Answer
Using Eq 42-20, we have:
$R$= $Nln(2) /T_{1/2}$ = $1 decay/y$
Work Step by Step
Density of water os $\rho= 1000 kg/m^3$ then the total mass of pool is :
$M$= $\rho V$ = $4.32 \times 10^5 kg$
Also, by counting the protons versus total nucleons in a water molecule, the fraction of mass made up by protons is $10/18$
So, the number of protons is :
$N$= $\frac {(10/18) M_{pool}} {m_p} $
= $\frac {(10/18) 4.32 \times 10^5 kg} {1.67 \times 10^{-27} kg }$ = $1.44 \times 10^{32} $
Also, we know half-life $T_{1/2} = 10^{32} y$
So, using the eq: $R$= $Nln(2) /T_{1/2}$ = $ \frac {1.44 \times 10^{32}ln(2) } {10^{32} y}$ = $1 decay/y$
