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


`\frac12=e^(14.28k)`

Solving for
`k` yields


`\frac12=e^(14.28k)`

`\ln\frac12=\lne^(14.28k)`

`-\ln2=14.28k`

`k=-(\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=(55)/(e^(57k))\approx874.889\text{ g}`