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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

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

See the step by step solution

## Step 1: Identifying the data given in the question

Given in the question

The half-life of a proton T1/2 = 1032 years

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

We know

The density of the water,

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

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

## Step 1: Concept used to solve the question

Most of the known nuclides are radioactive. they spontaneously decay at a rate proportional to the number N of 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.

## Step 2: Calculating the number of proton decays

The total mass of the pool

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

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" $\begin{array}{rcl}N& =& \frac{\left(10/18\right){M}_{pool}}{{m}_{p}}\\ & =& \frac{\left(10/18\right)\left(4.32×{10}^{5}kg\right)}{{m}_{p}=1.67×{10}^{-27}kg}\\ & =& 1.44×{10}^{32}\end{array}$

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

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

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