# How do you calculate the number of days required for 3/4 of a given amount of nuclide to decay if the half-life is 18 point 72 days?

May 29, 2014

You calculate the number of half-lives and multiply by the length of one half-life.

The number of half-lives is $n = \frac{t}{t} _ \left(\frac{1}{2}\right)$, so $t = n {t}_{\frac{1}{2}}$.

For each half-life, you divide the total amount of the isotope by 2, so

Amount remaining = $\frac{\text{original amount}}{2} ^ n$ or

$A = {A}_{0} / {2}^{n}$

You can rearrange this to

${A}_{0} / A = {2}^{n}$

If original amount was 1, and $\frac{3}{4}$ of the nuclide decayed, then $\frac{1}{4}$ of the nuclide remains undecayed.

$\frac{1}{\frac{1}{4}} = {2}^{n}$

4 = ${2}^{n}$

$n$ = 2

$t = n {t}_{\frac{1}{2}}$ = 2 × 18.72 days = 37.44 days