Q24.96P - Chemistry: Molecular Nature Of Matter And Change

Question

Plutonium "triggers" for nuclear weapons were manufactured at the Rocky Flats plant in Colorado. An $\mathbf{\text{85 - kg}}$ worker inhaled a dust particle containing $\mathbf{1}\mathbf{ }\mathbf{\text{μg}}$ of ${}_{\mathbf{\text{94}}}{}^{\mathbf{\text{239}}}\mathbf{\text{Pu}}$, which resided in his body for $\mathbf{\text{16h}}({\text{t}}_{\text{1/2}}$ of ${}^{\mathbf{\text{239}}}\mathbf{\text{Pu = 2.41×10}}^{\mathbf{\text{4}}}\mathbf{\text{ yr}}$, each disintegration released $\mathbf{\text{5.15 MeV}}$). 

(a) How many rads did he receive? 

(b) How many grays?

Step-by-Step Solution

Answer

1. Number of Rads (Dose) received by $\text{85 kg}$  worker $\text{16 h}$  is ${\text{1.28×10}}^{\text{ - 4}}\text{ rad}$.
2. Number of Gy (Dose) received by $\text{85 kg}$  worker in $\text{16 h}$  is  ${\text{1.28×10}}^{\text{ - 6}}\text{ Gy}$.

Step 1: Definition of Concept

Nuclear reaction: A nuclear reaction occurs when a lighter nucleus fuses together to form new stable nuclei, or when a heavier nucleus splits into stable daughter nuclei, releasing a large amount of energy.

Step 2: Find the Number of Rads (Dose) received by 85   kg worker in 16 h

(a)

Considering the given information:

Mass of worker $\text{ = } \text{85 kg}$ 

Energy released in each disintegration  $\text{ = } \text{5.15 MeV}$

Half-life time,  ${\text{t}}_{\text{1/2}}$ of  ${}_{\text{94}}{}^{\text{239}}\text{Pu} \text{ = } {\text{2.41×10}}^{\text{4}}\text{ yr}$

Mass of ${}_{\text{94}}{}^{\text{239}}\text{Pu}$  in the dust  $\text{ = } \text{1.00 μg = } {\text{1.00×10}}^{\text{ - 6}}\text{g}$

- Number of ${}^{\text{239}}\text{Pu}$ is determined as follows,

$\begin{array}{rcl}No.ofatoms & =& \frac{\text{ Mass }}{\text{ Molar mass }}\times Avogadro\text{'}snumber\\ & =& \frac{1\times {\displaystyle {{\displaystyle \text{10}}}^{\text{ - 6}}}{\displaystyle {{\displaystyle \text{ g }}}^{\text{239}}}{\displaystyle {{\displaystyle \text{P}}}_{\text{u}}}}{{\displaystyle {\text{239g/mol}}^{\text{2}}{}^{\text{29}}\text{Pu}}}\times {\text{6.022×10}}^{\text{23}}{\text{ atoms/mol }}^{\text{239}}\text{Pu}\\ & =& {\text{2.5196653×10}}^{\text{15}}{\text{ atoms }}^{\text{239}}\text{Pu}\end{array}$  

 

- The rate constant, k, is calculated as follows:

$\begin{array}{rcl}{\text{t}}_{\text{1/2}}& =& \frac{\text{ln 2}}{\text{k}}\\ \text{k}& =& \left(\frac{{\displaystyle \text{ln 2}}}{{\displaystyle {\text{2.41×10}}^{\text{4 }}\text{yr}}}\right)\left(\frac{\text{1 yr}}{\text{365.25 d}}\right)\left(\frac{\text{1 d}}{\text{24 h}}\right)\left(\frac{\text{1 h}}{\text{3600 s}}\right)\\ & =& {\text{9.113904×10}}^{\text{ - 13}}{\text{ yr}}^{\text{ - 1}}\end{array}$ 

- The following equation can be used to calculate the decay constant:

$\begin{array}{rcl}Deacyrate, A& =& \text{kn}\\ & =& \left({\text{9.111969×10}}^{\text{ - 13}}{\text{ yr}}^{\text{ - 1}}\right)\left({\text{2.5196653×10}}^{\text{15}}\text{ atoms}\right)\left(\frac{\text{1 disintegration }}{\text{1 atom }}\right)\\ & =& {\text{2.2963988×10}}^{\text{3}}\text{ dps}\\ & & \end{array}$ - The number of Rads (Dose) received by a  $\text{85 kg}$ human is calculated as follows:

$\begin{array}{rcl}\text{Dose }& =& \left(\frac{{\displaystyle {\text{2.2963988×10}}^{\text{3}}\text{ dps}}}{{\displaystyle \text{85 kg}}}\right)\text{(yr)}\left(\frac{\text{5.15 MeV}}{\text{ disintegration }}\right)\left(\frac{{\displaystyle {\text{1.602×10}}^{\text{ - 13}}\text{ J}}}{{\displaystyle \text{1 MeV}}}\right)\left(\frac{\text{1 rad}}{\text{0.01 J/kg}}\right)\left(\frac{\text{3600 s}}{\text{1 h}}\right)\text{(16 h)}\\ & =& {\text{1.2838686×10}}^{\text{ - 4}}\text{ rad}\\ & =& \text{1.28}{\times 10}^{\text{ - 4}} \text{rad}\end{array}$ 

Therefore, Number of Rads (Dose) received by $\text{85 kg}$  worker in $\text{16 h}$ is ${\text{1.28×10}}^{\text{ - 4}}\text{ rad}$.

Step 3: Find the energy released in joules

(b)

- Calculate the number of Rads in Gy.

$\begin{array}{rcl}\text{1rad}& =& \text{0.01Gy}\\ \text{Dose}& =& \left({\text{1.2838686×10}}^{\text{ - 4}}\text{ rad}\right)\text{(0.01 Gy/rad)}\\ & =& {\text{1.2838686×10}}^{\text{ - 6}}\mathrm{Gy}\\ & =& {\text{1.28×10}}^{\text{ - 6}}\mathrm{Gy}\end{array}$ 

Therefore, Number of Gy (Dose) received by  $\text{85 kg}$ in $\text{16 h}$ is $\begin{array}{rcl}& & {\text{1.28×10}}^{\text{ - 6}}\mathrm{Gy}\end{array}$.