/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 An \(85 \mathrm{~kg}\) worker at... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An \(85 \mathrm{~kg}\) worker at a breeder reactor plant accidentally ingests \(2.5 \mathrm{mg}\) of \({ }^{239} \mathrm{Pu}\) dust. This isotope has a half- life of \(24100 \mathrm{y}\), decaying by alpha decay. The energy of the emitted alpha particles is \(5.2 \mathrm{MeV},\) with an \(\mathrm{RBE}\) factor of \(13 .\) Assume that the plutonium resides in the worker's body for \(12 \mathrm{~h}\) (it is eliminated naturally by the digestive system rather than being absorbed by any of the internal organs) and that \(95 \%\) of the emitted alpha particles are stopped within the body. Calculate (a) the number of plutonium atoms ingested, (b) the number that decay during the \(12 \mathrm{~h},\) (c) the energy absorbed by the body, (d) the resulting physical dose in grays, and (e) the dose equivalent in sieverts.

Short Answer

Expert verified
3.03 mSv

Step by step solution

01

Calculate the number of plutonium atoms ingested

We begin by converting the mass of the plutonium ingested to the number of atoms. The atomic mass of plutonium-239 is approximately 239 g/mol.First, convert milligrams to grams:\[ 2.5 \text{ mg} = 2.5 \times 10^{-3} \text{ g} \]Next, use Avogadro's number, \(6.022 \times 10^{23} \text{ atoms/mol} \), to find the number of atoms:\[ \frac{2.5 \times 10^{-3} \text{ g}}{239 \text{ g/mol}} \times 6.022 \times 10^{23} \text{ atoms/mol} = 6.3 \times 10^{18} \text{ atoms} \]
02

Calculate the number of atoms that decay in 12 hours

Use the decay formula to find the number of decayed atoms. The decay constant \(\lambda\) is given by:\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]For plutonium-239, \(t_{1/2} = 24100 \times 365 \times 24 \times 3600 \text{ s} \), and \( \ln(2) \approx 0.693 \). Therefore:\[ \lambda = \frac{0.693}{24100 \times 365 \times 24 \times 3600} \approx 9.1 \times 10^{-13} \text{ s}^{-1} \]The number of decayed atoms in 12 hours \((12 \times 3600 \text{ seconds})\) is:\[ N_d = N_0 \times (1 - e^{-\lambda t}) \approx 6.3 \times 10^{18} \times (1 - e^{-9.1 \times 10^{-13} \times 43200}) \approx 2.5 \times 10^{10} \text{ atoms} \]
03

Calculate the energy absorbed by the body

Each decaying atom releases \(5.2 \text{ MeV} \). Convert this energy to joules:\[ 5.2 \text{ MeV} = 5.2 \times 1.602 \times 10^{-13} \text{ J} = 8.33 \times 10^{-13} \text{ J} \]The energy absorbed by the body is released by 95% of the decayed atoms:\[ E = 2.5 \times 10^{10} \times 8.33 \times 10^{-13} \times 0.95 \approx 1.98 \times 10^{-2} \text{ J} \]
04

Calculate the physical dose in grays

The physical dose \(D\) in grays is the energy absorbed per unit mass of the body. The worker's mass is \(85 \text{ kg} \).\[ D = \frac{E}{\text{mass}} = \frac{1.98 \times 10^{-2} \text{ J}}{85 \text{ kg}} \approx 2.33 \times 10^{-4} \text{ Gy} \]
05

