A radioactive isotope has an initial activity of . Its activity is reduced to exactly exactly half a — Nuclear Chemistry and Radioactivity Chemistry Question
Question
A radioactive isotope has an initial activity of $28\text{ dpm}$. Its activity is reduced to exactly $14\text{ dpm}$ exactly half an hour later. Based on the radioactive decay constant $\lambda$, calculate the initial number of undecayed nuclei present in the sample at $t=0$. (Use $\ln 2 = 0.693$)
Answer: B
💡 Solution & Explanation
Activity halves from $28$ to $14\text{ dpm}$ in $30\text{ min}$, so $T_{1/2} = 30\text{ min}$. Decay constant $\lambda = 0.693 / 30 = 0.0231\text{ min}^{-1}$. Initial activity $A_0 = \lambda N_0$. $28 = 0.0231 \times N_0 \Rightarrow N_0 = 28 / 0.0231 \approx 1212$ atoms.
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