A first order exponential decay can be written as
A(t)= Ae^(-t/r)
where A(t) is the amount (of substance) after time t, A is the initial amount at time t=0 and r is the decay time.
The fall time is defined as the time in which A(t) falls from 90% to 10% of its initial value. Find the relationship between the fall time of an exponential decay and r.
I will start it off
.9=e^t1/r
ln .9 = t1/r
likewise..
ln .1=t2/r
tfall=t2-t1, well you do that...
To find the relationship between the fall time of an exponential decay and the decay time (r), we need to determine how to express the fall time in terms of r.
Let's consider the amount (A) at time t during exponential decay. We can express it as:
A(t) = A * e^(-t/r)
Now, we want to find the time it takes for A(t) to fall from 90% to 10% of its initial value.
When A(t) falls to 90% of its initial value, it can be written as:
0.9A = A * e^(-t/r)
Similarly, when A(t) falls to 10% of its initial value, it can be written as:
0.1A = A * e^(-t/r)
To find the fall time, we need to solve for t, the time at which the above equations hold.
To do this, divide the equation for 0.9A by the equation for 0.1A:
(0.9A) / (0.1A) = (A * e^(-t/r)) / (A * e^(-t/r))
Simplifying, we get:
9 = e^(-t/r) / e^(-t/r)
Since exponential functions with the same base cancel each other out, we have:
9 = 1
This implies that the equation (0.9A) / (0.1A) = (A * e^(-t/r)) / (A * e^(-t/r)) is inconsistent and has no valid solutions.
Therefore, it can be concluded that the fall time does not depend on the decay time (r) in this specific case of an exponential decay. The fall time is independent of the initial decay rate and remains constant.