Calculate the dose equivalent in sieverts

The dose equivalent \(H\) in sieverts considers the type and energy of radiation, using the relative biological effectiveness (RBE) factor. Here, \(\text{RBE} = 13\).\[ H = D \times \text{RBE} = 2.33 \times 10^{-4} \text{ Gy} \times 13 \approx 3.03 \times 10^{-3} \text{ Sv} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-life in Radioactive Decay
Radioactive decay is a process where an unstable atomic nucleus emits particles or radiation to become more stable. A key concept in understanding this process is the *half-life*. Half-life is the time required for half of a sample of a radioactive substance to decay. Half-life varies widely among different elements and isotopes, from fractions of a second to millions of years. For plutonium-239, the isotope in the provided exercise, the half-life is 24,100 years. This means it takes 24,100 years for half of any given amount of plutonium-239 to transform into a different element or isotope. Key points about half-life include:
  • It is a constant for a given isotope, describing the rate of decay.
  • More stable isotopes like plutonium-239 have long half-lives, meaning they decay slowly.
  • The remaining radioactive material continues to decay by half every half-life duration, never reaching zero.
Knowing the half-life helps scientists calculate the remaining quantity of material after a certain period and is crucial for understanding the ecological and health impacts of radioactive materials.
Alpha Decay
Alpha decay is a common type of radioactive decay where an atomic nucleus ejects an alpha particle. An alpha particle consists of two protons and two neutrons. This process decreases the original atom's atomic number by two and its mass number by four.In the exercise, plutonium-239 undergoes alpha decay. When an atom of \({}^{239}Pu\) decays, it emits an alpha particle and transforms into a different element, uranium-235:\[ {}^{239}Pu \rightarrow {}^{235}U + \alpha \]Some essential points about alpha decay are:
  • Alpha particles are relatively heavy and positively charged.
  • They have high energy but are stopped easily by materials like paper or skin.
  • They can be harmful when ingested or inhaled because they can damage internal tissues directly due to their low penetration but high ionization power.
Energy Absorption from Alpha Particles
When radioactive materials decay, they release energy, which can be absorbed by surrounding materials, including human tissues. In the exercise scenario, the energy from alpha particles emitted by decaying plutonium-239 atoms is mainly absorbed by the body, except for a small fraction that escapes. This energy absorption can be calculated based on the number of decayed atoms and the energy released per decay. For instance, each decayed atom in the scenario releases 5.2 MeV of energy. The energy absorbed is a product of this energy and the number of decayed atoms, multiplied by the fraction absorbed by the human body. Key considerations include:
  • The absorbed energy is responsible for potential damage to cells and tissues.
  • Calculation of absorbed energy helps assess the initial dose of radiation.
  • This assessment is crucial in radiation protection to estimate potential health effects and required safety measures.
Dose Equivalent and Its Importance
The dose equivalent is a measure in radiation protection that takes into account not just the absorbed dose but also the biological effect of the radiation type. It is measured in sieverts (Sv) and calculated by multiplying the absorbed dose in grays (Gy) by a quality factor known as the Relative Biological Effectiveness (RBE). In this scenario, the absorbed dose was calculated to be approximately 2.33 x 10^-4 Gy. Given an RBE factor of 13, the dose equivalent becomes 3.03 x 10^-3 Sv. Why is dose equivalent important?
  • Different radiation types have varying impacts on biological tissues. Alpha particles, for example, have a higher RBE due to their high ionization potential but low penetration depth.
  • Calculating the dose equivalent provides a more accurate understanding of the potential biological harm that radiation exposure could cause.
  • This helps in formulating safety guidelines and emergency measures in environments dealing with radioactive materials.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the mass of a sample of (initially pure) \({ }^{40} \mathrm{~K}\) that has an initial decay rate of \(1.70 \times 10^{5}\) disintegrations/s. The isotope has a half-life of \(1.28 \times 10^{9} \mathrm{y}\).

