The half-life of strontium- 91, \(^{91}_{38} \mathrm{Sr}\) is \(9.70 \mathrm{h}\). Find (a) its decay constant and (b) for an initial 1.00 -g sample, the activity after 15 hours.

To find the decay constant, we can use the half-life formula: \[t_{1/2} = \frac{ln(2)}{\lambda}\] Rearranging to isolate the decay constant, we have: \[\lambda = \frac{ln(2)}{t_{1/2}}\] Now, we can substitute the given values into the formula: \[\lambda = \frac{ln(2)}{9.70}\] \[\lambda = 0.07146\, \mathrm{h}^{-1}\]

First, we need to find the number of atoms in an initial 1.00g strontium-91 sample: \[N_0 = \frac{m}{m_{atom}}\] Here, \(m = 1\, \mathrm{g}\) and \(m_{atom} = \frac{91\, \mathrm{g/mol}}{6.022 \times 10^{23}\, \mathrm{atoms/mol}} = 1.510 \times 10^{-22}\, \mathrm{g/atom}\). So, \[N_0 = \frac{1\, \mathrm{g}}{1.510 \times 10^{-22}\, \mathrm{g/atom}} = 6.62 \times 10^{22}\, \mathrm{atoms}\] Now, we can calculate the number of remaining particles after 15 hours using the decay formula: \[N = N_0 e^{\lambda t}\] Plugging in the values, we get: \[N = (6.62 \times 10^{22}\, \mathrm{atoms}) \times e^{(0.07146\, \mathrm{h}^{-1})(-15\, \mathrm{h})}\] \[N = 6.62 \times 10^{22} \times e^{-1.0719}\] \[N = 2.30 \times 10^{22}\, \mathrm{atoms}\]

Now that we have the number of remaining particles, we can calculate the activity using the activity formula: \[A = \lambda N\] Substitute the values into the formula: \[A = (0.07146\, \mathrm{h}^{-1})(2.30 \times 10^{22}\, \mathrm{atoms})\] \[A = 1.64 \times 10^{21}\, \mathrm{atoms/h}\] So, the activity of the 1.00g strontium-91 sample after 15 hours is \(1.64 \times 10^{21}\) atoms/h.