2:52 AM
Hi sir. Good morning :-)
Question:
> 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,
$$\lambda_p N_p dt=\lambda_d N_d dt \\ \lambda_p N_p=\lambda_d N_d$$
The answer I obtained is correct.
2.
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?

2 hours later…
5:37 AM
@JohnRennie: Hi sir. Good morning :-)

6:02 AM
@GuruVishnu hi :-)

@JohnRennie Once you're done with Aladdin's doubt on CodeClub could you please have a look at the above block of messages and clarify my doubt?
May I know how long will it take?

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.

@JohnRennie I understand. Is that the reason equating activities give incorrect result? I'm unable to see how?

You write:
> Thus, their activities must be equal at all time.
which is not true.
Suppose we start with $N_d = 0$, then as time goes by $N_p$ falls and $N_d$ rises.

Yes sir.

6:11 AM
$$A_{p0}e^{-\lambda_p t}=A_{d0}e^{-\lambda_d t}$$
The parent and daughter populations will look like this (this is a random graph I grabbed from the Internet):

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.

That's the case in your question as well.

The solution of the equation 2 gives the time $t_m$ in this graph. Am I right?

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 ...

6:23 AM
The second curve is easy to get because $dN_d/dt = - dN_p/dt - \lambda_d N_d$
And $dN_p/dt = -\lambda_pN_p = -\lambda_p N_{p0}e^{-\lambda_p t}$

7:04 AM
@JohnRennie: Hi sir :-)

@GuruVishnu hi

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?

You mean plot $N_p(t)$ and $N_d(t)$?

Yes sir. And also their activities.
I know to plot them individually but for constructing them with these constraints, I'm unable to find any clues.

OK. I'm working for a few minutes but I'll get back to you as soon as I'm free.

7:11 AM
Ok sir. No problem.

7:33 AM
@GuruVishnu I'm free for about half an hour

8:06 AM
I need to work now for around an hour