Physics · nuclei · NEET
Activity A = lambda N, where N is the number of undecayed nuclei present. As nuclei decay, N keeps falling, so there are fewer nuclei left to decay each second. Since lambda (the decay constant) is fixed, A must fall in step with N. That is why an old radioactive source clicks slower on a Geiger counter than a fresh one.
Start from N = N0 e^(-lambda t) (the decay law). Multiply both sides by lambda: lambda N = lambda N0 e^(-lambda t). Since activity A = lambda N and initial activity A0 = lambda N0, this gives A = A0 e^(-lambda t). So activity obeys the exact same exponential curve as the number of nuclei.
Yes. Both N and A contain the same factor e^(-lambda t), so both halve after the same time T = 0.693 / lambda. When N drops to N0/2, A also drops to A0/2 at the same instant. There is only one half-life for a given nuclide.
Use A = A0 / 2^n where n is the number of half-lives. After 3 half-lives n = 3, so A = A0 / 2^3 = A0 / 8. That is one-eighth, or 12.5 percent, of the starting activity. This 1/2^n trick is faster than the exponential for whole-number half-lives.
Number of nuclei N is how many undecayed atoms are still present. Activity A is how many decay per second, A = lambda N. N is a plain count (no unit); A has units becquerel (Bq = 1 decay/s) or curie (Ci = 3.7 x 10^10 Bq). But both fall on the identical exponential curve.
The SI unit is the becquerel (Bq), equal to 1 disintegration per second. The older unit is the curie (Ci), where 1 Ci = 3.7 x 10^10 Bq. Activity always has units of 'per second' because it counts decays each second.
No, for a simple single radioactive nuclide the activity only decreases with time because N keeps falling. It can appear to rise only when a parent feeds a daughter (a decay chain), but for one isotope the curve always drops.
The remaining fraction is (1/2)^n. So after 1, 2, 3, 4 half-lives you keep 1/2, 1/4, 1/8, 1/16 of the original activity respectively. Convert any time to n by n = t / T (half-life).
1/16 = (1/2)^4, so n = 4 half-lives have passed. If the half-life were, say, 5 days, the total time would be 4 x 5 = 20 days.
Yes. Activity A0 = lambda N0, and half-life T = 0.693 / lambda. A larger lambda means a shorter half-life and faster fall of activity. All three describe the same speed of decay.
