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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.

Respuesta :

LammettHash
Given that the half-life of the sample is 14.28 days, you are looking for a decay factor [tex]k[/tex] such that

[tex]\dfrac12=e^{14.28k}[/tex]

Solving for [tex]k[/tex] yields

[tex]\dfrac12=e^{14.28k}[/tex]
[tex]\ln\dfrac12=\lne^{14.28k}[/tex]
[tex]-\ln2=14.28k[/tex]
[tex]k=-\dfrac{\ln2}{14.28}\approx-0.0485[/tex]

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

[tex]55=Me^{57k}[/tex]

Solving for [tex]M[/tex] yields

[tex]M=\dfrac{55}{e^{57k}}\approx874.889\text{ g}[/tex]

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