Exponential decays typically start with a differential equation of the form:
(dN)/dt prop -N(t)dNdt∝−N(t)
That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time tt. So we can introduce a proportionality constant:
(dN)/dt=-alphaN(t)dNdt=−αN(t)
We will now solve the equation to find a function of N(t)N(t):
->(dN)/N=-alphadt→dNN=−αdt
->int(dN)/N=int-alphadt-> ln(N)=-alphat+C→∫dNN=∫−αdt→ln(N)=−αt+C
->N(t)=Ae^(-alphat)→N(t)=Ae−αt where AA is a constant.
This is the general form of the exponential decay formula and will typically have graphs that look like this:
graph{e^-x [-1.465, 3.9, -0.902, 1.782]}
Perhaps an example might help?
Consider a lump of plutonium 239 which initially has 10^241024 atoms. After one million years have elapsed years the plutonium now has 2.865times10^112.865×1011 atoms left. Work out, AA and alpha.α. When will the plutonium have only 5times10^85×108 atoms left and what is the decay rate here?
We are told the lump has 10^241024 atoms at t=0t=0 so:
N(0)=Ae^(0)=10^24-> A=10^24N(0)=Ae0=1024→A=1024
Now at 1 million years: 10^6106 years:
N(10^6) = 10^24e^(-alpha(10^6))=2.865times10^11N(106)=1024e−α(106)=2.865×1011
Rearrange to get:
alpha=-1/(10^6)ln((2.865times10^11)/10^24)~~2.888times10^(-5)yr^-1α=−1106ln(2.865×10111024)≈2.888×10−5yr−1
So N(t)=10^(24)e^(-2.888times10^(-5)t)N(t)=1024e−2.888×10−5t
For the next part:
N(t) =5times10^8=10^(24)e^(-2.888times10^(-5)t)N(t)=5×108=1024e−2.888×10−5t
Rearrange to get tt:
t=-1/(2.888times10^(-5))ln((5times10^8)/(10^24))~~1.22times10^6yrt=−12.888×10−5ln(5×1081024)≈1.22×106yr
Now for the last part, the decay rate is already defined a way back at the very start, simply evaluate it at the given time:
(dN)/dt=-alphat=-2.888times10^(-5)(1.22times10^6)dNdt=−αt=−2.888×10−5(1.22×106)
=-35.23=−35.23 atoms per year.
The idea is to start with differential equation above, which gives the decay rate, and solve it to get the population at any given time.
