# What happens if t is greater than half-life?

gabriellelee
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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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?

gabriellelee
That was a typo. I meant t=(t1/2)/3