Question #d4494

The thing to remember about an isotope's nuclear half-life, #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# to be the amount of a radioactive isotope that remains undecayed after a time #t# passes and #A_0# to be the initial amount of that isotope, you can say that you will have

• #A_t = A_0 * 1/2 = A_0/2 = A_0/2^color(red)(1) -># after #color(red)(1)# half-life
• #A_t = A_0/2 * 1/2 = A_0/4 = A_0/2^color(red)(2) -># after #color(red)(2)# half-lives
• #A_t = A_0/4 * 1/2 = A_0/8 = A_0/2^color(red)(3) -># after #color(red)(3)# half-lives
  #vdots#

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