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I got an exercise but I don't know the exact answer.

In the catalog note of the reference source of Polonium-210 radiation with a half-life of 138 days, it was stated that one day the activity was $\pu{2 \times 10^6 MBq}$. Calculate activity after 558 days.

After my calculations the final answer was ${126468.14}$, but my professor has told me that my answer should be about 1,7 times higher.

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    $\begingroup$ The number of significant figures in your answer should be cut in half. (I meant all figures, not just those after the dot.) As for the answer itself, well, now that's strange: mine is slightly different from yours. $\endgroup$
    – Ivan Neretin
    Commented Feb 17, 2021 at 22:38
  • $\begingroup$ Actually, I got only specified data from my professor, I used the half-life formula to calculate the final answer. $\endgroup$
    – zajkevv
    Commented Feb 17, 2021 at 22:46
  • $\begingroup$ 126468.14 +/- 30000 would not look so great, would it ? $\endgroup$
    – Poutnik
    Commented Feb 17, 2021 at 22:58
  • $\begingroup$ Why would there be as much as 30000 units approximation? $\endgroup$
    – zajkevv
    Commented Feb 17, 2021 at 23:17
  • $\begingroup$ Well, your answer and would be the answer are both wrong by the fact that after 552 days, the activity should be $\frac{2 \times \pu{10^6 MBq}}{16} = \pu{1.25 \times 10^5 MBq}$. Thus, answer should be lower than this number (disregard sig numbers). $\endgroup$
    – Mathew Mahindaratne
    Commented Feb 17, 2021 at 23:25

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Well, first of all, I should point out that your answer and the would be the answer ($1.7$ times your answer) are both wrong by the fact that after 552 days (4 half-lives), the activity should be $\displaystyle\frac{\pu{2E6 MBq}}{16} = \pu{1.25E5 MBq}.$ Since this is smaller than both your answer and the would be answer, they both would be incorrect. Accordingly, correct answer would be smaller than $\pu{1.25E5 MBq},$ because the sample would be further decays for another six days. Let’s do the proper calculations:

The half-life $\left(t_\frac{1}{2}\right)$ of a radioactive element is defined as

$$t_\frac{1}{2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}.\label{eqn:1}\tag{1}$$

The $t_\frac{1}{2} = \pu{138 d}$ in the equation \eqref{eqn:1} for your case. Thus from the equation \eqref{eqn:1}, you can calculate for $\lambda:$

$$\lambda = \frac{\ln 2}{t_\frac{1}{2}} = \frac{0.693}{\pu{138 d}} = \pu{5.02E-3 d-1}. \tag{2}$$

Also, the radioactive decay is according to the equation:

$$A_t = A_0\mathrm e^{-\lambda t}.\label{eqn:3}\tag{3}$$

where $A_0$ is initial activity and $A_t$ is the activity of the same same sample after time $t$ (Note: $t_\frac{1}{2}$ and $t$ should have same units). Thus, applying the $\lambda$ value on the equation \eqref{eqn:3}:

$$A_t = \pu{2E6 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{558 d}\right) = \pu{1.21E5 MBq}.$$

Note: I disregard the correct significant figures because I believe the initial activity should be given in three significant figures.

For fun, you can also checked the validity of the answer using my initial remark. You can use the activity of the sample after four consecutive $t_\frac{1}{2}$s (after $138 \times 4 = \pu{552 d})$ as its initial activity and find the activity after another six days using the equation \eqref{eqn:3}:

$$A_t = \pu{1.25E5 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{6 d}\right) = \pu{1.21E5 MBq}.$$

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