Derive radioactive decay equation
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A quantity is subject to exponential decay if it decreases at a rate proportional to its current value.
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From experimental observation we know that the decay rate is proportional to the number of radioactive nuclei (N) in a given sample.
That is, d N d t ∝ N ⇒ d N d t = − λ N dNdt∝N⇒dNdt=−λN
The negative sign indicates that it is a decay (i.e. decrease) rather than an increasing function. λ λ is the constant of proportionality and is called the decay constant. Now, the differential equation is first order and separable. I wouldn't imagine that you've solved any ODE's before so I'll give a fair amount of detail. First we divide by N, assuming of course that N ≠ 0 N≠0, which is a valid assumption since if N=0, there would be no radioactive nuclei and hence no decay, 1 N d N d t = − λ 1NdNdt=−λ Now we integrate with respect to t,
\int\frac{1}{N}{\frac{dN}{dt}dt = -\lambda\int dt Now, you can think of the next step as canceling the dt's on the RHS. This isn't rigorously correct, we actually make a change of variable, but it will do for illustrative purposes.
∫ d N N = − λ ∫ d t ∫dNN=−λ∫dt When we perform the integrals we obtain,
ln ( N ) = λ t + c o n s t .
ln(N)=λt+const.
We now exponentiate the equation,
N ( t ) = A e − λ t N(t)=Ae−λt
Where A is an arbitrary constant. We can fix this constant be saying that at t=0, we have a set number of radioactive nuclei, that is
N ( 0 ) = N 0 N(0)=N0.
Hence, we obtain the radioactive decay law,
N ( t ) = N 0 e − λ t .
Thanks.
That is, d N d t ∝ N ⇒ d N d t = − λ N dNdt∝N⇒dNdt=−λN
The negative sign indicates that it is a decay (i.e. decrease) rather than an increasing function. λ λ is the constant of proportionality and is called the decay constant. Now, the differential equation is first order and separable. I wouldn't imagine that you've solved any ODE's before so I'll give a fair amount of detail. First we divide by N, assuming of course that N ≠ 0 N≠0, which is a valid assumption since if N=0, there would be no radioactive nuclei and hence no decay, 1 N d N d t = − λ 1NdNdt=−λ Now we integrate with respect to t,
\int\frac{1}{N}{\frac{dN}{dt}dt = -\lambda\int dt Now, you can think of the next step as canceling the dt's on the RHS. This isn't rigorously correct, we actually make a change of variable, but it will do for illustrative purposes.
∫ d N N = − λ ∫ d t ∫dNN=−λ∫dt When we perform the integrals we obtain,
ln ( N ) = λ t + c o n s t .
ln(N)=λt+const.
We now exponentiate the equation,
N ( t ) = A e − λ t N(t)=Ae−λt
Where A is an arbitrary constant. We can fix this constant be saying that at t=0, we have a set number of radioactive nuclei, that is
N ( 0 ) = N 0 N(0)=N0.
Hence, we obtain the radioactive decay law,
N ( t ) = N 0 e − λ t .
Thanks.
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