If the amount of radioactive substance is increases three times, the number of disintegration per un β Nuclear Chemistry and Radioactivity Chemistry Question
Question
If the amount of radioactive substance is increases three times, the number of disintegration per unit time will be
π‘ Solution & Explanation
Step 1 - Law of Radioactive Decay The activity $A$ (disintegrations per unit time) of a radioactive sample is: $$A = \lambda N$$ where $\lambda$ = decay constant (property of the isotope) and $N$ = number of active nuclei present. Step 2 - Effect of Tripling the Amount Let initial amount $= N_1$, initial activity $= A_1 = \lambda N_1$. If amount is tripled: $N_2 = 3N_1$ $$A_2 = \lambda N_2 = \lambda (3N_1) = 3(\lambda N_1) = 3A_1$$ Step 3 - Analysis of Options - (A) doubled: incorrect β would require doubling the amount - (B) one-third: incorrect β would occur if amount were reduced to one-third - (C) triple: correct β $A \propto N$ (first-order kinetics), so $3 \times$ amount $\Rightarrow$ $3 \times$ activity - (D) unchanged: incorrect β activity depends directly on the number of nuclei present $$\boxed{C}$$