A patient is given #0.050# #mg# of technetium-99m, a radioactive isotope with a half-life of about #6.0# hours. How long until the radioactive isotope decays to #6.3 * 10^-3# #mg#?

1 Answer
Dec 5, 2015

Answer:

#"44 h"#

Explanation:

As you know, a radioactive isotope's nuclear half-life tells you exactly how much time must pass in order for an initial sample of this isotope to be halved.

The equation that establishes a relationship between the initial mass of a radioactive isotope, the mass that remains undecayed after a given period of time, and the isotope's half-life looks like this

#color(blue)(A = A_0 * 1/2^n)" "#, where

#A# - the initial mass of the sample
#A_0# - the mass remaining after a given time
#n# - the number of half-lives that pass in the given period of time

In your case, the initial amount of technetium-99 is said to be equal to #"0.050 mg"#. You are asked to find the amount of time needed to reduce this initial sample to #6.3 * 10^(-3)"mg"#.

Use the above equation to find the value of #n#

#6.3 * 10^(-3) color(red)(cancel(color(black)("mg"))) = 0.050 * 10^(-3)color(red)(cancel(color(black)("mg"))) * 1/2^n#

This is equivalent to

#(6.3 * 10^(-3))/0.050 = 1/2^n#

Take the natural log of both sides of the equation to get

#ln((6.3 * 10^(-3))/0.050) = ln( (1/2)^n)#

#ln((6.3 * 10^(-3))/0.050) = n * ln(1/2)#

#n = ln(1/2)/ln((6.3 * 10^(-3))/0.050) = 7.31#

Since #n# represens the number of half-lives that pass in a given period of time, you can say that

#n = t/t_"1/2" implies t = n * t_"1/2"#

In your case, #t_"1/2" = "6.0 h"#, which means that

#t = 7.31 * "6.0 h" = "43.86 h"#

Rounded to two sig figs, the answer will be

#t = color(green)("44 h")#