If we start with 1 mg of strontium, 0.953 mg will remain after 2.0 years. What is the half life of strontium-90?

1 Answer
Jun 26, 2016

Answer:

#t_"1/2" = "29 years"#

Explanation:

The nuclear half-life of a radioactive nuclide, #t_"1/2"#, is defined as the time needed for a sample of this nuclide to decay to half of its initial mass.

The first thing to notice here is that it took #2.0# years for only

#"1 mg " - " 0.953 mg" = "0.047 mg"#

of strontium-90 to decay. This tells you that the half-life of the nuclide is significantly longer than #2.0# years, since you need half of the original sample, the equivalent of #"0.50 mg"#, to decay in order to mark one half-life.

Your tool of choice here will be the equation

#color(blue)(|bar(ul(color(white)(a/a)A_"t" = A_0 * 1/2^ncolor(white)(a/a)|)))#

Here

#A_"t"# - the amount of the nuclide that remains undecayed after a period of time #t#
#A_0# - the initial mass of the nuclide
#n# - the number of half-lives that pass in the period of time #t#, calculated as

#color(purple)(|bar(ul(color(white)(a/a)color(black)(n = t/t_"1/2")color(white)(a/a)|)))#

Use this equation to find how many half-lives passed in #20# years

#A_"t" = A_0 * 1/2^n#

#0.953 color(red)(cancel(color(black)("mg"))) = 1color(red)(cancel(color(black)("mg"))) * 1/2^n#

#2^n = 1/0.953#

This will be equivalent to

#ln(2^n) = ln(1/0.953)#

#n * ln(2) = ln(1/0.953) implies n = ln(1/0.953)/ln(2)#

This will get you

#n = 0.06945#

This means that only #0.06945# of a half-life passed in #2.0# years. The half-life of the nuclide, #t_"1/2"#, will thus be

#n = "2.0 years"/0.06945 = color(green)(|bar(ul(color(white)(a/a)color(black)("29 years")color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the initial mass of the sample.