91Ó°ÊÓ

Given in the question

The half-life of a proton T_1/2 = 10³² years

The volume of the pool, V = 4.32 x 10⁵L

We know

The density of the water,

$\begin{array}{rcl}\mathrm{ÒÏ}& =& 1000kg/m^3\\ & =& 1kg/L\end{array}$

Mass of the proton,$m_p=1.67x{10}^{-27}kg$

Most of the known nuclides are radioactive. they spontaneously decay at a rateproportional to the numberNof radioactive atoms present.

for the decay rate R(or the number N) in a sample to

drop to half of its initial value.

The total mass of the pool

$\begin{array}{rcl}{M}_{pool}& =& \mathrm{ÒÏ}\mathrm{ν}\\ & =& (4.32×{10}^{5}L)(1kg/L)\\ & =& 4.32×{10}^{5}kg\end{array}$

Since we know, by counting the protons versus total nucleons in a water molecule the fraction of that mass made up by the protons is$\frac{10}{18}$.

Therefore

the number of particles susceptible to decay is

localid="1663253177640" $\begin{array}{rcl}N& =& \frac{(10/18)M_{pool}}{m_p}\\ & =& \frac{(10/18)(4.32×{10}^{5}kg)}{m_p=1.67×{10}^{-27}kg}\\ & =& 1.44×{10}^{32}\end{array}$

We know the rate of radioactive decay can be given as

$R=\frac{NIn2}{T_{1/2}}$

Where R is the rate of decay, N is the number of particles susceptible to decay, $T_{1/2}$and the half-life.

Now,

substituting the values into the formula.

localid="1663253193460" $\begin{array}{c}R=\frac{\left(1.44×{10}^{32}\right) \ln 2}{{10}^{32}  \mathrm{years}}\\ =  1   \frac{\mathrm{decay}}{\mathrm{year}}\end{array}$

Hence the number of protons decay is$ 1   \frac{\mathrm{decay}}{\mathrm{year}}$