Question

A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 days, what was the mass of the original sample?
Note: the correct answer is 880 g. I need someone to show me how to get this answer.

Answer 1

Given that the half-life of the sample is 14.28 days, you are looking for a decay factor $k$ such that

$\dfrac12=e^{14.28k}$

Solving for $k$ yields

$\dfrac12=e^{14.28k}$
$\ln\dfrac12=\lne^{14.28k}$
$-\ln2=14.28k$
$k=-\dfrac{\ln2}{14.28}\approx-0.0485$

Now, after 57 days, you're told that a sample of unknown mass decayed to 55g, which means if $M$ was the starting mass of the sample, then

$55=Me^{57k}$

Solving for $M$ yields

$M=\dfrac{55}{e^{57k}}\approx874.889\text{ g}$