B Calculating activity of a radionuclide

[Andrew1949]

Activity (in Bq) of 1 mol of a radionuclide is given by formula:
A = N_A x ln(2) / t_1/2 = 0.693 x N_A / t_1/2
where N_A is Avogadro number and t_1/2 the half-life (in seconds)

Why don't we use simply A = 0.5 x N_A / t_1/2 ?

After all, t_1/2 means that, after that time, half of atoms will have decayed...

 

[phyzguy]

If I have N0 atoms in the sample, with a decay rate of λ, the number of atoms in the sample will decay as follows:
N = N_0 e^{-\lambda t}
The activity is the number of decays per second, which is given by:
A = -\frac{dN}{dt} = \lambda N_0 e^{-\lambda t} = \lambda N
The half-life is the time when 1/2 of the atoms have decayed, which from the first equation is given by:
\frac{N}{N_0} = 1/2 = e^{-\lambda t_{1/2}} ;\,\,\, t_{1/2} = \frac{\log(2)}{\lambda}
So the activity is given by A = \frac{\log(2)}{t_{1/2}}N. If we used "1/e-life" instead of "half-life", we wouldn't have this complication.

 

[Andrew1949]

So, calculating A = 0.5 x N_A / t_1/2 would give us the mean activity from t = 0 (present time) up to t = t_1/2 ..., because during the length time t_1/2, exactly 0.5 x N_A atoms decay.

But calculating A = N_A x ln(2) / t_1/2 will give the instant activity, when the number of atoms involved is the present number N_A.

In that case, OK, I understand the difference.

 

[phyzguy]

Yes, what you said is correct.