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Answer :
Answer:
3 half-lives; 3.125 mCi
Step-by-step explanation:
A half-life of carbon-11 is 20min, so there are 60÷20=3half-lives in 1 hour.Let x=the number of half-livesin 1 hour, 3.Let y=the sample size after 1hour.Let a=the starting sample size,25 mCi.Let b=the decay factor orhalf-life, 0.5.y ====a•bx25•0.5325•0 . 1253 . 125 A half life of carbon 11 is 20 minutes, so there are 60 divided by 20 equals 3 half lives in 1 hour. Let x equal the number of half lives in 1 hour, 3. Let y equal the sample size after 1 hour. Let a equal the starting sample size, 25 millicuries. Let b equal the decay factor or half life, 0.5. Y equals a multiplied by b raised to the power of x. Equals 25 multiplied by 0.5 cubed. Equals 25 multiplied by 0.125. Equals 3.125.
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Answer:
The answer to your question is:
a) 4 half-life
b) 2.5 mCi
Step-by-step explanation:
Half-life of iodine 124 is 4 days
# of half- lives in 16 days = ?
Process
0 days ----------------- 0 half-life
4 days ------------------ 1 half life
8 days ------------------ 2 halves lives
12 days ----------------- 3 halves lives
16 days ----------------- 4 halves lives
b) 40 mCi ---------------- initially
20 mCi ---------------- 1 half life
10 mCi ---------------- 2 half life
5 mCi ---------------- 3 half life
2.5 mCi ---------------- 4 half life
