SOLUTION: 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? could

Question 553517: 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? could someone please help?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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?
:
Using the radioactive decay formula: A = Ao*2^(-t/h), where
A = resulting amt after t time
Ao = initial amt (t=0)
t = time
h = half-life of substance
:
We want to find Ao:
Ao*2^(-57/14.28) = 55
using a calc to find 2^(-57/14.28)
.062865Ao = 55
Ao =
Ao = 875 grams was the initial amt
:
:
We can check this solution on our calculator; enter: 875*2^(-57/14.28) results: 55.00 grams

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