> Suppose, the daughter nucleus in a nuclear decay is itself radioactive. Let $\lambda_p$ and $\lambda_d$ be the decay constants of the parent and the daughter nuclei. Also, let $N_p$ and $N_d$ be the number of parent and daughter nuclei at time $t$. Find the condition for which the number of daughter nuclei becomes constant.
My approaches:
1.
We know that $dN=N\lambda dt$ where $dN$ is the number of nuclei which disintegrate in time $dt$ out of $N$ nuclei. We are asked to find the condition under which the number of daughter nuclei becomes constant. This is possible only when it's rate of formation is equal to the rate of disintegration. Hence,
Activity of a radioactive substance denotes the number of disintegrations per second. So for the number of daughter nuclei to become constant, the rate at which parent nuclei disintegrate must be equal to the rate at which daughter nuclei disintegrate. Thus, their activities must be equal at all time. Hence we have,
$$A_{p0}e^{-\lambda_p t}=A_{d0}e^{-\lambda_d t}$$
where $A_{p0}$ and $A_{d0}$ are the initial activities of the product and daughter nuclei respectively.
I stopped proceeding. There is an exponential term which is dependant on time and doesn't cancel on both sides.
So, what is the reason for this method (equating activities) to turn incorrect?
The number of daughter nuclei can never become constant as it will always change with time. What you've found is the time at which $dN_d/dt = 0$ i.e. you've found the maximum in the curve of $N_d(t)$ against time.
I see. I suspect even in this graph, both parent and daughter nuclei are radioactive as the product nuclei's count decreases after attaining a maximum.
Yes. Though note that this curve is unusual in that the concentrations are equal at the point where $N_d$ is a maximum. I don't think that is always the case.
Actually i should probably check that - maybe it is the case ...
Now, I understood why my second method failed. If possible, could you tell how the graph of concentration as well as activity look for the first case? Any ideas on how to plot this in a graphing calculator, sir?
The rate of decay of $N$ is $\lambda N$, and at the start $N=0$ so the decay rate is zero. That means at the start the total rate of change of $N$ is just the (constant) production rate $R$.
@JohnRennie Now I can see if we put $t>>t_{1/2}$ the equation turns out to be the final answer. Could you tell what this corresponds to intuitively, sir?
The power of $e$ depends on the negative of $t/t_{1/2}$ and hence as we approach infinity, the dependence on time vanishes to zero. And hence $t>>t_{1/2}$.
@JohnRennie I don't think we can ascertain a particular value of time after which we would observe constancy. Just the difference reduces as time progresses until it becomes zero.
Your approach just short cut the full derivation and jumped straight to the final result i.e. that the rate of production had to be equal to the rate of decay. That's a perfectly reasonable way to do it.
Although you had to do it in full to understand why times much greater than the half life are required.