Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $1 \frac{decay}{year}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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,

$\begin{array}{rcl} ρ & = & 1000 k g / m^{3} \\ = & 1 k g / L \end{array}$

Mass of the proton, $m_{p} = 1.67 x 10^{- 27} k g$

Step 1: Concept used to solve the question

Radioactive Decay

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

$\begin{array}{rcl} M_{p o o l} & = & ρ ν \\ = & (4.32 \times 10^{5} L) (1 k g / L) \\ = & 4.32 \times 10^{5} k g \end{array}$

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" $\begin{array}{rcl} N & = & \frac{(10 / 18) M_{p o o l}}{m_{p}} \\ = & \frac{(10 / 18) (4.32 \times 10^{5} k g)}{m_{p} = 1.67 \times 10^{- 27} k g} \\ = & 1.44 \times 10^{32} \end{array}$

We know the rate of radioactive decay can be given as

$R = \frac{N I n 2}{T_{1 / 2}}$

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" $\begin{matrix} R = \frac{(1.44 \times 10^{32}) \ln 2}{10^{32} years} \\ = 1 \frac{decay}{year} \end{matrix}$

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

	q	S		q	S
$∆^{-}$	-1	0	$\sum_{}^{* 0}$	0	-1
$∆^{0}$	0	0	$\sum_{}^{* +}$	+1	-1
$∆^{+}$	+1	0	$\equiv^{* 0}$	-1	-2
$∆^{+ +}$	+2	0	$\equiv^{* 0}$	0	-2
$\sum_{}^{* -}$	-1	-1	$Ω^{-}$	-1	-3

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Fundamentals Of Physics

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Short Answer

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.

Step by Step Solution

Step 1: Identifying the data given in the question

Step 1: Concept used to solve the question

Step 2: Calculating the number of proton decays

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Fundamentals Of Physics

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Short Answer

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.

Step by Step Solution

Step 1: Identifying the data given in the question

Step 1: Concept used to solve the question

Step 2: Calculating the number of proton decays

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Turning Points in Physics

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.