Question

The plutonium isotope 239 Pu has a half-life of 24,000 years and decays by the emission of a 5.2 MeV alpha particle. Plutonium is not especially dangerous if handled because the activity is low and the alpha radiation doesn’t penetrate the skin. However, there are serious health concerns if even the tiniest particles of plutonium are inhaled and lodge deep in the lungs. This could happen following any kind of fire or explosion that disperses plutonium as dust. Let’s determine the level of danger. a. Soot particles are roughly 1 mm in diameter, and it is known that these particles can go deep into the lungs. How many atoms are in a 1.0@mm@diameter particle of 239 Pu? The density of plutonium is 19,800 kg/m3 . b. What is the activity, in Bq, of a 1.0@mm@diameter particle? c. The activity of the particle is very small, but the penetrating power of alpha particles is also very small. The alpha particles are all stopped, and each deposits its energy in a 50@mm@diameter sphere around the particle. What is the dose, in mSv/year, to this small sphere of tissue in the lungs? Assume that the tissue density is that of water. d. Is this exposure likely to be significant? How does it compare to the natural background of radiation exposure?

Short answer

Part a

The number of ${}^{239}Pu$in $1.00\mu m$diameter particles is $2.6\times {10}^7$.

Part b

The activity of $1.0\mu m$particle is $2.4\times {10}^{-5}$.

Part c

The dose in $mSv$is $1.9\times {10}^5$.

Part d

This is a very high dose to a very small volume of body mass. The exposure to this tissue is much higher than the background level.

Step-by-step solution

Given information

The half-life of the ${}^{239}Pu$is $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=24000yrs$

The radius of soot particles is $r=\frac{1\times {10}^{-6}m}{2}=0.5\times {10}^{-6}m$

The density of Plutonium is$\rho =19,800kg/m^3$

Part a

The number of ${}^{239}Pu$atoms in a $1.00\mu m$diameter particle is

$$\begin{gathered} N=\left(\frac{m}{M_A}\right)N_A \\ =\frac{\rho 4\pi r^3{N}_A}{3M_A} \end{gathered}$$

Substitute the given values

$$\begin{gathered} N=\frac{19.8\times 4\mathrm{\pi }\times {\left(0.5\times {10}^{-6}\right)}^3\times 6.02\times {10}^{23}}{3\times 239\times {10}^{-3}} \\ =2.6\times {10}^7 \end{gathered}$$

Therefore, the number of${}^{239}Pu$atoms in$1.00\mu m$diameter particle is$2.6\times {10}^7$.

Part b

The activity of the particles is given by

$$\begin{gathered} R=rN \\ =\frac{\ln 2}{t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}N \end{gathered}$$

Substitute the given values

$$\begin{gathered} R=\frac{0.693}{24000\times 3.17\times {10}^7}\times 2.6\times {10}^7 \\ \approx 2.4\times {10}^{-5} \end{gathered}$$

Therefore, the activity of$1.0\mu m$diameter particle is$2.4\times {10}^{-5}$.

Part c

The volume of $50\mu m$diameter sphere of tissue around the particle is

$$\begin{gathered} V=\frac{4\mathrm{\pi }}{3}{\left(\frac{50\times {10}^{-6}}{2}\right)}^3 \\ =6.545\times {10}^{-14}{m}^3 \end{gathered}$$

This volume of tissue has a mass of

$$\begin{gathered} m=\rho V \\ =1000\times 6.545\times {10}^{-14} \\ =6.545\times {10}^{-11}kg \end{gathered}$$

The number of decays per year is

$$\begin{gathered} \left|\frac{dN}{dt}\right|∆t=2.39\times {10}^{-5}\times 3.15\times {10}^7 \\ =7.529\times {10}^2 \end{gathered}$$

As each decay creates an $\alpha$particle with energy 5.2 MeV, the total energy received per year by the tissue is

$7.529\times {10}^2\times 5.2\times {10}^6\times 1.6\times {10}^{-19}=6.264\times {10}^{-10}J$

The dose received by the tissue is

$$\begin{gathered} \frac{6.264\times {10}^{-10}}{6.545\times {10}^{-11}}=9.586J/kg \\ =9.586Gy \end{gathered}$$

The dose per year in mSv is

$$\begin{gathered} 9.586Gy\left(RBE\right)=9.586Gy\times 20 \\ \approx 1.9\times {10}^{5}mSv \end{gathered}$$

Therefore, the dose in$mSv$is$1.9\times {10}^5$.

Part d

This is a very high dose to a very small volume of body mass. The background radiation from various natural-occurring sources is about $3mSv/yr$. The exposure to this tissue is much higher than the background level.