Question

Certain theories predict that the proton is unstable, with a half-life of about 10³² years. Assuming that this is true, calculate the number ofproton decays you would expect to occur in one year in the water ofan Olympic-sized swimming pool holding 4.32 x 10⁵L of water.

Short answer

The number of proton decays is $1 \frac{\mathrm{decay}}{\mathrm{year}}$

Step-by-step solution

Identifying the data given in the question

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,

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

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

Concept used to solve the question 

• Radioactive Decay

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

• The half-life of a radioactive nuclide is the time required

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

drop to half of its initial value.

Calculating the number of proton decays

The total mass of the pool

$\begin{array}{rcl}{M}_{pool}& =& \rho \nu \\ & =& (4.32\times {10}^{5}L)(1kg/L)\\ & =& 4.32\times {10}^{5}kg\end{array}$

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

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

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