Chapter 24, Problem 24.94P

(a)

Interpretation Introduction

Interpretation:

Number of $Rads$ (Dose) received by $85\text{ kg}$ worker for $16\text{ h}$ has to be determined using the given data.

Concept Introduction:

The half–life period for first order reaction is as follows:

  $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\ln \text{ 2}}{\text{k}}\\ \text{Where,}\\ \text{k = Rate constant}\\ {\text{t}}_{\text{1/2}}\text{= half-life}\end{array}$

Decay rate can be calculated using the given equation as follows,

  $\begin{array}{c}\text{Deacy rate =kN}\\ \\ \text{where, }\\ \text{k = Deacy constant}\\ \text{N = No}\text{. of radioactive nuclei}\end{array}$

Mass of a substance can be determined from molar mass as given,

  $Mass\text{ = No}\text{. of moles}\times \text{Molar mass}$

Radiation Dose is the dose quantity that explains the effects of ionizing radiation. It describes the quantity of radiation energy that is absorbed by a matter. The units commonly used to measure dose are $\text{gray }\left(Gy\right)\text{ and rad}.$

  $\begin{array}{l}\text{1 Gy = 1 J/kg}\\ \\ \text{1 rad = 0}\text{.01 J/kg = 0}\text{.01 Gy}\end{array}$

Unit conversion:

  $1\text{ MeV = }\left({\text{10}}^6\text{ eV}\right)\left(\frac{1.602\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)$

Explanation of Solution

Given information is shown below,

$\begin{array}{c}Mass\text{ of worker = 85 kg}\\ \\ \text{Energy released in each disintegration = 5}\text{.15 MeV}\\ \\ {\text{Half-life time, t}}_{\text{1/2}}{\text{ of }}_{94}^{239}\text{Pu = 2}\text{.41}\times {\text{10}}^4\text{ yr}\\ \\ Mass{\text{ of }}_{94}^{239}\text{Pu in the dust = 1}\text{.00 }\mu \text{g = 1}\text{.00}\times {\text{10}}^{-6}\text{ g}\end{array}$

• Number of ${}^{239}Pu$ is determined as follows,

$\begin{array}{c}No.of\text{ atoms = }\frac{Mass}{\text{Molar mass}}\times Avogadro\text{'}snumber\\ \\ =\frac{1\times {10}^{-6}{\text{ g }}^{239}Pu}{239{\text{ g/mol}}^{239}Pu}\times 6.022\times {10}^{23}{\text{ atoms/mol }}^{239}Pu\\ \\ =\text{ 2}\text{.5196653}\times {10}^{15}{\text{ atoms }}^{239}Pu\end{array}$

• Rate constant, k is determined as shown,

  $\begin{array}{c}{\text{t}}_{\text{1/2}}\text{ = }\frac{\ln \text{ 2}}{k}\\ \\ k\text{ = }\left(\frac{\ln \text{ 2}}{2.41\times {10}^4\text{ yr}}\right)\left(\frac{1\text{ yr}}{365.25\text{ d}}\right)\left(\frac{1\text{ d}}{24\text{ h}}\right)\left(\frac{1\text{ h}}{3600\text{ s}}\right)\\ \\ =\text{ 9}\text{.113904}\times {\text{10}}^{-13}{\text{ yr}}^{-1}\end{array}$

• Decay constant can be calculated using the given equation as follows,

$\begin{array}{c}\text{Deacy rate, A}=\text{ kN}\\ \\ \text{ = }\left(\text{ 9}\text{.111969}\times {\text{10}}^{-13}{\text{ yr}}^{-1}\right)\left(\text{2}\text{.5196653}\times {10}^{15}\text{ atoms }\right)\left(\frac{1\text{ disintegration}}{1\text{ atom}}\right)\\ \\ \text{ = 2}\text{.2963988}\times {10}^3\text{ d}ps\end{array}$

• Number of $Rads$ (Dose) received by $85\text{ kg}$ human is determined as given below,

$\begin{array}{c}Dose\text{ = }\left(\frac{\text{2}\text{.2963988}\times {10}^3\text{ d}ps}{85\text{ kg}}\right)\left(yr\right)\left(\frac{\text{5}\text{.15 MeV}}{\text{disintegration}}\right)\left(\frac{1.602\times {10}^{-13}\text{ J}}{1\text{ MeV}}\right)\left(\frac{1\text{ rad}}{0.01\text{ J/kg}}\right)\left(\frac{3600\text{ s}}{1\text{ h}}\right)\left(16\text{ h}\right)\\ \\ =\text{ 1}\text{.2838686}\times {\text{10}}^{-4}rad\\ \\ =\text{ 1}\text{.28}\times {\text{10}}^{-4}\text{ rad}\end{array}$

Number of $Rads$ (Dose) received by $85\text{ kg}$ worker receive for $16\text{ h}$ is $1.28\times 10^{-4}rad$.

(b)

Interpretation Introduction

Interpretation:

Number of $Rads$ (Dose) received by $65\text{ kg}$ worker for $16\text{ h}$ has to be determined using the given data.

