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?

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Answer 1 · 2014-05-29T02:04:40

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

The number of half-lives is $n = \frac{t}{t_{\frac{1}{2}}}$ $n = t/t_(1/2)$ , so $t = n t_{\frac{1}{2}}$ $t = nt_(1/2)$ .

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

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

$A = \frac{A_{0}}{2^{n}}$ $A = A_0/2^n$

You can rearrange this to

$\frac{A_{0}}{A} = 2^{n}$ $A_0/A = 2^n$

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

$\frac{1}{\frac{1}{4}} = 2^{n}$ $1/(1/4) = 2^n$

4 = $2^{n}$ $2^n$

$n$ $n$ = 2

$t = n t_{\frac{1}{2}}$ $t = nt_(1/2)$ = 2 × 18.72 days = 37.44 days