What happens if t is greater than half-life?

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When t exceeds the half-life, the remaining quantity of a substance can be calculated using the equation A(t) = 0.125A0, indicating that only 12.5% of the original amount remains after three half-lives. The discussion highlights the challenge of determining the initial activity of a source when the half-life is significantly shorter than the counting time, as most activity may originate from daughter isotopes. The conversation also addresses a misunderstanding regarding the time variable, clarifying that the correct relationship is t = (t1/2)/3. Participants emphasize the importance of accurately interpreting time intervals in decay calculations. Overall, the complexities of radioactive decay and its implications for measuring initial activity are central to the discussion.
gabriellelee
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Homework Statement
What would be the difficulty in determining the activity of a source if the half-life is 1/3 shorter than the counting time,t?
Relevant Equations
A(t) = A0e^(-lambda*t)
I tried plugging into the equation t = 3t1/2 and I got A(t) = 0.125A0 since -lambda*t = - ln(2)*t/(t/3) = -ln(2)*3.
So I understand that there will be 12.5% left of the original value since three half-lives have passed.
But then I realized this won't help me answer the question much.
So I'm thinking maybe it's because since the half-life is too short compared to the counting time, the majority of the activity will come from the daughter so it will be difficult to determine the initial activity of the source.
But I'm not sure if my thought process is right. Any help on this would be very much appreciated!
 
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gabriellelee said:
I tried plugging into the equation t = 3t1/2
Why would you plug that in?
Do you understand what "1/3 shorter" means? What time is 1/3 shorter than 12 hours?
 
That was a typo. I meant t=(t1/2)/3
 
gabriellelee said:
That was a typo. I meant t=(t1/2)/3
Still wrong. Please answer my question in post #2.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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