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

Jan 6, 2018

$10$ mg

#### Explanation:

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

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

$\left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = {\left(\frac{1}{2}\right)}^{3} = \frac{1}{8}$

$\frac{1}{8} \cdot 80$mg = $10$ mg

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

Jan 6, 2018

Consider the radioactive decay to be a first order reaction,

$\ln {\left[A\right]}_{t} = - k t + \ln {\left[A\right]}_{0} \text{ } \left(1\right)$

$\implies \frac{\ln {\left[A\right]}_{0}}{\ln {\left[A\right]}_{t}} / k = t \text{ } \left(2\right)$

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

${t}_{\frac{1}{2}} = \frac{0.693}{k}$

Hence,

$8 d = \frac{0.693}{k}$
$\therefore k \approx 8.66 \cdot {10}^{-} 2 {d}^{-} 1$

, and

$\ln \left(\frac{{\left[A\right]}_{t}}{80 m g}\right) = - 8.66 \cdot {10}^{-} 2 {d}^{-} 1 \cdot 24 d$

$\therefore \ln {\left[A\right]}_{t} \approx 10 \text{mg}$ where $t = 24 d$