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43.38. In an industrial accident a \(65-\mathrm{kg}\) person receives a lethal whole-body equivalent dose of 5.4 Sv from \(\mathrm{x}\) rays. (a) What is the equivalent dose in rem? (b) What is the absorbed dose in rad? (c) What is the total energy absorbed by the person's body? How does this amount of energy compare to the amount of energy required to raise the temperature of 65 \(\mathrm{kg}\) of water \(0.010^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) 540 rem. (b) 540 rad. (c) 351 J. The energy absorbed is much less than the energy needed to raise the temperature of 65 kg of water by 0.010°C.

Step by step solution

01

Convert Sieverts to Rems

The equivalent dose in rem can be found by converting 5.4 Sv to rems. The conversion factor is 1 Sv = 100 rem. Thus, the equivalent dose in rem is obtained by multiplying by 100.\[5.4 \, \text{Sv} \times 100 = 540 \, \text{rem}\]
02

Relationship between Rad and Rem

To find the absorbed dose in rad, we need to know the relationship between rad and rem. For x-rays, the quality factor (QF) is 1, so absorbed dose in rad is equal to the equivalent dose in rem.\[\text{Absorbed dose in rad} = \text{Equivalent dose in rem} = 540 \, \text{rad}\]
03

Calculate Total Energy Absorbed

The total energy absorbed can be found using the formula:\[E = \text{absorbed dose in rad} \times \text{mass in kg} \times \frac{0.01 \, \text{J/kg}}{\text{rad}}\]Substituting the known values (absorbed dose = 540 rad, mass = 65 kg):\[E = 540 \, \text{rad} \times 65 \, \text{kg} \times 0.01 \, \text{\frac{J}{kg\, rad}} = 351 \, \text{J}\]
04

Compare to Energy Required to Heat Water

The energy required to raise 65 kg of water by \(0.010^{\circ} \text{C}\) can be calculated using the specific heat of water (4.186 J/g°C) and the mass of the water (converted to grams):\[\text{Energy} = 65,000 \, \text{g} \times 4.186 \, \text{\frac{J}{g°C}} \times 0.010^{\circ} \text{C} = 2,721 \, \text{J}\]The absorbed energy from the radiation (351 J) is much smaller compared to the energy required to raise the water's temperature by 0.010°C.

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Key Concepts

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

Equivalent Dose
In radiation dosimetry, the equivalent dose is a critical concept that helps us understand the risk produced by exposure to radiation. It is expressed in units of Sieverts (Sv) or rems and considers the biological effect that ionizing radiation has on humans. The conversion between the two units is straightforward: 1 Sievert is equal to 100 rem.

This factor of 100 is used because Sieverts and rems reflect similar concepts but in different measurement systems. To convert from Sieverts to rem, simply multiply the number of Sieverts by 100. For instance, if someone is exposed to a dose of 5.4 Sv, this translates to 5.4 × 100 = 540 rem.

The equivalent dose takes into account the type of radiation we're dealing with. Different types of radiation have different potentials for biological damage. X-rays, for example, have a quality factor of 1, meaning their potential for biological harm is the baseline for comparison with other types of radiation.
Absorbed Dose
The absorbed dose is the amount of radiation energy absorbed per unit mass of an object or tissue, and it is measured in rads or Grays (Gy). While the rad and Gray measure the same concept, they use different scales: 1 rad equals 0.01 Gy.

For the specific case of x-rays, the quality factor is 1. This means that the equivalent dose in rem is numerically the same as the absorbed dose in rad, when dealing with x-rays. Hence, our previous conversion from Sieverts to rem also gives us the absorbed dose in rads. If the absorbed dose is needed in our scenario of a 5.4 Sv exposure, we just convert the equivalent dose of 540 rem directly to 540 rad.

It's important to note that while the absorbed dose tells us how much energy is deposited in tissue, it doesn't convey biological impact—that's where equivalent dose comes into play, by accounting for the type of radiation.
Energy Absorption
When discussing energy absorption, we refer to the actual energy taken up by a material or tissue as a result of radiation exposure. The energy absorbed by a human body can be calculated once the absorbed dose is known. In this instance, the absorbed dose was calculated as 540 rad.

The formula to compute the total energy absorbed (\[E\]) involves the absorbed dose, mass of the tissue, and the conversion factor of 0.01 J/kg per rad:
\[E = ext{absorbed dose in rad} \times ext{mass in kg} \times 0.01 \, \text{\frac{J}{kg rad}}\]
Substituting the values for a 65 kg person yields:\[E = 540 \, \text{rad} \times 65 \, \text{kg} \times 0.01 \, \text{\frac{J}{kg rad}} = 351 \, \text{J}\]

This shows the total energy absorbed by the body. To put this number into perspective, compare it to the energy required to heat up water. It takes significantly more energy to raise the temperature of water slightly—a common benchmark when considering energy quantities. In this case, heating 65 kg of water by 0.010°C takes much more energy, specifically 2,721 J. This demonstrates that even a lethal dose of radiation corresponds to a surprisingly small amount of energy.

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Most popular questions from this chapter

43.72. An Oceanographic Tracer. Nuclear weapons tests in the 1950 s and 1960 s released significant amounts of radioactive tritium \((3 \mathrm{H}, \text { half-life } 12.3 \text { years) into the atmosphere. The tritium }\) atoms were quickly bound into water molecules and rained out of the air, most of them ending up in the ocean. For any of this tritium-tagged water that sinks below the surface, the amount of time during which it has been isolated from the surface can be calculated by measuring the ratio of the decay product, \(\frac{3}{2} \mathrm{He},\) to the remaining tritium in the water. For example, if the ratio of \(\frac{3}{1}\) He to \(\frac{3}{1} \mathrm{H}\) in a sample of water is \(1 : 1,\) the water has been below the surface for one half-life, or approximately 12 years. This method has provided oceanographers with a convenient way to trace the movements of subsurface currents in parts of the ocean. Suppose that in a particular sample of water, the ratio of \(\frac{3}{2} \mathrm{He}\) to \(\frac{3}{1} \mathrm{H}\) is 4.3 to 1.0 . How many years ago did this water sink below the surface?

43.31. The radioactive nuclide "Pt has a half-life of 30.8 minutes. A sample is prepared that has an initial activity of \(7.56 \times 10^{11} \mathrm{Bq}\) . (a) How many 199 \(\mathrm{Pt}\) nuclei are initially present in the sample? (b) How many are present after 30.8 minutes? What is the activity at this time? (c) Repeat part (b) for a time 92.4 minutes after the sample is first prepared.

43.12. The most common isotope of copper is \(\frac{63}{29} \mathrm{Cu}\) . The measured mass of the neutral atom is 62.929601 u. (a) From the measured mass, determine the mass defect, and use it to find the total binding energy and the binding energy per nucleon. (b) Calculare the binding energy from Eq. ( 43.11\()\) . (Why is the fifth term zero? Compare to the result you obtained in part (a). What is the percent difference? What do you conclude about the accuracy of Eq. \((43.11) ?\)

43.49. Use conservation of mass-energy to show that the energy released in alpha decay is positive whenever the mass of the original neutral atom is greater than the sum of the masses of the final neutral atom and the neutral "He atom. (Hint: Let the parent nucleus have atomic number \(Z\) and nucleon number \(A\) . First write the reaction in terms of the nuclei and particles involved, and then \(n\) then \(n\) add \(Z\) electron masses to both sides of the reaction and allot them as needed to arrive at neutral atoms.)

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