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An \(80.0\) -kg man receives a short-lived radioactive isotope that decays via the emission of gamma-ray photons with an \(\mathrm{RBE}\) of 0.90. If the isotope leads to the absorption of \(0.30 \mathrm{~J}\) of radiant energy, what is the biological equivalent dose he receives? [Hint: Take the dose to be to the full body.]

Short Answer

Expert verified
The man receives a biological equivalent dose of 0.003375 Sv.

Step by step solution

01

Identify the Given Values

First, let's identify the values given in the problem:- The mass of the man: \(80.0\, \text{kg}\)- The absorbed energy: \(0.30 \, \text{J}\)- The RBE (Relative Biological Effectiveness): \(0.90\)
02

Understand the Key Formulas

To find the biological equivalent dose, use the formula: \[ \text{Equivalent dose} = \text{Absorbed dose} \times \text{RBE} \] where \(\text{Absorbed dose (Gy)} = \frac{\text{Absorbed energy (J)}}{\text{mass (kg)}}\).
03

Calculate the Absorbed Dose

Calculate the absorbed dose in grays (Gy) using the formula for absorbed dose.\[ \text{Absorbed dose} = \frac{0.30 \, \text{J}}{80.0 \, \text{kg}} = 0.00375 \, \text{Gy} \]
04

Calculate the Equivalent Dose

Now calculate the equivalent dose using the RBE value.\[ \text{Equivalent dose} = 0.00375 \, \text{Gy} \times 0.90 = 0.003375 \, \text{Sv} \]
05

Final Answer

The biological equivalent dose that the man receives is \(0.003375 \, \text{Sv}\).

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Key Concepts

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

Radiation Dosimetry
Radiation dosimetry is a crucial concept in understanding how radiation affects the human body. It involves measuring, calculating, and assessing the dose uptake in matter, specifically biological entities like humans. Dosimetry helps in ensuring that the radiation levels received are within safe limits.
Radiation doses are commonly expressed in units of gray (Gy) and sievert (Sv). While a gray measures the absorbed dose—the amount of radiation energy absorbed per kilogram of tissue—a sievert considers the biological impact of that energy, factoring in the type of radiation and its effect on different organs. This is where the concept of dose equivalence becomes important, as it adjusts for the potential biological harm caused, not just the raw energy absorbed.
By understanding dosimetry, medical professionals and radiologists can better manage patient exposure to radiation, ensuring treatments are both safe and effective.
  • Measures the radiation dose absorbed by human tissue
  • Uses gray (Gy) and sievert (Sv) for dose measurement
  • Key in protecting individuals from harmful radiation effects
Relative Biological Effectiveness
Relative Biological Effectiveness (RBE) is a measure used to compare the biological effectiveness of different types of radiation. It is particularly essential when dealing with radiations like alpha particles, beta particles, gamma rays, and neutrons, as each type has a different capacity to cause biological damage.
RBE is defined as the ratio of the dose of a reference radiation (usually gamma or X-rays) needed to produce the same biological effect as a given dose of another type of radiation. In simpler terms, it equates the effectiveness of different radiations in causing harm to living tissues.
When calculating the equivalent dose, RBE is used to adjust the absorbed dose to reflect the biological damage potential of the radiation type involved. This is why, in our exercise, we use the RBE of 0.90 for gamma-ray photons, which indicates it is somewhat less damaging than the standard reference radiation.
  • Helps in comparing biological impacts of different radiations
  • Essential for accurate assessment of radiation risk
  • RBE is often less than or greater than 1, reflecting its relative effectiveness
Absorbed Dose Calculation
The calculation of absorbed dose is a fundamental part of radiation dosimetry. It involves determining how much energy from radiation is absorbed per unit mass of tissue. This calculation is critical for assessing the potential health risks posed by radiation exposure.
The absorbed dose is calculated using the formula:\[ \text{Absorbed dose (Gy)} = \frac{\text{Absorbed energy (J)}}{\text{mass (kg)}} \]Here, the energy absorbed is measured in joules (J), and the mass is measured in kilograms (kg). This computation tells us the amount of energy deposited, which provides the base for further calculations of potential biological effects, especially when used alongside RBE to determine equivalent dose.
In the context of the given exercise, converting the absorbed energy of 0.30 J for an 80 kg man results in an absorbed dose of 0.00375 Gy. This dose is then modified by the RBE to calculate the biological equivalent dose, demonstrating the pathway from energy absorption to potential biological impact.
  • Quantifies energy absorbed per mass unit
  • Serves as the base for further biological impact assessment
  • Key for understanding exposure intensity and managing radiation safety

