Question #95320

1 Answer
Dec 15, 2015

Answer:

#"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")#