Activity Of A Radioactive Sample

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A radioactive sample initially registers an activity of 8000 disintegrations per second on a detector. If the isotope has a half-life of 15 days, what activity will the same sample show after 45 days have elapsed?

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The activity of a radioactive sample, equal to $\lambda N$, is directly proportional to the number of surviving nuclei and therefore decays by the same exponential law, halving every half-life. The elapsed time of 45 days spans $\dfrac{45}{15} = 3$ half-lives, so the activity falls by a factor of $\left(\dfrac{1}{2}ight)^3 = \dfrac{1}{8}$. Thus the final activity is $\dfrac{8000}{8} = 1000$ disintegrations per second. The value 2000 dps corresponds to only two half-lives (a factor of $1/4$) and undercounts the elapsed time. The value 4000 dps reflects just one half-life. The value 500 dps assumes four half-lives, overcounting by one. Activity is measured in becquerel, equal to one disintegration per second, or in the older unit the curie, and because it tracks the surviving population it can never fall faster than the nuclei themselves disappear, with each successive half-life removing half of whatever activity remained. This applies the standard NCERT rule that activity follows the same half-life decay as the nuclei count. A plausibility check confirms the progression $8000 	o 4000 	o 2000 	o 1000$ over three successive 15-day intervals, exactly reaching 1000 dps after 45 days.

Activity Of A Radioactive Sample

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