Chapter 19, Problem 69AE

Interpretation Introduction

Interpretation: The mass percent of carbon in a typical human body and mass percent of ${}^{\text{14}}\text{C}$ in natural carbon is given. The decay events per second in a $180-\text{lb}$ person are to be calculated.

Concept introduction: Nuclei of radioactive element decompose in various ways. There are two major categories. One involves a change in mass number of the decaying nucleus while others do not. Types of radioactive processes include $\text{α}$ particle production, $\text{β}$ particle production, $\text{γ}$ ray production, electron capture and many others.

Answer to Problem 69AE

Answer

The decay events per second in a $180-\text{lb}$ person is $\underset{\_}{4.525\times 10^{3}decaysper\sec ond}$.

Explanation of Solution

Explanation

To determine: The decay events per second in a $180-\text{lb}$ person.

The number of atoms of ${}^{\text{14}}\text{C}$ in $180-\text{lb}$ person is $\underset{\_}{1.18\times 10^{15}gatoms}$.

Weight of person is $180-\text{lb}$.

Mass percent of carbon in typical body is $18\%$.

Therefore, weight of carbon in person’s body $\begin{array}{l}=\frac{18\times 180}{100}\\ =32.\text{4 lb}\end{array}$

Mass percent of ${}^{\text{14}}\text{C}$ in natural carbon is $1.6\times {10}^{-10}\%$.

Weight of $32.\text{4 lb}$ ${}^{\text{14}}\text{C}$ in natural carbon $\begin{array}{l}=\frac{1.6\times {10}^{-10}\times 32.4}{100}\\ =5.184\times {10}^{-11}\text{ lb}\end{array}$

The conversion of $\text{lb}$ to $\text{g}$ is done as,

$\text{1 lb}=453.5\text{9 g}$

Therefore, the conversion of $5.184\times {10}^{-11}\text{ lb}$ to $\text{g}$ is,

$5.184\times {10}^{-11}\text{ lb}=5.184\times {10}^{-11}\times 453.5\text{9 g}$

Molar mass of ${}^{\text{14}}\text{C}$ is $\text{12 g}$.

Number of atoms of ${}^{\text{14}}\text{C}$ in $\text{12 g}=6.023\times {10}^{\text{23}}\text{ atoms}$.

Therefore, number of atoms of ${}^{\text{14}}\text{C}$ in $\begin{array}{c}5.184\times {10}^{-11}\times 453.5\text{9 g}\\ =\frac{6.022\times {10}^{\text{23}}\times 5.184\times {10}^{-11}\times 453.5\text{9}}{12}\\ =\underset{\_}{1.18\times 10^{15}gatoms}\end{array}$

Explanation

The value of decay constant is $\underset{\_}{3.835\times 10^{-12}\sec ond{s}^{-1}}$.

The decay constant is calculated by the formula,

$\lambda =\frac{0.693}{t_{1/2}}$

Where

• $t_{1/2}$ is the half life.
• $\lambda$ is decay constant.

The half life of ${}^{\text{14}}\text{C}$ is $\text{5730 years}$.

The conversion of years to days is done as,

$\text{1 year}=36\text{5 days}$

Therefore, the conversion of $\text{5730 years}$ to days is as follows,

$\begin{array}{c}\text{5730 years}=5730\times 3\text{65 days}\\ =\text{2091450 days}\end{array}$

The conversion of days to hours is done as,

$\text{1 day}=2\text{4 hours}$

Therefore, the conversion of $\text{2091450 days}$ to hours is as follows,

$\begin{array}{c}\text{2091450 days}=\text{2091450 }\times \text{ 24 hours}\\ =\text{50194800 hours}\end{array}$

The conversion of hours to seconds is done as,

$1\text{ hour}=3\text{600 seconds}$

Therefore, the conversion of $\text{50194800 hours}$ to seconds is as follows,

$\begin{array}{c}\text{50194800 hours}=\text{50194800}\times 3\text{600 seconds}\\ =180701280000\text{ seconds}\end{array}$

Substitute the value of half life in the above formula.

$\begin{array}{c}\lambda =\frac{0.693}{180701280000\text{ seconds}}\\ =\underset{\_}{3.835\times 10^{-12}\sec ond{s}^{-1}}\end{array}$.

Therefore the value of decay constant is $\underset{\_}{3.835\times 10^{-12}\sec ond{s}^{-1}}$.

Explanation

The rate of decay of ${}^{\text{14}}\text{C}$ in the person is $\underset{\_}{4.525\times 10^{3}decayper\sec ond}$.

The decay events per second is the rate of decay of ${}^{\text{14}}\text{C}$ and it is calculated by the formula,

$\text{Rate}=\lambda \times {\text{Number of atoms of }}^{\text{14}}\text{C}$

Substitute the value of $\lambda$ and number of atoms of ${}^{\text{14}}\text{C}$ in the above equation.

$\begin{array}{c}\text{Rate}=3.835\times {10}^{-12}\times 1.18\times {10}^{\text{15}}\\ =\underset{\_}{4.525\times 10^{3}decayper\sec ond}\end{array}$

The rate of decay of ${}^{\text{14}}\text{C}$ in the person is $\underset{\_}{4.525\times 10^{3}decayper\sec ond}$.

Conclusion

Conclusion

The rate of decay of ${}^{\text{14}}\text{C}$ in the person is $\underset{\_}{4.525\times 10^{3}decayper\sec ond}$.