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(II) Ionizing radiation can be used on meat products to reduce the levels of microbial pathogens. Refrigerated meat is limited to \(4.5 \mathrm{kGy}\). If \(1.2-\mathrm{MeV}\) electrons irradiate \(5 \mathrm{~kg}\) of beef, how many electrons would it take to reach the allowable limit?

Short Answer

Expert verified
Approximately \(1.171 \times 10^{17}\) electrons are needed.

Step by step solution

01

Understand Energy Units

We need to convert the energy limit from kilograys (kGy) to more familiar units like joules (J). The gray (Gy) is a unit of absorbed energy, which is the amount of energy absorbed by one kilogram of a substance. Specifically, 1 Gy = 1 J/kg. Therefore, 1 kGy = 1000 J/kg.
02

Calculate Total Energy Allowance

Using the radiation limit of 4.5 kGy, find the total energy in joules that can be absorbed by the meat. Since the meat weighs 5 kg: \[ 4.5 kGy = 4.5 imes 1000 J/kg = 4500 J/kg \] The total energy allowance is: \[ 4500 ext{ J/kg} imes 5 ext{ kg} = 22500 ext{ J} \] This is the energy that can be absorbed by the meat.
03

Convert Electron Energy

The energy of each electron is given as 1.2 MeV. Convert this energy into joules. Since 1 eV = 1.602 imes 10^{-19} J, \[ 1.2 ext{ MeV} = 1.2 imes 10^6 imes 1.602 imes 10^{-19} ext{ J} = 1.9224 imes 10^{-13} ext{ J/electron} \] This is the energy of each electron.
04

Calculate Number of Electrons Required

To find the number of electrons required, divide the total energy allowance by the energy per electron:\[ \text{Number of electrons} = \frac{22500 ext{ J}}{1.9224 imes 10^{-13} ext{ J/electron}} \]Performing the calculation gives:\[ \text{Number of electrons} \approx 1.171 imes 10^{17} \]
05

Final Review and Verification

Ensure that all conversions and calculations are correct and reflect that the number of electrons calculated can indeed provide the specified energy limit in the exercise problems' context.

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Key Concepts

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

Energy Conversion
In the realm of ionizing radiation and its applications, understanding energy conversion is crucial. This involves converting between various units of energy to ensure precision and clarity. For instance, in the exercise, the energy absorbed by the meat is measured in kilograys (kGy), a less familiar unit for many. One Gray (Gy) is equivalent to one joule of energy absorbed per kilogram of matter. Therefore, 1 kGy equals 1000 joules per kilogram.
In practice, this conversion is essential as it helps identify how much energy can be safely applied to food products, ensuring compliance with safety standards. This conversion bridges the gap between scientific measurements and practical applications, allowing industries to harness ionizing radiation effectively without compromising safety.
Electron Energy
Electron energy is key to understanding the process of irradiation in food products. In our given exercise, each electron has a defined energy of 1.2 Mega-electron Volts (MeV). Mega-electron Volts are a unit of energy commonly used in atomic and subatomic processes.
Converting this energy into joules - the SI unit of energy - makes calculations more straightforward. Since one electron volt (eV) is equivalent to a minuscule 1.602 x 10^-19 joules, converting 1.2 MeV, which equals 1.2 million eV, results in approximately 1.9224 x 10^-13 joules per electron.
This conversion is pivotal as it determines how many electrons are necessary to reach a given energy threshold, in this case, the maximum energy absorbed by meat during irradiation. Understanding electron energy in such contexts is crucial for precise dosing in radiation applications.
Radiation Dosage
Radiation dosage refers to the total amount of ionizing radiation energy absorbed by a substance. In our example, the dosage limit is set at 4.5 kilograys for refrigerated meat products. It is vital to adhere to these limits to ensure food safety and reduce microbial activity without damaging the product.
By multiplying the mass of the product by the allowed absorbed energy per kilogram, one can determine the total energy dosage a given volume of meat can safely absorb. This ensures that the meat remains safe for consumption while effectively lowering the levels of microbial pathogens. Calculating the appropriate radiation dosage is fundamental in the food industry, ensuring the balance between sterilization and quality maintenance.
Microbial Pathogens
Microbial pathogens are tiny organisms that can cause diseases. In food products, especially meat, ionizing radiation is used as a means to reduce their levels. Irradiation disrupts their DNA, effectively neutralizing their ability to reproduce and cause infections.
This process is particularly useful in maintaining food safety standards and prolonging the shelf life of perishable goods. By understanding the effects of ionizing radiation on microbial pathogens, industries can utilize this method as a non-chemical alternative to preservatives and antibiotics.
Thus, ionizing radiation serves as a powerful tool in food safety, making it essential to comprehend its implications and applications in the context of microbial pathogen control.

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Most popular questions from this chapter

(II) Assume a liter of milk typically has an activity of \(2000 \mathrm{pCi}\) due to \({ }_{19}^{40} \mathrm{~K} .\) If a person drinks two glasses \((0.5 \mathrm{~L})\) per day, estimate the total effective dose (in Sv and in rem) received in a year. As a crude model, assume the milk stays in the stomach \(12 \mathrm{hr}\) and is then released. Assume also that very roughly \(10 \%\) of the \(1.5 \mathrm{MeV}\) released per decay is absorbed by the body. Compare your result to the normal allowed dose of 100 mrem per year. Make your estimate for (a) a \(60-\mathrm{kg}\) adult, and \((b)\) a \(6-\mathrm{kg}\) baby.

(II) If \({ }_{3}^{6} \mathrm{Li}\) is struck by a slow neutron, it can form \({ }_{2}^{4} \mathrm{He}\) and another isotope. ( \(a\) ) What is the second isotope? (This is a method of generating this isotope.) \((b)\) How much energy is released in the process?

(II) A \(0.035-\mu\) Ci sample of \({ }_{15}^{32} \mathbf{P}\) is injected into an animal for tracer studies. If a Geiger counter intercepts \(25 \%\) of the emitted \(\beta\) particles, what will be the counting rate, assumed \(85 \%\) efficient?

(II) (a) Suppose for a conventional X-ray image that the X-ray beam consists of parallel rays. What would be the magnification of the image? (b) Suppose, instead, that the X-rays come from a point source (as in Fig. 21\()\) that is 15 \(\mathrm{cm}\) in front of a human body which is 25 \(\mathrm{cm}\) thick, and the film is pressed against the person's back. Determine and discuss the range of magnifications that result.

A shielded \(\gamma\) -ray source yields a dose rate of \(0.052 \mathrm{rad} / \mathrm{h}\) at a distance of \(1.0 \mathrm{~m}\) for an average-sized person. If workers are allowed a maximum dose of 5.0 rem in 1 year, how close to the source may they operate, assuming a \(35-\mathrm{h}\) work week? Assume that the intensity of radiation falls off as the square of the distance. (It actually falls off more rapidly than \(1 / r^{2}\) because of absorption in the air, so your answer will give a better-than-permissible value.)

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