Question 43.9: Are bananas safe to eat? Potassium K is an important mineral......
Are bananas safe to eat?
Potassium K is an important mineral in the human body. Your muscles need potassium to operate, potassium keeps the sodium levels in your body under control, and all of your cells need potassium to function normally. The average adult needs 4700 mg of potassium per day. Many people try to meet this need by eating bananas. A single typical banana has a mass of 125 g, about 450 mg of which is potassium, nearly 10% of your daily requirement. However, potassium-40 is a radioactive isotope of potassium with a natural abundance of 0.012%. Potassium-40 has a half-life of 1.25 \times 10^9 years, and it radiates gamma and beta rays. The average energy released with each decay is about 0.5 MeV. The body takes around 50 hours to expel digested food after it has been consumed. Assume the radiation from the potassium is mostly absorbed by the tissue and organs along the digestive tract: the esophagus, liver, salivary glands and stomach. Further assume the combined mass of all of these tissues is about 2 kg. Estimate the absorbed dose, the equivalent dose, and the effective dose delivered to these tissues. Report your estimates to one significant figure.
INTERPRET and ANTICIPATE
The absorbed dose D is the energy delivered to tissue per unit mass of the tissue. We know the mass of the tissue, so we just need to find the energy delivered. We also know the energy released by each decay. So we must find the number of decays in 50 hours to find the total energy delivered to the tissues. Once we know the absorbed dose, the effective dose D_{eff} is found from the equivalent dose and the tissue weighting factor.
SOLVE
To find the number of decays, we must first find the initial number N_0 of potassium-40 nuclei when the banana first entered the body. The banana contains 0.45 g of potassium. The molar mass of potassium is 39 g, and the natural abundance of potassium-40 is 0.012%.
From the half-life, we find the number of decays that occur in the 50 hours in which the potassium is in the body. The number of parent nuclei as a function of time is given by Equation 43.8. The number of decays is the initial number of parent nuclei minus the number of parent nuclei remaining after 50 hours. The half-life must be converted from years to hours (1.09 \times 10^{13} h); the units have been left off the exponent for clarity.
\begin{aligned}& N=N_0\left(\frac{1}{2}\right)^{t / T_{1 / 2}}\quad \quad (43.8) \\& N_{\text {decays }}=N_0-N \\& N_{\text {decays }}=N_0\left[1-\left(\frac{1}{2}\right)^{t / T_{1 / 2}}\right] \\ & N_{\text {decays }}=8.3 \times 10^{17}\left[1-\left(\frac{1}{2}\right)^{50 / 1.09 \times 10^{13}}\right] \\& N_{\text {decays }}=2.7 \times 10^6\end{aligned}Each decay releases 0.5 MeV of energy. So we find the total energy delivered by multiplying the number of decays by 0.5 MeV.
\begin{aligned}& E=(0.5 MeV ) N_{\text {decays }}=(0.5 MeV )\left(2.7 \times 10^6\right) \\ & E=1.3 \times 10^6 MeV =2 \times 10^{-7} J\end{aligned}The absorbed dose D is found by dividing the delivered energy by the mass of the tissue.
D=\frac{E}{m}=\frac{2 \times 10^{-7} J }{2 kg }=1 \times 10^{-7} GyThe radiation weighting factor Q is 1 Sv/Gy for both photons and beta particles with energy above 30 keV (Table 43.3). So the equivalent dose (Eq. 43.23)
H=Q D \quad \quad (43.23)is numerically equal to the absorbed dose.
\begin{aligned}& H=Q D=(1 S\nu / Gy )\left(1 \times 10^{-7} Gy \right) \\& H=1 \times 10^{-7} S\nu\end{aligned}The effective dose takes into account the sensitivity of the particular organs and tissues that absorb the radiation. We add the tissue weighting factors for these organs and tissues (Table 43.4). Then find D_{eff} using Equation 43.24. (The subscripts stand for esophagus, liver, salivary glands and stomach.)
\begin{aligned}& w=w_{\text {esoph }}+w_{\text {liver }}+w_{\text {sal gld }}+w_{\text {stmch }} \\ & w=0.04+0.04+0.01+0.12=0.21 \\& D_{\text {eff }}=w H \quad \quad (43.24) \\& D_{\text {eff }}=(0.21)\left(1 \times 10^{-7} S\nu \right)=2 \times 10^{-8} S\nu \end{aligned}CHECK and THINK
If you search the Internet, you are likely to find the vague statement that the radiation dose of a banana is 0.1 μSv. Perhaps the statement is referring to the equivalent dose H, in which case the statement exactly matches what we found. Your search will also reveal an informal unit for measuring a radiation dose—the banana equivalent dose (BED). The idea of this informal unit is to help make radiation doses more understandable. For example, the maximum permitted radiation leakage for a nuclear power plant is 2500 BED. We all feel safe eating a banana, but how about 2500 bananas?
| Table 43.3 Some radiation weighting factors | |
| Type of radiation | Q (Sv/Gy) |
| α particles | 20 |
| Neutrons (in range 100 keV – 2 MeV) | 20 |
| Neutrons (in range 10-100 keV or 2 MeV – 20 MeV) | 10 |
| Neutrons (of less than 10 keV or greater than 20 MeV) | 5.0 |
| Photons (of 30 keV or more) | 1.0 |
| β particles (of 30 keV or more) | 1.0 |
| β particles (of less than 30 keV) | 1.7 |
| Table 43.4 Tissue weighting factors | |
| Organ or tissue | w |
| Bladder | 0.04 |
| Bone surface | 0.01 |
| Brain | 0.01 |
| Breast | 0.12 |
| Colon | 0.12 |
| Esophagus | 0.04 |
| Gonads | 0.08 |
| Liver | 0.04 |
| Lung | 0.12 |
| Red bone marrow | 0.12 |
| Salivary glands | 0.01 |
| Skin | 0.01 |
| Stomach | 0.12 |
| Thyroid | 0.04 |
| Remainder of the body | 0.12 |
| Total | 1.00 |
| The data is from ICRP Publication 103 37 (2007). | |