Q.The half-life of $_{38}^{90}$Sr is 28 years. Determine the disintegration rate of its 5 mg sample.

[!TLDR] $\lambda=0.693/28$ yr $=7.85\times10^{-10}$ s$^{-1}$; $N=\frac{5\times10^{-3}\text{g}}{90\text{ g/mol}}\times6.022\times10^{23}\approx3.35\times10^{19}$; Activity $=\lambda N\approx2. …

First convert the half-life to a decay constant: $\lambda=\frac{0.693}{T_{1/2}}=\frac{0.693}{28\text{ yr}}=0.02475$ yr$^{-1}$, then to seconds: $\lambda=\frac{0.02475}{3.156\times10^7\text{ s/yr}}\approx7.84\times10^{-10}$ s$^{-1}$. Next find the number of atoms in a 5 mg sample of strontium-90 (molar mass approximately 90 g/mol): $N=\frac{m}{M}\times N_A=\frac{5\times10^{-3}\text{ g}}{90\text{ g/mol}}\times6.022\times10^{23}\approx3.346\times10^{19}$ atoms. The disinte …

Find $\lambda$ from the half-life, find N from the sample mass and Avogadro's number using the isotope's approximate …

Using the wrong molar mass (e.g. a generic strontium atomic weight rather than the isotope's own mass number, ~90 g/mol for Sr-90), w …