Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The Number of $^{239}\text{Pu}$ Atoms in a 1.0-µm Diameter Particle is given by
$$
N = \frac{m_{\rm sample}N_A}{M }
$$
where $M$ is the molar mass of the element under study, $m$ is the mass of the given sample, and $N_A$ is Avogadro's number.
$$
N = \frac{\rho V N_A}{M_\text{Pu} }= N = \frac{\frac{4}{3}\pi r^3 \rho N_A}{M_\text{Pu} }
$$
Plug the known;
$$
N = \frac{\frac{4}{3}\pi (0.5\times 10^{-6})^3(19800 ) (6.022\times 10^{23})}{ 239}
$$
$$
N = \color{red}{\bf 2.61 \times 10^7} \, \text{atoms}
$$
$$\color{blue}{\bf [b]}$$
The Activity $ R $ of a particle is given by:
$$
R = rN\tag 1
$$
- The decay constant $ r$ is related to the half-life $ t_{1/2} $ by:
$$
r = \frac{\ln(2)}{t_{1/2}}\tag 2
$$
Plug (2) into (1);
$$
R = \frac{\ln(2)}{t_{1/2}}N
$$
Plug the known, where For $^{239}\text{Pu}$,
$ t_{1/2} = 24,000 \, \text{years} $
$$
R = \frac{\ln(2)}{756864000000 } (2.61 \times 10^7)
$$
$$
R = \color{red}{\bf 2.39 \times 10^{-5}} \, \text{Bq}
$$
$$\color{blue}{\bf [c]}$$
Now we need to calculate the mass of the Tissue Sphere.
$$
m=\rho V= \frac{4}{3} \pi r^3 \rho
$$
Plug the known; where $\rho_{\rm tissue}=\rho_{\rm water}$
$$
m_{\rm tissue}= \frac{4}{3} \pi (25 \times 10^{-6})^3(1000) =\bf 6.54 \times 10^{-11} \, \rm {kg}
$$
Now we can calculate the total energy Absorption per year.
Total decays per year is given by
$$
\text{Decays per year}=\left|\dfrac{dN}{dt}\right|\Delta t = R \Delta t
$$
$$
\left|\dfrac{dN}{dt}\right|\Delta t = (2.39 \times 10^{-5})(365\times 24\times 3600)$$
$$
\left|\dfrac{dN}{dt}\right|\Delta t =\bf 754\, \rm {decays/year}
$$
We know that each decay releases 5.2 MeV, so the total energy released in a year is given by
$$
E_{\rm yr}= 754 \times 5.2 \times 10^6\times 1.6 \times 10^{-19} \, \text{J}
$$
$$
E_{\rm yr} = \bf 6.27\times 10^{-10} \, \rm {J}
$$
Now let's calculate the Dose received by the tissue.
$$ D = \frac{\text{Energy absorbed}}{\text{Tissue Mass}} =\dfrac{ E_{\rm yr} }{m_{\rm tissue}}$$
$$
D = \frac{6.27 \times 10^{-10}}{6.54 \times 10^{-11}} =\bf 9.587\, \rm {Gy}
$$
Convert to Dose in mSv:
- Using an RBE of 20 for alpha particles:
$$
D = 9.587\times 20 = 191.7 \, \text{Sv/year}=\color{red}{\bf 1.92\times 10^5}\, \text{mSv/year}
$$
$$\color{blue}{\bf [d]}$$
This dose (1.92\times 10^5 mSv/year) is much higher than the typical background radiation level of approximately 3 mSv/year. This indicates that even a small particle of inhaled $^{239}\text{Pu}$ lodged in lung tissue could significantly increase the radiation dose, potentially increasing the risk of lung damage.
Therefore, this exposure level is indeed significant and much higher than normal background radiation, suggesting serious health risks associated with inhaling plutonium dust.
