Question

Question #95320

Answer 1

`"10.8 days"`

Explanation:

Here's a quick way to solve this problem - divide the mass of the initial sample by the mass of the final sample, then find a number that when used as an exponent for `2` gives you this ratio.

Once you have that number, multiply it by the half-life of the isotope and you have your answer.

In this case,

`(180 color(red)(cancel(color(black)("g"))))/(11.25color(red)(cancel(color(black)("g")))) = 16`

`2^x = 16 implies x = 4`

Therefore,

`t = 4 * "2.69 days" = "10.76 days"`

*Now, let me explain what's going on here. *

As you know, an isotope's nuclear half-life tells you the time needed for a sample of this isotope to decay to half of its initial value.

So, if your initial sample is `A_0`, you can say that you'll be left with

• `A_0 * 1/2 = A_0/2 ->` after one half-life**
• `A_0/2 * 1/2 = A_0/4 ->` after two half-lives**
• `A_0/4 * 1/2 = A_0/8 ->` after three half-lives**
• `A_0/8 * 1/2 = A_0/16 ->` after four half-lives**
  `vdots`

and so on.

This of course allows you to find a relationship between the initial sample, `A_0`, and the remaining sample, `A`, by using the number of half-lives that pass

`color(blue)(A = A_o * 1/2^n)" "`, where

`n` - how many half-lives pass in a given period of time

In your case, you know that the initial sample of `""^198"Au"` had a mass of `"180 g"`, and that after a period of time, the sample is down to `"11.25 g"`.

Plug this into the above equation to get

`11.25 color(red)(cancel(color(black)("g"))) = 180 color(red)(cancel(color(black)("g"))) * 1/2^n`

This is equivalent to

`1/2^n = 11.25/180`

`1/2^n = 1/16 implies 2^n = 16 <=> n = 4`

Since `n` is equal to

`color(blue)(n = "period of time"/"half-life")`

you can say that

`4 = "period of time"/"2.69 days"`

Therefore, the initial sample will be reduced to `"11.25 g"` in

`"period of time" = 4 xx "2.69 days" = "10.76 days"`

You should round this off to two sig figs, the number of sig figs you have for the initial sample, but I'll leave it rounded to three sig figs

`t = color(green)("10.8 days")`