Asked by Victoria
A mosquito fed on a solution containing phosphorous-32 is released. Phosphorous-32 has a half-life of 14 days. When the mosquito is recaptured 28 days later, what percentage of the original Phosphorus-32 will remain?
Answers
Answered by
DrBob222
28 days is 2 half lives.
Start P-32 = 100%
1st half life = 1/2 that or 50% left
2nd half life = 1/2 that or 25% left.
Start P-32 = 100%
1st half life = 1/2 that or 50% left
2nd half life = 1/2 that or 25% left.
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