0
$\begingroup$

I am trying to understand how do I get approximation for the activity for a radioactive nuclide that produced during a chain while the activity of the source is given at time $t_0$. For example lets take Sr90 wich decay to yittirium90 through $\beta ^-$ decay and the yittirium90 decay to zirconium-90 through a $\beta ^-$ decay.

Lets say that the activity of sr90 at its time production $t_0$ is given as $B(t_0)[Bq]$ Now, if I want to find the actvity of sr90 at time $t$ is just equal to $$B(t)= \frac{N(t)}{N_A}\cdot M $$ while M is the molar mass of sr90, $N_A $ is avogadro's number and $N(t)$ is the number of sr90 atoms in the material at time $t$, which given by $N(t)=N_0 \cdot e^{-\lambda t}$, $\quad\lambda $ is the probability per unit time of a single sr90 atoms to decay and can be taken from internet.

The above is clear for me, what I would want to know now is how do I get an estimate for the activity of yittirium90 at time $t$.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Growth & decay of radioactivity page 129 -> $\endgroup$
    – Farcher
    Sep 14, 2022 at 21:52
  • $\begingroup$ Thank you very much $\endgroup$
    – Sagigever
    Sep 15, 2022 at 6:23
  • $\begingroup$ The multiple decay equation is called the Bateman equation $\endgroup$
    – Farcher
    Sep 15, 2022 at 6:53
  • $\begingroup$ @Farcher I am getting in my calculations that sometimes the activity of one of the daughters radioactive nuclide is higher than the parent radionuclide activity. I understand that there is a relation to the half life time but Im not sure I can understand how is it possible intuitively. Do you have any nice explanation? $\endgroup$
    – Sagigever
    Sep 15, 2022 at 7:59
  • 1
    $\begingroup$ Imagine short lived parent atoms which decay into long lived daughter atoms. After a time the short lived parent atoms all but die out leaving the long lived parent atoms still decaying. $\endgroup$
    – Farcher
    Sep 15, 2022 at 8:05

1 Answer 1

Reset to default
0
$\begingroup$

In the limit where one half life is much shorter than the other, you can ignore that isotope completely, and approximate the decay chain with being just one single decay of the isotope with the longest half life.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.