The half life of a specific radionuclide is 8 days. How much of an 80 mg sample will be left after 24 days?

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Answer 1 · 2018-01-06T22:45:09

#10# mg

#8# days #=1# half-life
#24# days #=3# half-lives

in #3# half-lives, the mass of the radionuclide will be halved #3# times.

#(1/2) * (1/2) * (1/2) = (1/2)^3 = 1/8#

#1/8 * 80#mg = #10# mg

after #3# half-lives, #10#mg will be left.

Answer 2 · 2018-01-06T23:20:42

Consider the radioactive decay to be a first order reaction,

#ln[A]_t = -kt + ln[A]_0 " " (1)#

#=> (ln[A]_0)/(ln[A]_t)/k = t " "(2)#

Moreover, consider we want half of a substance at one unit of concentration at #t=0#,

#t_(1/2) = 0.693/k#

Hence,

#8d = (0.693)/k#
#therefore k approx 8.66*10^-2d^-1#

, and

#ln(([A]_t)/(80mg)) = -8.66 * 10^-2d^-1 * 24d#

#therefore ln[A]_(t) approx 10"mg"# where #t = 24d#