Chapter 18, Problem 84CP

 (a)

Interpretation Introduction

Interpretation: The number of $\text{α}and\text{β}$ particles produced and the nuclei produced during the complete decay series is to be stated.  Reason for the concern over inhaling radon gas, the solid produced during its decay and its balanced chemical equation is to be stated. $\text{4}\text{pCi}$ per liter of air into concentration units of radon per liter of air and moles of radon per liter of air is to be converted.

Concept introduction: A process through which an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.

To determine: The number of $\text{α}$ and $\text{β}$ particles produced during the complete decay series.

Answer to Problem 84CP

Answer

The number of $\text{α}$ and $\text{β}$ particles produced are $\underset{\_}{4}$ and $\underset{\_}{2}$.

Explanation of Solution

Explanation

The number of $\text{α}$ and $\text{β}$ particles produced are $\underset{\_}{4}$ and $\underset{\_}{2}$.

The mass number of ${}_{92}^{238}\text{U}$ is $238$ and the mass number of ${}_{86}^{222}\text{Rn}$ is 222. The difference in mass number is $238-222=16$

The number of alpha particles is calculated by the formula,

$\text{Alpha}\text{particles}=\frac{\text{Difference}\text{in}\text{mass}\text{number}}{\text{4}}$

Therefore, the alpha particles are $\frac{16}{4}=4$.

The atomic number of ${}_{92}^{238}\text{U}$ is $92$ and the mass number of ${}_{86}^{222}\text{Rn}$ is $86$. Difference in the atomic number is $6$.

The number of beta particles is calculated by the formula,

$\text{Beta}\text{particles}=\frac{\text{2×}\left(\text{Number}\text{of}\text{alpha}\text{particles}\text{produced}\right)\text{-Difference}\text{in}\text{atomic}\text{number}}{\text{4}}$

Therefore, the number of beta particles is

$\begin{array}{l}2\times 4-\left(92-86\right)\\ =8-6\\ =2\end{array}$

The number of $\text{α}$ and $\text{β}$ particles produced are $\underset{\_}{4}$ and $\underset{\_}{2}$.

(b)

Interpretation Introduction

Interpretation: The number of $\text{α}and\text{β}$ particles produced and the nuclei produced during the complete decay series is to be stated.  Reason for the concern over inhaling radon gas, the solid produced during its decay and its balanced chemical equation is to be stated. $\text{4}\text{pCi}$ per liter of air into concentration units of radon per liter of air and moles of radon per liter of air is to be converted.

Concept introduction: A process through which an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.

To determine: The reason for the concern over inhaling radon gas.

Answer to Problem 84CP

Answer

Since, radon gas can cause effective damage if it is ingested, therefore there is a concern over inhaling it.

Explanation of Solution

Explanation

Since, radon gas can cause effective damage if it is ingested, therefore there is a concern over inhaling it.

Radon gas increases the risk of lung cancer. Therefore, there is a concern over inhaling it.

(c)

Interpretation Introduction

Interpretation: The number of $\text{α}and\text{β}$ particles produced and the nuclei produced during the complete decay series is to be stated.  Reason for the concern over inhaling radon gas, the solid produced during its decay and its balanced chemical equation is to be stated. $\text{4}\text{pCi}$ per liter of air into concentration units of radon per liter of air and moles of radon per liter of air is to be converted.

Concept introduction: A process through which an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.

To determine: The solid produced during the decay of radon and its balanced chemical equation.

Answer to Problem 84CP

Answer

The solid particle is polonium and the balanced chemical equation for the given decay process is

${}_{86}^{222}\text{Rn}\to {}_{84}^{218}{\text{Po+}}_2^4\text{He}$

Explanation of Solution

Explanation

The solid particle is polonium and the balanced chemical equation for the given decay process is

${}_{86}^{222}\text{Rn}\to {}_{84}^{218}{\text{Po+}}_2^4\text{He}$

The radon decay is characterized by the production of alpha particle. In the alpha particle production, the resultant nuclide has a mass number less by $4$ units and atomic number less by $2$ units. Therefore, the solid particle produced is polonium.

