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We have been learning a lot on the topic, and my professor introduced a couple of formulas that can help me evaluate specific activity:

$$a = \frac{A}{m};$$ $$a=\frac{\lambda}{m}N_0e^{-\lambda t}$$

Knowing this, it is obvious that $a$ is not constant and that it changes exponentially, just like radioactive activity $A$. However, this formula is in use as well:

$$a = \frac{\lambda N_A}{M};$$ With $M$ being the molar mass of an element

The formula can be derived by setting: $m = \frac{N}{N_A}M$

What seems illogical to me is the fact that both $\lambda$ and $M$ are constant values, which means that $a$ is constant as well... I wasn't able to find an answer online and this isn't really talked about so I am probably wrong. Of course, I would very much appreciate and explanation. Thanks in advance!

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  • $\begingroup$ If you take a specific example, e.g. a lump of $^{238}U$, then the radioactivity will decrease as it decays but so will the mass of $^{238}U$. So, it is reasonable that $A/m$ will be constant. This is just a comment since I am guessing here. $\endgroup$
    – badjohn
    Commented Apr 8, 2018 at 16:28
  • $\begingroup$ Thank you. I think that is reasonable as well, but the first two formulas show that specific activity becomes smaller exponentially so I don't know what to think... $\endgroup$
    – Luka Duranovic
    Commented Apr 8, 2018 at 16:40
  • $\begingroup$ The activity is proportional to the number of undecayed unstable nuclei which can be written as $A=-\lambda N$ $\endgroup$
    – Farcher
    Commented Apr 8, 2018 at 17:59
  • $\begingroup$ @Farcher That's what I guessed. So, if you regard both $A$ and $m$ as functions of $t$ and $m$ is the mass of the undecayed original atom, $^{238}U$ in my example, then $A/m$ would be constant. However, this will not be the mass of the sample since various daughter products, e.g. $^{206}Pb$, will be present. $\endgroup$
    – badjohn
    Commented Apr 8, 2018 at 18:09
  • $\begingroup$ Also, Farcher's observation, with the same assumption on $m$, explains why the second formula is also constant. Since $N_0e^{-\lambda t}$ is the number of undecayed atoms of the original substance. $\endgroup$
    – badjohn
    Commented Apr 8, 2018 at 18:13

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I think it is easier to understand if you write the variables as functions of time. Activity is defined as $$A(t)=\lambda N(t) =\lambda \frac{m(t) N_a}{M}$$ where $N(t)$ is the number density as a function of time, $\lambda$ is the decay constant, $m(t)$ is the mass as a function of time, $N_a$ is Avogodro's number, and $M$ is the molar mass.

Specific activity is defined as: $$a=\frac{A(t)}{m(t)}$$

Since both $A(t)$ and $m(t)$ decay at the same rate, $a$ is constant in time.

It might be easier to visualize if you write the equation using the solution for a single radioactive isotope with a constant value and no source $$a=\frac{A(t)}{m(t)}=\frac{A_0 e^{\lambda t}}{m_0 e^{\lambda t}}$$

Note, however, that the same results still holds up if the time dependence is more complicated.

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