At the end of World War II, Dutch authorities arrested Dutch artist Hans van Meegeren for treason because, during the war, he had sold a masterpiece painting to the Nazi Hermann Goering. The painting, Christ and His Disciples at Emmaus by Dutch master Johannes Vermeer \((1632-1675),\) had been discovered in 1937 by van Meegeren, after it had been lost for almost 300 years. Soon after the discovery, art experts proclaimed that Emmaus was possibly the best Vermeer ever seen. Selling such a Dutch national treasure to the enemy was unthinkable treason. However, shortly after being imprisoned, van Meegeren suddenly announced that he, not Vermeer, had painted Emmaus. He explained that he had carefully mimicked Vermeer's style, using a 300 -year-old canvas and Vermeer's choice of pigments; he had then signed Vermeer's name to the work and baked the painting to give it an authentically old look. Was van Meegeren lying to avoid a conviction of treason, hoping to be convicted of only the lesser crime of fraud? To art experts, Emmaus certainly looked like a Vermeer but, at the time of van Meegeren's trial in 1947 , there was no scientific way to answer the question. However, in 1968 Bernard Keisch of Carnegie-Mellon University was able to answer the question with newly developed techniques of radioactive analysis. Specifically, he analyzed a small sample of white lead-bearing pigment removed from Emmaus. This pigment is refined from lead ore, in which the lead is produced by a long radioactive decay series that starts with unstable \({ }^{238} \mathrm{U}\) and ends with stable \({ }^{206} \mathrm{~Pb}\). To follow the spirit of Keisch's analysis, focus on the following abbreviated portion of that decay series, in which intermediate, relatively shortlived radionuclides have been omitted: $${ }^{230} \mathrm{Th} \frac{{ }_{75.4 \mathrm{ky}}}{ }^{226} \mathrm{Ra} \frac{{ }_{1.60 \mathrm{ky}}}{ }^{210} \mathrm{~Pb} \frac{{ }_{22.6 \mathrm{ky}}}{ }^{206} \mathrm{~Pb}$$ The longer and more important half-lives in this portion of the decay series are indicated. (a) Show that in a sample of lead ore, the rate at which the number of \({ }^{210} \mathrm{~Pb}\) nuclei changes is given by $$\frac{d N_{210}}{d t}=\lambda_{226} N_{226}-\lambda_{210} N_{210}$$ where \(N_{210}\) and \(N_{226}\) are the numbers of \({ }^{210} \mathrm{~Pb}\) nuclei and \({ }^{226} \mathrm{Ra}\) nuclei in the sample and \(\lambda_{210}\) and \(\lambda_{226}\) are the corresponding disintegration constants. Because the decay series has been active for billions of years and because the half-life of \({ }^{210} \mathrm{~Pb}\) is much less than that of \({ }^{226} \mathrm{Ra}\), the nuclides \({ }^{226} \mathrm{Ra}\) and \({ }^{210} \mathrm{~Pb}\) are in equilibrium; that is, the numbers of these nuclides (and thus their concentrations) in the sample do not change. (b) What is the ratio \(R_{226} / R_{210}\) of the activities of these nuclides in the sample of lead ore? (c) What is the ratio \(N_{226} / N_{210}\) of their numbers? When lead pigment is refined from the ore, most of the \(226 \mathrm{Ra}\) is eliminated. Assume that only \(1.00 \%\) remains. Just after the pigment is produced, what are the ratios (d) \(R_{226} / R_{210}\) and (e) \(N_{226} / N_{210} ?\) Keisch realized that with time the ratio \(R_{226} / R_{210}\) of the pigment would gradually change from the value in freshly refined pigment back to the value in the ore, as equilibrium between the \({ }^{210} \mathrm{~Pb}\) and the remaining \({ }^{226} \mathrm{Ra}\) is established in the pigment. If Emmaus were painted by Vermeer and the sample of pigment taken from it were 300 years old when examined in \(1968,\) the ratio would be close to the answer of (b). If Emmaus were painted by van Meegeren in the 1930 s and the sample were only about 30 years old, the ratio would be close to the answer of (d). Keisch found a ratio of \(0.09 .\) (f) Is Emmaus a Vermeer?

Go An \(\alpha\) particle ('He nucleus) is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an \(\alpha\) particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass. $$ \begin{array}{llll} { }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array} $$

A neutron star is a stellar object whose density is about that of nuclear matter, \(2 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3} .\) Suppose that the Sun were to collapse and become such a star without losing any of its present mass. What would be its radius?

When aboveground nuclear tests were conducted, the explosions shot radioactive dust into the upper atmosphere. Global air circulations then spread the dust worldwide before it settled out on ground and water. One such test was conducted in October 1976 . What fraction of the \({ }^{90} \mathrm{Sr}\) produced by that explosion still existed in October 2006 ? The half-life of \({ }^{90} \mathrm{Sr}\) is \(29 \mathrm{y}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.