So, I am pretty new to that concept, which i am learning about in the context of physical chemistry. I am a mathematician, so a physicists approach can be confusing at times. I have two questions :

In my reference book, it is said that activity $$A$$ is the number of decays per second $$A = -\frac{dN}{dt} \tag{1}$$ where $$N$$ is the number of particles. It is also said that $$A = kN. \tag{2}$$ Yet it is said that $$(2)$$ is always true, whereas $$(1)$$ isn't.
But, $$(1)$$ is the mathematical definition of number of decays per second. So isn't it always true, if you look at the activity of one particular particle and not for example a system ?

And now, this is what confuses me the most. For some exercise, I am supposed to look at a Room, where some decay is happening (222Rn) and you want it's activity to remain constant, by adding some amount of the decaying particle (Rn) in the Room at the end of each day. So, mathematically speaking, the decay is exponential, so although it is true that asymptotically, it can be said to be constant, every time you add some amount, you ruin that behaviour.
So is it meant that the average over the whole day is supposed to be constant ?

• physicist consider a single radon atom (or nucleus). It has a fixed rate of decay, which is an infinitesimal probability per infinitesimal time period that never changes nor is influenced by anything else, and then construct an ensemble of many of them which can then be treated statistically.
– JEB
Oct 14, 2019 at 18:11
• @JEB is your name a reference to one of the Kerman's ? On a more serious note. I get that $N(t) = N(t_0)exp(-k(t-t_0))$ for that exact reason. And also that $N(t_0)$ is what the value number of particles you know are there at $t_0$ . But how on Earth are you supposed, by adding once a day some Radon quantity, get a constant rate ? If you add what you have lost during your day, then on the span of the whole day, i believe the average activity would indeed be constant. Oct 14, 2019 at 18:16