91影视

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.

(a)

Considering the given information:

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

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

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

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

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

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

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

$\begin{array}{rcl}{\text{t}}_{\text{1/2}}& =& \frac{\text{ln2}}{\text{k}}\\ \text{k}& =& \left(\frac{{\displaystyle \text{ln2}}}{{\displaystyle {\text{2.41脳10}}^{\text{4}}\text{yr}}}\right)\left(\frac{\text{1yr}}{\text{365.25d}}\right)\left(\frac{\text{1d}}{\text{24h}}\right)\left(\frac{\text{1h}}{\text{3600s}}\right)\\ & =& {\text{9.113904脳10}}^{\text{- 13}}{\text{yr}}^{\text{- 1}}\end{array}$

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

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

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

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

(b)

- Calculate the number of Rads in Gy.

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

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