A worker at a breeder reactor plant accidentally ingests 2.5 mg of ${}^{\mathbf{239}}\mathit{P}\mathit{u}$dust. This isotope has a half-life of 24 100y , decaying by alpha decay. The energy of the emitted alpha particles is 5.2 MeV , with an RBE factor of 13. Assume that the plutonium resides in the worker’s body for (it is eliminated naturally by the digestive system rather than being absorbed by any of the internal organs) and that 95% of the emitted alpha particles are stopped within the body. Calculate (a) the number of plutonium atoms ingested, (b) the number that decay during the 12h , (c) the energy absorbed by the body, (d) the resulting physical dose in grays, and (e) the dose equivalent in sieverts.

Mass of the worker, m = 85 kg

Mass of the Plutonium isotope${}^{239}Pu$ingested,$m_{Pu}=2.5mg$

Half-life of${}^{239}Pu$ ,$T_{1/2}=24100y$

Energy of emitted alpha particles,$E_{\propto }=5.2MeV$

RBE factor is 13.

The time of decay, t = 12h

The percentage of alpha particles absorbed by the body is 95%.

As per the problem, the number of Plutonium atoms ingested can be given by the stochiometric calculations for the given ingested mass of the isotope. From this value, we can get the decayed quantity of atoms and the energy absorbed due to these absorbed emitted atoms by the body.

The equivalent dose of radiation is a measure of the biological damage to the human body due to the ionizing radiation in the radioactive decay processes. The new international system (SI) unit of radiation dose, expressed as absorbed energy per unit mass of tissue is the SI unit "gray". 1 Gy = 1 Joule/kilogram or 1 Gy = 1 Sv. RBE (relative biological effectiveness) is a relative measure of the damage done by a given type of radiation per unit of energy deposited in biological tissues.

Formulae:

The undecayed sample remaining after a given time, $N=N_0{e}^{\left(\frac{\ln 2}{T_{1/2}}\right)t}..............\left(1\right)$

The number of atoms in a given mass of a substance,

$N=\frac{m}{A}{N}_A..............\left(2\right)where{N}_A=6.022\times {10}^{23}atom/mol$

Where, A is the molar mass of the substance.

The dose absorbed of a radiation source,

$absorbeddose=\frac{Totalabsorbedenergy}{Massofthesample}............\left(3\right)$

The dose equivalent of a radiation source, $D.E=RBE\times Absorbeddose...........\left(4\right)$

Using the given data in equation (2), the number of ingested atoms of the plutonium isotope${}^{239}Pu$can be given as follows:

localid="1661927400775" $$\begin{gathered} N_0=\frac{2.5\times {10}^{-3}g}{239g/mol}\left(6.022\times {10}^{23}atoms/mol\right) \\ =6.3\times {10}^{18}mol \end{gathered}$$

Hence, the number of atoms islocalid="1661927404979" $6.3\times {10}^{18}mol$.

Now, using the above value and equation (1), the number of decayed plutonium${}^{239}Pu$atoms can be calculated as follows:

$$\begin{gathered} ∆N=N_0\left(1-e^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}\right) \\ =\left(6.3\times {10}^{18}\right)\left(1-e^{\left(\frac{\ln 2}{24100y\times 8760h/y}\right)12h}\right) \\ =2.5\times {10}^{11} \end{gathered}$$

Hence, the value of decayed atoms is $2.5\times {10}^{11}$.

According to the problem, only 95% of the emitted alpha particles is absorbed. Thus, the total energy absorbed by the body can be given as:

$$\begin{gathered} E=\left(0.95\right)\left(∆N\right)E_{\propto } \\ =\left(0.95\right)\left(2.5\times {10}^{11}\times 5.2MeV\right)\left(1.6\times {10}^{-19}J/MeV\right) \\ =0.20J \end{gathered}$$

Hence, the value of the energy absorbed is 0.20J .

Using the value of absorbed energy from part (c), the absorbed dose by the body in gray unit can be given using equation (3) as:

$$\begin{gathered} Absorbeddose=\frac{0.20J}{85kg} \\ =2.3\times {10}^{-3}J/kg \\ =2.3mGy \end{gathered}$$

Hence, the value of the energy is 2.3 mGy .

Using the above absorbed dose from part (d), the dose equivalent in sieverts can be given using equation (4) as follows:

$$\begin{gathered} D.E=13\times 2.3mGy \\ =29.9mSv \\ \approx 30mSv \end{gathered}$$

Hence, the value of dose equivalent is 30 mSv .