Problem 88 - Chemistry: The Central Science

Question

Tests on human subjects in Boston in 1965 and 1966, following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits \(8 \times 10^{-13} \mathrm{~J}\) of energy and if the average person weighs \(75 \mathrm{~kg}\), calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.

Step-by-Step Solution

Answer

The number of disintegrations per second due to plutonium radioactivity in an average person is \(7.4 \times 10^{-2} \ \mathrm{disintegrations/s}\). Within one year, the average person experiences approximately \(2.49 \times 10^{-8} \ \mathrm{rads}\) and \(4.98 \times 10^{-7} \ \mathrm{rems}\) of radiation from this level of plutonium.

Step 1: Convert radioactivity to disintegrations per second

We have the radioactivity level \(2 \mathrm{pCi}\) (picocuries). To convert this to disintegrations per second, we need to use the conversion factor \(1 \mathrm{pCi}= 3.7 \times 10^{-2} \ \mathrm{disintegrations \ per \ second}\). So, \( Disintegrations \ per \ second = 2 \mathrm{pCi} \cdot 3.7 \times 10^{-2} \ \mathrm{\frac{disintegrations}{s \ pCi}} = 7.4 \times 10^{-2} \ \mathrm{disintegrations/s} \)

Step 2: Calculate energy deposited per second

We are given that each alpha particle deposits \(8 \times 10^{-13} \ \mathrm{J}\) of energy. To find the energy deposited per second, we multiply the energy per alpha particle with the disintegrations per second. \( Energy \ deposited \ per \ second = 7.4 \times 10^{-2} \ \mathrm{\frac{disintegrations}{s}} \cdot 8 \times 10^{-13} \ \mathrm{\frac{J}{disintegration}} = 5.92 \times 10^{-14} \ \mathrm{\frac{J}{s}} \)

Step 3: Calculate energy deposited in one year

To find the energy deposited in one year, multiply the energy deposited per second by the number of seconds in a year. \( Energy \ deposited \ per \ year = 5.92 \times 10^{-14} \ \mathrm{\frac{J}{s}} \cdot 3.1536 \times 10^{7} s = 1.869 \times 10^{-6} \ \mathrm{J} \)

Step 4: Calculate dose in rads and rems

One rad is defined as an absorbed dose of \(0.01 \mathrm{\frac{J}{kg}}\). We are given that the average person weighs \(75 \mathrm{~kg}\). To find the absorbed dose in rads, we divide the total absorbed energy per year by the mass of the person and convert the result to rads. \( Absorbed \ dose \ (rads) = \frac{1.869 \times 10^{-6} \ \mathrm{J}}{75 \mathrm{kg}} \cdot \frac{100 \ rads}{1 \mathrm{J/kg}} \approx 2.49 \times 10^{-8} \ \mathrm{rads} \) Since plutonium decays by alpha particles, the relative biological effectiveness (RBE) assumed is 20. To convert rads to rems, we multiply by the RBE. \( Absorbed \ dose \ (rems) = 2.49 \times 10^{-8} \ \mathrm{rads} \cdot 20 \approx 4.98 \times 10^{-7} \ \mathrm{rems} \) So, the number of rads experienced by an average person is approximately \(2.49 \times 10^{-8} \ \mathrm{rads}\) and the number of rems is approximately \(4.98 \times 10^{-7} \ \mathrm{rems}\) in one year.