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 $t = 24d$