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Chapter 21: Problem 83

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.

Short Answer

Expert verified
The disintegration rate due to plutonium radioactivity in an average person is \(2 \times 3.7 \times 10^{-2}\) disintegrations/s. The energy deposited in a person's body in a year is \((8 \times 10^{-13} J) \times (2 \times 3.7 \times 10^{-2}) disintegrations/s \times (3.154 \times 10^7) s\). The number of rads in a year is \(\frac{(8 \times 10^{-13} J) \times (2 \times 3.7 \times 10^{-2}) disintegrations/s \times (3.154 \times 10^7) s}{75kg}\), and the number of rems in a year is \(\frac{(8 \times 10^{-13} J) \times (2 \times 3.7 \times 10^{-2}) disintegrations/s \times (3.154 \times 10^7) s}{75kg}\) × 20.

Step by step solution

01

Determine disintegration rate

We are given the level of activity in each person as 2 pCi (picocuries), which is a measure of radioactivity. We know that 1 curie (Ci) is equal to 3.7 × 10^10 disintegrations per second. To convert pCi to Ci, we have: 1 pCi = 10^-12 Ci Thus, we can find the total disintegration rate of plutonium in each person: Disintegration rate = (2 pCi) × (1 Ci / 10^12 pCi) × (3.7 × 10^10 disintegrations/s per Ci) = \(2 \times 3.7 \times 10^{-2} disintegrations/s\)
02

Calculate energy per year

Now, we will find the total energy deposited in a person's body in a year. We are given that each alpha particle deposits \(8 \times 10^{-13} J\) of energy. There are a total of 2 × 3.7 × 10^(-2) disintegrations per second, and each disintegration results in one alpha particle. Energy per year = (Energy per alpha particle) × (Number of alpha particles per second) × (Number of seconds in a year) = \((8 \times 10^{-13} J) \times (2 \times 3.7 \times 10^{-2}) disintegrations/s \times (3.154 \times 10^7) s\)
03

Calculate rads

To calculate the number of rads, we need to divide the total energy deposited by the mass of the person in kilograms. We are given that the average person weighs 75 kg. Number of rads = (Energy per year) / (Mass of the person in kg) = \(\frac{(8 \times 10^{-13} J) \times (2 \times 3.7 \times 10^{-2}) disintegrations/s \times (3.154 \times 10^7) s}{75kg}\)
04

Calculate rems

To calculate the number of rems, we need to multiply the number of rads by the quality factor (QF) of the radiation type. For alpha particles, the QF is 20. Number of rems = (Number of rads) × (Quality Factor) = \(\frac{(8 \times 10^{-13} J) \times (2 \times 3.7 \times 10^{-2}) disintegrations/s \times (3.154 \times 10^7) s}{75kg}\) × 20 Now you have the number of rads and rems of radiation in 1 year from such a level of plutonium in an average person.

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Most popular questions from this chapter

A \(65-\mathrm{kg}\) person is accidentally exposed for \(240 \mathrm{~s}\) to a \(15-\mathrm{mCi}\) source of beta radiation coming from a sample of \({ }^{90}\) Sr. (a) What is the activity of the radiation source in disintegrations per second? In becquerels? (b) Each beta particle has an energy of \(8.75 \times 10^{-14} \mathrm{~J},\) and \(7.5 \%\) of the radiation is absorbed by the person. Assuming that the absorbed radiation is spread over the person's entire body, calculate the absorbed dose in rads and in grays. (c) If the \(\mathrm{RBE}\) of the beta particles is 1.0, what is the effective dose in mrem and in sieverts? (d) How does the magnitude of this dose of radiation compare with that of a mammogram ( \(300 \mathrm{mrem}\) )?

(a) What is the function of the moderator in a nuclear reactor? (b) What substance acts as the moderator in a pressurized water generator? (c) What other substances are used as a moderator in nuclear reactor designs?

Why are nuclear transmutations involving neutrons generally easier to accomplish than those involving protons or alpha particles?

Based on the following atomic mass values \(-1 \mathrm{H}\), 1.00782 amu; \({ }^{2} \mathrm{H}, 2.01410 \mathrm{amu} ;{ }^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;{ }^{3} \mathrm{He}\) 3.01603 amu; \({ }^{4}\) He, 4.00260 amu- and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: (a) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (c) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{1}^{1} \mathrm{H}\)

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