Question

Question #cc5db

Answer 1

Here's what I got.

Explanation:

The nuclear half-life, `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` to be the initial mass of the radioactive substance, you can say that you will have

• `B_0 * 1/2 = B_0 * (1/2)^color(red)(1)->` after `color(red)(1)` half-life
• `B_0/2 * 1/2 = B_0 * 1/4 = B_0 * (1/2)^color(red)(2) ->` after `color(red)(2)` half-lives
• `B_0/4 * 1/2 = B_0 * 1/8 = B_0 * (1/2)^color(red)(3) ->` after `color(red)(3)` half-lives
• `B_0/8 * 1/2 = B_0 * 1/16 = B_0 * (1/2)^color(red)(4) ->` after `color(red)(4)` half-lives
  `vdots`

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

`B_t = B_0 * (1/2)^color(red)(n)`

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

`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" " "` and `" " B_0 = "3 g"`

so

`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.