A radioactive element has atomic mass 90 amu and a half-life of 28 years. The number of disintegrati — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Activity Formula The activity (disintegration rate) of a radioactive sample: $$A = \lambda N$$ where $\lambda$ = decay constant (s$^{-1}$) and $N$ = number of radioactive atoms in the sample. Step 2 - Find $N$ (Number of Atoms in 1.0 g) Atomic mass = 90 amu → molar mass $M = 90$ g/mol $$N = \frac{m}{M} \times N_A = \frac{1.0\ \text{g}}{90\ \text{g/mol}} \times 6.022 \times 10^{23}\ \text{mol}^{-1}$$ $$N = 6.691 \times 10^{21}\ \text{atoms}$$ Step 3 - Convert $t_{1/2}$ to Seconds $$t_{1/2} = 28\ \text{yr} \times 365\ \frac{\text{d}}{\text{yr}} \times 24\ \frac{\text{h}}{\text{d}} \times 3600\ \frac{\text{s}}{\text{h}}$$ $$t_{1/2} = 8.83 \times 10^8\ \text{s}$$ Step 4 - Find Decay Constant $\lambda$ $$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8.83 \times 10^8\ \text{s}} = 7.848 \times 10^{-10}\ \text{s}^{-1}$$ Step 5 - Compute Activity $$A = \lambda N = (7.848 \times 10^{-10}\ \text{s}^{-1}) \times (6.691 \times 10^{21})$$ $$A = \boxed{5.24 \times 10^{12}\ \text{dps}}$$ Step 6 - Evaluate Options - **(A) $5.24 \times 10^{10}$**: Off by factor 100 — likely a unit conversion error in computing seconds per year. - **(B) $5.24 \times 10^{8}$**: Off by factor $10^4$. - **(C) $5.24 \times 10^{-10}$**: This is of the order of $\lambda$ alone — forgot to multiply by $N$ (conceptual error). - **(D) $5.24 \times 10^{12}$**: Matches our calculation. **Correct.** $$\boxed{\text{Answer: D}}$$