A radioisotope has a half-life of 10 h . Determine the total number...

Initial activity $A_0 = 1 \text{ Ci} = 3.7 \times 10^{10} \text{ disintegrations/s}$.

Half-life $T_{1/2} = 10 \text{ h} = 36000 \text{ s}$.

Decay constant $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{\ln 2}{10} \text{ h}^{-1}$.

The tenth hour corresponds to the time interval from $t = 9 \text{ h}$ to $t = 10 \text{ h}$. The total number of disintegrations $\Delta N$ in this interval is given by integrating the activity $A(t)$:

$\Delta N = \int_{9}^{10} A(t) dt = \int_{9}^{10} A_0 e^{-\lambda t} dt$

$\Delta N = \dfrac{A_0}{\lambda} \left(e^{-9\lambda} - e^{-10\lambda} \right)$

Since $e^{-\lambda t} = 2^{-t / T_{1/2}}$, we can rewrite this as:

$\Delta N =$\dfrac{A_0 T_{1/2}}{\ln 2} \left( 2^{-9/10} - 2^{-10/10} \right)$<br><br>Substituting the values (with$T_{1/2}$in seconds to match$A_0$):<br><br>\dfrac{A_0 T_{1/2}}{\ln 2} = \dfrac{3.7 \times 10^{10} \times 36000}{0.693} \approx$1.922 \times 10^{15}$<br><br>Calculating the fractional part:<br><br>$2^{-0.9} - 2^{-1} \approx 0.5359 - 0.5000 = 0.0359$<br><br>Therefore, the total number of disintegrations is:<br><br>$\Delta N \approx 1.922 \times 10^{15} \times 0.0359 \approx 6.89 \times 10^{13}$<br><br>Rounding to one decimal place, we get$6.9 \times 10^{13}$.<br><br>Answer:$6.9 \times 10^{13}