amount by $2$ $2$ . This means that if $n$ $color(red)(n)$ half-lives pass in a given period of tiem $t$ $t$ , you will divide the initial amount by $2$ $2$ a total of $n$ $color(red)(n)$ times.

$A_t =A_0/(underbrace(2 * 2 * ... * 2)_(color(black)(color(red)(n)color(white)(.)"times"))) = A_0/2^color(red)(n)$

This is equivalent to

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

with

$color(red)(n) = t/t_"1/2"$

In your case, you start with $"200 g"$ of this radioactive isotope and end up with $"25 g"$ . This means that you have

$25 color(red)(cancel(color(black)("g"))) = 200color(red)(cancel(color(black)("g"))) * (1/2)^color(red)(n)$

Divide both sides by $200$ to get

$25/200 = 1/2^color(red)(n)$

This is equivalent to

$1/4 = 1/2^color(red)(n)$

$1/2^2 = 1/2^color(red)(n) implies color(red)(n) = 2$

So you can say that in order for your sample to be reduced from $"200 g"$ to $"25 g"$ , two half-lives must pass.

This means that you have

$t = 2 * t_"1/2"$

$color(darkgreen)(ul(color(black)(t = 2 * "8.4 days" = "16.8 days")))$

I'll leave the answer rounded to three sig figs, but a more accurate answer would be

$t = "20 days"$

because you have only one significant figure for the initial mass of the sample.

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Question #d4494

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