The initial-value problem dA/dt = kA, A(0) = A₀ is used to represent the radioactive decay. The radioactive substance having half-life T= -(ln 2)/k,will take 2.5 T for the substance to decay from A₀ to A₀ / 6 .
a.) To solve this, we have the following differential equation:
dA/dt = kA
With the initial condition A(0) = A₀
Rewriting the differential equation like this:
dA/A = kdt
And if we integrate both sides we get:
ln |A| = kt + c₁
Where is a constant. If we apply exponential for both sides we get:
Using the initial condition A(0) = A₀ we get ,
A₀ = C
So the solution for the differential equation is given by:
For the half life we know that we need to find the value of t for where we have,
A(t) = 1/2 (A₀)
Using this we get ,
Cancel A₀ on both sides and applying log on both sides we get ,
ln(1/2) = kt
t = {ln(1/2) / k } ----(1)
And using the fact that ln(1/2) = -ln(2) we get,
t = - { ln(2) / k }
b.) To solve this we consider ,
Replacing k with value obtained from 1 we get,
k = - {ln(2) / T}
Cancel the exponential with the natural log, we get,
c.)For this case we find the value of t when we have remaining A₀/6
So we can use the following equation:
A₀/6 =
Simplifying we got:
1/6 =
We can apply natural log on both sides and we got:
ln (1/6) = -t/T{ln(2)}
And if we solve for t we got:
t = T { ln(6) / ln(2) }
We can rewrite this expression like this:
t = T { ln(2²°⁵) / ln(2) }
Using properties of natural logs we got:
t = 2.5T { ln(2) / ln(2) }
t = 2.5T
Thus it will take 2.5 T for the substance to decay from A₀ to A₀/6
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