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Answer 1 · 2017-04-08T01:46:14

Here's what I got.

Explanation:

The nuclear half-life, $t_{1/2}$ $t_"1/2"$ , tells you the amount of time needed in order for an initial sample of a radioactive nuclide to decay to half of its original mass.

If you take $B_{0}$ $B_0$ to be the initial mass of the radioactive substance, you can say that you will have

$B_{0} \cdot \frac{1}{2} = B_{0} \cdot {(\frac{1}{2})}^{1} \to$ $B_0 * 1/2 = B_0 * (1/2)^color(red)(1)->$ after $1$ $color(red)(1)$ half-life

$\frac{B_{0}}{2} \cdot \frac{1}{2} = B_{0} \cdot \frac{1}{4} = B_{0} \cdot {(\frac{1}{2})}^{2} \to$ $B_0/2 * 1/2 = B_0 * 1/4 = B_0 * (1/2)^color(red)(2) ->$ after $2$ $color(red)(2)$ half-lives

$\frac{B_{0}}{4} \cdot \frac{1}{2} = B_{0} \cdot \frac{1}{8} = B_{0} \cdot {(\frac{1}{2})}^{3} \to$ $B_0/4 * 1/2 = B_0 * 1/8 = B_0 * (1/2)^color(red)(3) ->$ after $3$ $color(red)(3)$ half-lives

$\frac{B_{0}}{8} \cdot \frac{1}{2} = B_{0} \cdot \frac{1}{16} = B_{0} \cdot {(\frac{1}{2})}^{4} \to$ $B_0/8 * 1/2 = B_0 * 1/16 = B_0 * (1/2)^color(red)(4) ->$ after $4$ $color(red)(4)$ half-lives
$⋮$ $vdots$

and so on. Now, if you take $B_{t}$ $B_t$ to be the amount that remains undecayed after a given period of time $t$ $t$ , you can say that $B_{t}$ $B_t$ will be equal to

$B_{t} = B_{0} \cdot {(\frac{1}{2})}^{n}$ $B_t = B_0 * (1/2)^color(red)(n)$

Here $n$ $color(red)(n)$ represents the number of half-lives that pass in the given period of time $t$ $t$ . You can express the number of half-lives that pass in the given period of time $t$ $t$ by using the half-life of the substance

$n = \frac{t}{t_{1/2}} (a \frac{\leftarrow the total time that passes}{\leftarrow one half-life a a a a a a a a a a a}$ $color(red)(n) = t/t_"1/2" color(white)((acolor(black)( larr " the total time that passes")/(color(black)( larr " one half-life")aaaaaaaaaaa)$

In your case, you have

$t_{1/2} = 40 years$ $t_"1/2" = "40 years" " "$ and $B_{0} = 3 g$ $" " B_0 = "3 g"$

so

$n = \frac{t}{40 years}$ $color(red)(n) = t/"40 years"$

and the exponential equation will look like this

$color(darkgreen)(bar(|ul(color(black)(color(white)(a/a)B_t = "3 g" * (1/2)^(t/"40 years")color(white)(a/a)))|))$

To determine how much $B$ is left after $8$ years, simply calculate the number of half-lives that will pass in this amount of time

$(8 color(red)(cancel(color(black)("years"))))/(40color(red)(cancel(color(black)("years")))) = 1/5$

and plug the result into the equation

$B_"8 years" = "3 g" * (1/2)^(1/5) = color(darkgreen)(ul(color(black)("2.61 g B")))$

Finally, to determine the number of years needed for the initial sample to be reduced to $"0.5 g"$ , rearrange the equation as

$0.5 color(red)(cancel(color(black)("g"))) = 3 color(red)(cancel(color(black)("g"))) * (1/2)^(t/"40 years")$

$(1/2)^(t/"40 years") = 0.5/3$

Take the natural log from both sides

$ln[(1/2)^(t/"40 years")] = ln(1/6)$

This will be equivalent to

$t/"40 years" * ln(1/2) = ln(1/6)$

You will end up with

$t = ln(1/6)/ln(1/2) * "40 years" = color(darkgreen)(ul(color(black)("103 years")))$

I'll leave the values rounded to three sig figs.

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Question #cc5db

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