Is there an explanation for the unexpected increase in activity of Nuclide A?

  • Thread starter Thread starter songoku
  • Start date Start date
  • Tags Tags
    Activity
AI Thread Summary
The discussion revolves around the unexpected increase in the activity of Nuclide A, which initially appears to contradict the principles of radioactive decay. The decay constant for Nuclide A is calculated, leading to an initial activity of 2.84 Bq, yet the activity later rises to 10,000 Bq. Participants suggest potential errors in the problem, such as incorrect half-lives or an underestimated initial number of nuclei. They introduce the concept of "Transient Equilibrium," where the decay of a parent nuclide leads to an increase in the activity of its daughter nuclide. The thread emphasizes the need for clarity in the problem's parameters to resolve the apparent contradiction.
songoku
Messages
2,475
Reaction score
389
Homework Statement
Nuclide A decays to nuclide B. Initially, there are ##1.29 \times10^9## nuclei of A and after some time the activity of A is 10 000 Bq. If the half life of A and B is 10 years and 10 hours respectively, find activity of B
Relevant Equations
##A=\lambda N##

##t_{\frac 1 2}=\frac{ln~2}{\lambda}##
I found something I think does not make sense.

Decay constant of A:
$$\lambda_{A}=\frac{ln~2}{t_{\frac 1 2}}$$
$$=\frac{ln~2}{10\times 365 \times 24 \times 3600}$$
$$=2.2\times 10^{-9} / s$$

Initial activity of A = ##\lambda_{A} N_{\text{initial}}## = 2.2 x 10-9 x 1.29 x 109 = 2.84 Bq

Then after some time the activity becomes 10 000 Bq. How can the activity increase instead of decrease?

Is there something wrong with the question or something wrong with me?

Thanks
 
Physics news on Phys.org
songoku said:
Is there something wrong with the question or something wrong with me?
There's nothing wrong with you! The question appears to have one or more mistakes.

Some possibilities are:
- the half-lives of A and B are the wrong way round;
- ##1.29 \times10^9## is a very small number for a number of nuclei in this context; maybe it should be ##1.29 \times10^{19}## for example.
 
You could at least obtain the general form of the solution.
 
Thank you very much Steve4Physics and haruspex
 
songoku said:
Homework Statement:: Nuclide A decays to nuclide B. Initially, there are ##1.29 \times10^9## nuclei of A and after some time the activity of A is 10 000 Bq. If the half life of A and B is 10 years and 10 hours respectively, find activity of B
Relevant Equations:: ##A=\lambda N##

##t_{\frac 1 2}=\frac{ln~2}{\lambda}##

Then after some time the activity becomes 10 000 Bq. How can the activity increase instead of decrease?
One should research "Transient Equilibrium", where t1/2(parent) > t1/2(daughter), or λ(parent) < λ(daughter). This is observed for the natural decay series of radionuclides 232Th, 235U, 238U and others.

https://en.wikipedia.org/wiki/Transient_equilibrium
Also see related Secular equilibrium.
 
Astronuc said:
One should research "Transient Equilibrium", where t1/2(parent) > t1/2(daughter), or λ(parent) < λ(daughter). This is observed for the natural decay series of radionuclides 232Th, 235U, 238U and others.

https://en.wikipedia.org/wiki/Transient_equilibrium
Also see related Secular equilibrium.
At this stage it is not clear from @songoku's posts whether he has any difficulty in solving a correctly posed version of the question. The thread centres on the impossible combination of given facts.
 
  • Like
Likes songoku and Orodruin
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top