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?

@JohnRennie: Hi sir. Good morning :-)

@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.

That contradicts your equation:

$$ A_{p0}e^{-\lambda_p t}=A_{d0}e^{-\lambda_d t} $$

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.

Yes, both are radioactive.

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?

Actually i should probably check that - maybe it is the case ...

Ok sir. I'll think about this for some time.

The second curve is easy to get because $dN_d/dt = - dN_p/dt - \lambda_d N_d$

And $dN_p/dt = -\lambda_pN_p = -\lambda_p N_{p0}e^{-\lambda_p t}$

@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.

Ok sir. No problem.

@GuruVishnu I'm free for about half an hour

I need to work now for around an hour

Question:

I solved it by using $dN=Rdt$ and $dn=\ln2 N dt/t_{1/2}$ and obtained $$N=\frac{Rt_{1/2}}{\ln2}$$

The answer is correct.

But, what is the significance of $t>>t_{1/2}$ here?

@JohnRennie: Hi. Could you ping me once you're done with your discussion in two other rooms, sir? Thanks!

@GuruVishnu hi

@JohnRennie hi

What happens here is that we start with $N=0$, then the nucleide starts being produced at a constant rate $dN/dt = R$ so $N$ increases with time. Yes?

Yes 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 $R$.

Yes sir.

But as time goes by, and $N$ increases, that means the decay rate $\lambda N$ also increases.

So the rate of change of $N$ will be:

$$ \frac{dN}{dt} = R - \lambda N $$

Yes sir.

This is going to end up giving us an equation for $N$ that is something like:

$$ N(t) = \frac{R}{\lambda}\left( 1 - e^{-\lambda t} \right) $$

(I kind of guessed that so I won't guarantee the equation is correct, but it will have a form something like that).

OK so far?

Ok sir. I think we need to solve the previous differential equation under proper limits. Is that how you got the final equation?

Yes.

@JohnRennie Ok sir.

But I already know what the solution should look like so it was an informed guess.

:-)

You could of course just crunch through the algebra. It's actually very straightforward.

Ok sir.

$$ \int \frac{dN}{R -\lambda N} = \int dt $$

That's easy sir. I know. I just asked how you solved it so quickly.

And to determine the constant of integration use $N(0) = 0$

Yes sir.

Anyhow, the question is asking for the value of $N$ once it has settled to a constant value. Yes?

@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?

@JohnRennie Yes sir.

So at what time $t$ does the value of $N$ become constant?

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.

Well, $N$ becomes constant as $t \to \infty$

Yes sir.

Which is somewhat same as $t>>t_{1/2}$

And that's why your question specifies times much greater than the half life.

Actually, we should rewrite the equation terms of the half life to make this clear.

@GuruVishnu Shall I do this or will you do it?

@JohnRennie That's easy sir. I can understand it. However, I have a different doubt.

Yes?

Could you tell how my method worked? Was it a luck? It was so short and I think I missed some details which we covered now.

I just used two equations I learnt in this chapter and equated them as the rate of formation must match the rate of disintegration.

I started my analysis by writing the differential equation:

$$ \frac{dN}{dt} = R - \lambda N $$

Yes sir. I can see the two terms in your equation are the individual components of the two equations I used.

And if the value of $N$ is constant that means $dN/dt = 0$ and therefore $R = \lambda N$. Yes?

@JohnRennie Ultra Cool! Got it :-)

Thank you sir.

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.

Yes sir. It was short. But only after seeing your approach I was able to understand the significance of $t>>t_{1/2}$

@JohnRennie Yes sir :-)

Please ignore this: HCV-46-Ex-38 and also 39 [^^^]