@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

I know you're busy in a different room. May I ask the question now and could you please reply after some time?

Question:

> The half life of a radioisotope is 10 hours. Find the total number of disintegrations in the tenth hour measured from a time when the activity was 1 Ci.

My approach:

The activity after 10 hours is half of the initial activity i.e., 0.5 Ci.

I considered the activity to be constant in one full hour. And so the number of disintegrations in this period is $0.5\times3.7\times10^{10}\times3600=6.66\times10^{13}$

There is a far easier way to do this ...

But the answer according to the book is $6.91\times10^{13}$.

I don't understand how the value could be higher than my answer as activity must clearly decrease with time.

The tenth hour is from t = 9 hours to t = 10 hours. So you're actually considering the eleventh hour.

Remember that the first hour is from t=0 to t=1 hour

Ah. Ok sir. Let me try my method with this change in mind.

@JohnRennie If possible, could you tell the easier way? Mine involves using a calculator to find difficult powers of $e$ which is not allowed in exams.

The number of atoms at a time $t$ is given by:

$$ N(t) = N_0 e^{-t/\tau} $$

Where $\tau$ is the time constant for the decay (NB $\tau$ is not the half life).

OK so far?

@JohnRennie Is $\tau$ the average time? I haven't learnt about that notation.

Because I've seen, $\lambda=1/t_{avg}$

To find the half life just set $N(t)/N_0 = \tfrac{1}{2}$

Then we get:

$$ \tfrac12 = e^{-t/\tau} $$

Ok sir.

$$ -\ln 2 = -t/\tau $$

$$ t_{1/2} = \tau\ln2 $$

@JohnRennie Yes sir. Understood till this.

So we can replace $\tau$ with $t_{1/2}/\ln2$ to get:

$$ N(t) = N_0 \exp(-t\ln2/t_{1/2}) $$

And we are told that $t_{1/2} = 10$ hours.

Yes sir.

The number of disintegrations in the tenth hour is $N_{10} = N(9) - N(10)$. So now we have an exact equation for the result we need.

But I have to admit I don't see a way to do this without a calculator.

@JohnRennie Ok sir.

@JohnRennie I think this form is more accurate than my method.

Because we also take the non-constancy of the activity in this window of time.

Thank you sir.

:-)

May I know how did you think of this method instead of the method I chose within a short period of time? I think it might help me make good decisions on which path to choose under a given circumstance.

Once you've learned radioactive decay you know it's an exponential decay i.e. has the form $N(t) = N_o e^{-t/\tau}$ for some constant $\tau$.

Yes sir. I already knew it. Even my approach is based on the formulae I learnt in this chapter.

You can get this by solving the differential equation for the rate of decay if you want, but I just remember it.

OK, in that case it seems obvious that each hour $N(t)$ decreases by the number of decays that occur in that hour.

Yes sir.

Thank you sir. Got the correct answer. But I used calculator to find $e^{-0.9\ln 2}$.

Yes, I assume you wouldn't be asked a question like this in the JEE unless the numbers were chosen to make the arithmetic easy.

Yes sir. I too think so.