What is the half life of the isotope?

AI Thread Summary
The discussion revolves around determining the half-life of an isotope given that its rate has decreased to one-eighth of its initial value in 18 days. The initial calculation suggested a half-life of 72 days, which was corrected to 6 days after applying the correct formula. The formula R(t) = R(0) * (2^(-t/th)) was used to clarify that three half-lives occurred in 18 days. The confusion stemmed from the interpretation of the phrase "dropped by one eighth." Ultimately, the consensus is that the half-life of the isotope is 6 days.
mike2007
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If the rate of an isotope in 18 days has dropped by one eight of its initial value. what is the half life of the isotope?

My answer
In 18 days the rate has dropped by 1/8 so therefore the half life is 4/8 which is 18*4 = 72days
 
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That's wrong. What's the formula for the rate as a function of initial rate, elapsed time and half life?
 
Or even if you haven't studied the formula.
After one half-life what is the rate, after two, after three ...?
 
T1/2 = ln(2)/lambda
Thats the formula i think will work.
 
1/2*1/2*1/2 = 1/8
so there will be three half lives after 18 days. So therefore one half life is 6 days?
 
Here's a better one:

R(t)=R(0)*(2^(-t/th)). Where R(t) is the rate at time t, and th is the half life time.
 
mike2007 said:
1/2*1/2*1/2 = 1/8
so there will be three half lives after 18 days. So therefore one half life is 6 days?

Yes, if you mean the final rate is 1/8 of the initial rate. I thought by saying 'dropped by one eighth' you meant that the final rate was 7/8 of the initial rate.
 
Yes that is what it means, sorry about the mistake.
Thank you very much for the clarification.
 

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