Problem
A detector measures the activity of a radioactive sample at regular intervals, obtaining the table below.
(hours) (kBq)
- (a) Verify that the data follow an exponential decay law.
- (b) Determine the half-life and identify the isotope between technetium-99m () and iodine-123 ().
Solution
(a) Verifying the exponential trend. In an exponential decay, the ratio between activities at equal time intervals is constant. Let us calculate the ratio every hours:
The ratio is constant ( every hours): this confirms the exponential law .
(b) Half-life. From a factor of every hours we obtain by imposing , i.e.:
Check against the last data point. At with : , in perfect agreement with the table.
Identification. Of the two options proposed, the measured value is closer to technetium-99m () than to iodine-123 (). The isotope is therefore technetium-99m.
Note on the data
The values in the table strictly give , not as stated in the textbook’s answer key. The numerical data are therefore slightly inconsistent with the nominal value for technetium-99m (): although technetium-99m remains the closer isotope of the two options, the half-life extracted from the data does not exactly match the tabulated value.
Collegamenti
Argomenti: Fisica nucleare Concetti: Decadimento radioattivo Competenze: Lettura dei grafici · Uso delle scale logaritmiche