Question #bb66f

A radioactive nuclide's nuclear half-life is defined as the time needed for half of an initial sample to undergo radioactive decay.

If you take #"N"_0# to be the initial concentration of your radioactive nuclide #"X"#, you can say that you will be left with

#1/2 * "N"_0 = "N"_0/2 -> # after the passing of one half-life

#1/2 * "N"_0/2 = "N"_0/4 -># after the passing of two half-lives

#1/2 * "N"_0/4 = "N"_0/8 -># after the passing of three half-lives
#vdots#

and so on.

This means that you can express the amount of the nuclide that remains undecayed after a period of time #t# like this

#color(blue)(|bar(ul(color(white)(a/a)"N"_t = "N"_0 * 1/2^ncolor(white)(a/a)|)))" " " "color(orange)("(*)")#

Here

#n# - the number of half-lives that pass in the given period of time #t#

All you have to do now is pick a period of time #t# and use the known ratio that exists between #"N"_t# and #"N"_0# to find #n#.

For example, for #t=0 -> t="5 h"# you have

#"N"_t/"N"_0 = 1/2^n = 0.6484#

This is equivalent to

#2^n = 1/0.6484#

#ln(2^n) = ln(1/0.6484)#

#n * ln(2) = ln(1/0.6484) implies n = ln(1/0.6484)/ln(2) = 0.625#

So, you know that #0.625# half-lives pass in #5# hours, which means that the half-life of the nuclide, #t_"1/2"#, is

#t_"1/2" = "5 h"/0.625 = color(green)(|bar(ul(color(white)(a/a)color(black)("8 h")color(white)(a/a)|)))#

I recommend using different time periods to find #n#, the half-life must come out the same in every case. That is the case because radioactive decay is a first-order reaction, meaning that its half-life is constant throughout the decay process.

For example, for #t=0 -> t = "18 h"# you have

#"N"_t/"N"_0 = 0.2102#

This time you get

#n = ln(1/0.2102)/ln(2) = 2.25#

Once again, you have

#t_"1/2" = "18 h"/2.25 = "8 h"#

For the second part, use the half-life of the reaction to find the value of #n#

#n = (64 color(red)(cancel(color(black)("h"))))/(8color(red)(cancel(color(black)("h")))) = 8#

then use equation #color(orange)("(*)")# to find the mass of the nuclide that remains undecayed

#"N"_t = "2.500 mg" * 1/2^8#

#"N"_t = "0.009766 mg"#

This means that the amount of the nuclide that decayed is

#"N"_"decayed" = "2.500 mg" - "0.009766 mg" = color(green)(|bar(ul(color(white)(a/a)color(black)("2.490 mg")color(white)(a/a)|)))#

I'll leave the answer rounded to four sig figs.

O.K., this is actually one that requires almost no math to solve!

As you know, or are learning, a half-life is the amount of time it takes for half of a radioactive element to decay. So, in the language of your question Nt/N0 = 0.5 at one half life.

Looking at your chart, we see that that happened somewhere between 5 and 10 hours.

Now we also know that after two half lives Nt/N0 will be 0.25.

Woa!!!! THAT's ON THE CHART!!!! Awesome!

Two half-lives = 16 hours.. Therefore one half-life = 8 hours.

Wait! Does that check with the previous observation that the half-life will be between 5 and 10 hours? Yup! O.K.! We are in business!

On to part B! 64 hours is 64/8 half-lives = 8 half-lives. After 8 half-lives there will be #1/2^8# of the original substance remaining. or 0.00390625 * original mass remaining. So we have lost (2.500 - (0.00390625*2.500)) grams, or 2.4902 grams of the original material.