Since, polonium is a highly radioactive element, therefore, it is a more potent alpha particle producer.

(d)

Interpretation Introduction

Interpretation: The number of $\text{α}and\text{β}$ particles produced and the nuclei produced during the complete decay series is to be stated.  Reason for the concern over inhaling radon gas, the solid produced during its decay and its balanced chemical equation is to be stated. $\text{4}\text{pCi}$ per liter of air into concentration units of radon per liter of air and moles of radon per liter of air is to be converted.

Concept introduction: A process through which an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.

To determine: The conversion of $\text{4}\text{pCi}$ per liter of air into concentration units of radon per liter of air and moles of radon per liter of air.

Answer to Problem 84CP

Answer

The concentration units of ${}^{222}\text{Rn}$ atoms per liter of air is $\underset{\_}{7\times 10^{4}Rnatomsperlitreair}$. The concentration units of moles of ${}^{222}\text{Rn}$ atoms per liter of air is $\underset{\_}{1.2\times 10^{-19}molesRnatomsperlitreair}$

Explanation of Solution

Explanation

The decay constant for alpha decay of ${}^{222}\text{Rn}$ is $3.713\times 10^{-3}decayse{c}^{-1}$.

The half life is $\text{3}\text{.11}\text{minutes}$

The conversion of minutes into seconds is done as,

$\text{1}\text{minute}=\text{60}\text{seconds}$

Therefore, the conversion of $3.11\text{minutes}$ into seconds is,

$3.11\text{minutes}=3.11\times 60\text{seconds}=186.6\text{seconds}$

The decay constant can be calculated by the formula given below.

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

Where,

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

Substitute the value of half life in the above equation.

$\begin{array}{c}k=\frac{0.693}{186.6}\text{decay}{\text{sec}}^{\text{-1}}\\ =\underset{\_}{3.713\times 10^{-3}decayse{c}^{-1}}\end{array}$

The concentration units of ${}^{222}\text{Rn}$ atoms per liter of air is $\underset{\_}{39.85Rnatomsperlitreair}$.

The conversion of $\text{pCi}$ to $\text{Ci}$ is done as,

$1\text{pCi}=1\times {10}^{-12}\text{Ci}$

Therefore, the conversion of $\text{4}\text{pCi}$ to $\text{Ci}$ is,

$\begin{array}{c}\text{4}\text{pCi}=\text{4}\times 1\times {10}^{-12}\text{Ci}\\ =\text{4}\times {10}^{-12}\text{Ci}\end{array}$

The conversion of $\text{Ci}$ to decay ${\text{s}}^{-1}$ is done as,

$1\text{Ci}=3.7\times {10}^{10}\text{decay}{\text{s}}^{\text{-1}}$

Therefore, the conversion of $\text{4}\times {10}^{-12}\text{Ci}$ to decay ${\text{s}}^{-1}$ is,

$\text{4}\times {10}^{-12}\text{Ci}=4\times {10}^{-12}\times 3.7\times {10}^{10}\text{decay}{\text{s}}^{\text{-1}}$

Therefore, the concentration units of ${}^{222}\text{Rn}$ atoms per liter of air is

$\frac{4\times {10}^{-12}\times 3.7\times {10}^{10}}{3.713\times {10}^{-3}}=\underset{\_}{39.85Rnatomsperlitreair}$

The concentration units of moles of ${}^{222}\text{Rn}$ atoms per liter of air is $\underset{\_}{6.6174\times 10^{-23}mole/LRn}$

The concentration units of moles of ${}^{222}\text{Rn}$ atoms per liter of air is calculated as,

$\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{Rn}\text{atoms}\text{in}\text{1}\text{mole}\\ 39.85\text{Rn}\text{atoms}\text{in}\frac{39.85}{\text{6}{\text{.022×10}}^{\text{23}}}\text{mole}=\underset{\_}{6.6174\times 10^{-23}mole/LRn}\end{array}$