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Most popular questions from this chapter

The binding energy per nucleon for \({ }^{238} \mathrm{U}\) is about \(7.6 \mathrm{MeV}\), while it is about \(8.6 \mathrm{MeV}\) for nuclei of half that mass. If a \({ }^{238} \mathrm{U}\) nucleus were to split into two equal-size nuclei, about how much energy would be released in the process? There are 238 nucleons involved. Each nucleon will release about \(8.6-7.6=1.0 \mathrm{MeV}\) of energy when the nucleus undergoes fission. The total energy liberated is therefore about \(238 \mathrm{MeV}\) or \(2.4 \times 10^{2} \mathrm{MeV}\)

Rubidium-87 has a half-life of \(4.9 \times 10^{10}\) years and decays to strontium-87, which is stable. In an ancient rock, the ratio of \({ }^{87} \mathrm{Sr}\) to \({ }^{87} \mathrm{Rb}\) is \(0.0050 .\) If we assume all the strontium came from rubidium decay, about how old is the rock? Repeat if the ratio is \(0.210\)

An alpha-particle beam enters a charge collector and is measured to carry \(2.0 \times 10^{-14} \mathrm{C}\) of charge into the collector each second. The beam has a cross-sectional area of \(150 \mathrm{~mm}^{2}\), and it penetrates human skin to a depth of \(0.14 \mathrm{~mm}\). Each particle has an initial energy of \(4.0 \mathrm{MeV}\). The \(Q\) for such particles is about \(15 .\) What effective dose, in Sv and in rem, does a person's skin receive when exposed to this beam for 20 s? Take \(\rho=900 \mathrm{~kg} / \mathrm{m}^{3}\) for skin.

The luminous dial of an old watch gives off 130 fast electrons each minute. Assume that each electron has an energy of \(0.50 \mathrm{MeV}\) and deposits that energy in a volume of skin that is \(2.0 \mathrm{~cm}^{2}\) in area and \(0.20 \mathrm{~cm}\) thick. Find the dose (in both Gy and rad) that the volume experiences in \(1.0\) day. Take the density of skin to be \(900 \mathrm{~kg} / \mathrm{m}^{3}\).

Neutrons produced by fission reactions must be slowed by collisions with moderator nuclei before they are effective in causing further fissions. Suppose an 800 -keV neutron loses 40 percent of its energy on each collision. How many collisions are required to decrease its energy to \(0.040 \mathrm{eV} ?\) (This is the average thermal energy of a gas particle at \(35^{\circ} \mathrm{C}\).) After one collision, the neutron energy is down to \((0.6)(800 \mathrm{keV})\). After two, it is \((0.6)(0.6)(800 \mathrm{keV})\); after three, it is \((0.6)^{3}(800\) \(\mathrm{keV}\) ). Therefore, after \(n\) collisions, the neutron energy is \((0.6) n(800 \mathrm{keV})\). We want \(n\) large enough so that $$ (0.6) n\left(8 \times 10^{5} \mathrm{eV}\right)=0.040 \mathrm{eV} $$ Taking the logarithms of both sides of this equation yields $$ \begin{array}{c} n \log _{10} 0.6+\log _{10}\left(8 \times 10^{5}\right)=\log _{10} 0.04 \\ (n)(-0.222)+5.903=-1.398 \end{array} $$ from which we find \(n\) to be \(32.9 .\) So 33 collisions are required.

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