A radioactive isotope A is continuously produced in a nuclear reaction at a constant rate of 800 nuclei·s⁻¹. At the same time A decays to a stable product. The half-life of A is 10 s, and initially no A nuclei are present. If the production continues for 20 s, find the activity of A in Bq at that instant.
Physics · Nuclear physics · radioactivity and law of decay
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Question

A radioactive isotope A is continuously produced in a nuclear reaction at a constant rate of 800 nuclei·s⁻¹. At the same time A decays to a stable product. The half-life of A is 10 s, and initially no A nuclei are present. If the production continues for 20 s, find the activity of A in Bq at that instant.

Physics · Nuclear physics · radioactivity and law of decay

G1 Physics · Accepted Answer

With constant production rate R and radioactive decay constant λ, dN ÷ dt = R − λN. Starting from N=0, the activity grows as A(t) = R[1−exp(−λt)]. The saturation activity is therefore A(sat) = R = 800 Bq. The irradiation lasts 20 s, equal to two half-lives. After two half-lives: exp(−λt) = 1 ÷ 2² = 1 ÷ 4. Therefore, A(t) = 800[1−1 ÷ 4] = 800×3 ÷ 4 = 600 Bq. Hence N = 600.