Concept Introduction:

The half–life period for first order reaction is as follows:

  $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\ln \text{ 2}}{\text{k}}\\ \text{Where,}\\ \text{k = Rate constant}\\ {\text{t}}_{\text{1/2}}\text{= half-life}\end{array}$

Decay rate can be calculated using the given equation as follows,

  $\begin{array}{c}\text{Deacy rate =kN}\\ \\ \text{where, }\\ \text{k = Deacy constant}\\ \text{N = No}\text{. of radioactive nuclei}\end{array}$

Mass of a substance can be determined from molar mass as given,

  $Mass\text{ = No}\text{. of moles}\times \text{Molar mass}$

Radiation Dose is the dose quantity that explains the effects of ionizing radiation. It describes the quantity of radiation energy that is absorbed by a matter. The units commonly used to measure dose are $\text{gray }\left(Gy\right)\text{ and rad}.$

  $\begin{array}{l}\text{1 Gy = 1 J/kg}\\ \\ \text{1 rad = 0}\text{.01 J/kg = 0}\text{.01 Gy}\end{array}$

Unit conversion:

  $1\text{ MeV = }\left({\text{10}}^6\text{ eV}\right)\left(\frac{1.602\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)$

Explanation of Solution

Given information is shown below,

$\begin{array}{c}Mass\text{ of worker = 85 kg}\\ \\ \text{Energy released in each disintegration = 5}\text{.15 MeV}\\ \\ {\text{Half-life time, t}}_{\text{1/2}}{\text{ of }}_{94}^{239}\text{Pu = 2}\text{.41}\times {\text{10}}^4\text{ yr}\\ \\ Mass{\text{ of }}_{94}^{239}\text{Pu in the dust = 1}\text{.00 }\mu \text{g = 1}\text{.00}\times {\text{10}}^{-6}\text{ g}\end{array}$

• Number of ${}^{239}Pu$ is determined as follows,

$\begin{array}{c}No.of\text{ atoms = }\frac{Mass}{\text{Molar mass}}\times Avogadro\text{'}snumber\\ \\ =\frac{1\times {10}^{-6}{\text{ g }}^{239}Pu}{239{\text{ g/mol}}^{239}Pu}\times 6.022\times {10}^{23}{\text{ atoms/mol }}^{239}Pu\\ \\ =\text{ 2}\text{.5196653}\times {10}^{15}{\text{ atoms }}^{239}Pu\end{array}$

• Rate constant, k is determined as shown,

  $\begin{array}{c}{\text{t}}_{\text{1/2}}\text{ = }\frac{\ln \text{ 2}}{k}\\ \\ k\text{ = }\left(\frac{\ln \text{ 2}}{2.41\times {10}^4\text{ yr}}\right)\left(\frac{1\text{ yr}}{365.25\text{ d}}\right)\left(\frac{1\text{ d}}{24\text{ h}}\right)\left(\frac{1\text{ h}}{3600\text{ s}}\right)\\ \\ =\text{ 9}\text{.113904}\times {\text{10}}^{-13}{\text{ yr}}^{-1}\end{array}$

• Decay constant can be calculated using the given equation as follows,

$\begin{array}{c}\text{Deacy rate, A}=\text{ kN}\\ \\ \text{ = }\left(\text{ 9}\text{.111969}\times {\text{10}}^{-13}{\text{ yr}}^{-1}\right)\left(\text{2}\text{.5196653}\times {10}^{15}\text{ atoms }\right)\left(\frac{1\text{ disintegration}}{1\text{ atom}}\right)\\ \\ \text{ = 2}\text{.2963988}\times {10}^3\text{ d}ps\end{array}$

• Number of $Rads$ (Dose) received by $85\text{ kg}$ human is determined as given below,

  $\begin{array}{c}Dose\text{ = }\left(\begin{array}{l}\left(\frac{\text{2}\text{.2963988}\times {10}^3\text{ d}ps}{85\text{ kg}}\right)\left(yr\right)\left(\frac{\text{5}\text{.15 MeV}}{\text{disintegration}}\right)\\ \left(\frac{1.602\times {10}^{-13}\text{ J}}{1\text{ MeV}}\right)\left(\frac{1\text{ rad}}{0.01\text{ J/kg}}\right)\left(\frac{3600\text{ s}}{1\text{ h}}\right)\left(16\text{ h}\right)\end{array}\right)\\ \\ =\text{ 1}\text{.2838686}\times {\text{10}}^{-4}rad\end{array}$

Number of $Rads$ (Dose) received by $85\text{ kg}$ worker receive for $16\text{ h}$ is $\text{1}\text{.2838686}\times {\text{10}}^{-4}rad$.

• Convert Number of $Rads$ to $Gy$,

  $\begin{array}{c}1\text{ rad = 0}\text{.01 Gy}\\ \\ \text{Dose = }\left(\text{1}\text{.2838686}\times {\text{10}}^{-4}rad\right)\left(0.01\text{ Gy/rad}\right)\\ \\ =\text{ 1}\text{.2838686}\times {10}^{-6}\\ \\ =\text{ 1}\text{.28}\times {\text{10}}^{-6}\text{ Gy}\end{array}$

Number of $Gy$ (Dose) received by $85\text{ kg}$ worker receive for $16\text{ h}$ is $1.28\times 10^{-6}Gy$.