Answered

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.

Verified Answer :

LammettHash

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

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

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