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43.40. A person ingests an amount of a radioactive source with a very long lifetime and activity 0.72\(\mu \mathrm{Ci}\) . The radioactive material lodges in the lungs, where all of the 4.0 -MeV \(\alpha\) particles emitted are absorbed within a \(0.50-\mathrm{kg}\) mass of tissue. Calculate the absorbed dose and the equivalent dose for one year.

Short Answer

Expert verified
Absorbed dose: Convert energy ejected over a year by \(\alpha\) particles to joules; Equivalent dose: Multiply absorbed dose by 20.

Step by step solution

01

Convert Activity to SI Units

Activity is given as 0.72 \(\mu \mathrm{Ci}\). First, convert this to becquerels (Bq) since \(1 \mu \mathrm{Ci} = 3.7 \times 10^4 \mathrm{Bq}\). Thus, the activity \(A = 0.72 \times 3.7 \times 10^4 \mathrm{Bq} = 2.664 \times 10^4 \mathrm{Bq}\). This means 26640 disintegrations per second.
02

Calculate the Total Energy Released in a Year

The energy per \(\alpha\) particle is 4.0 MeV. Convert this to joules: \(1 \mathrm{MeV} = 1.602 \times 10^{-13} \mathrm{J}\), so \(4.0 \mathrm{MeV} = 6.408 \times 10^{-13} \mathrm{J}\). The total number of \(\alpha\) particles emitted in one year is \(2.664 \times 10^4 \times 365 \times 24 \times 3600\). Multiply this by the energy per \(\alpha\) particle to find the total energy, \(E = 6.408 \times 10^{-13} \times 2.664 \times 10^4 \times 365 \times 24 \times 3600\).
03

Calculate the Absorbed Dose

The absorbed dose (D) is the energy absorbed per unit mass. Substitute the total energy calculated in Step 2 and divide by the mass of the tissue (0.50 kg) to find D. \(D = \frac{E}{0.50}\).
04

Convert Absorbed Dose to Grays

The absorbed dose from Step 3 will be in joules per kilogram, which are also known as grays (Gy). The numerical value of D from Step 3 is already in grays.
05

Calculate the Equivalent Dose

The equivalent dose (H) accounts for the biological effect, depending on the type of radiation. Use the radiation weighting factor for \(\alpha\) particles, which is 20. So, \(H = 20 \times D\). The unit for equivalent dose is sieverts (Sv).

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Key Concepts

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

Radioactivity
Radioactivity refers to the process by which unstable atomic nuclei release energy by emitting radiation. This occurs naturally in certain types of elements and isotopes, which are called radioactive. The three most common types of radiation are alpha (\(\alpha\)), beta (\(\beta\)), and gamma (\(\gamma\)) radiation, each with distinct properties.
- \(\alpha\) particles are positively charged and consist of two protons and two neutrons. They are relatively heavy and have low penetration power, being stopped by a sheet of paper or the outer layer of human skin.
- \(\beta\) particles are electrons or positrons and are lighter than alpha particles, with a moderate ability to penetrate materials. They can penetrate skin but are generally stopped by a few millimeters of plastic or glass.
- \(\gamma\) rays are high-energy photons with no mass or charge. They have a high penetration power and require dense materials like lead for shielding.
Understanding radioactivity is crucial for handling radioactive materials safely and for applications in medicine, energy, and more.
Absorbed Dose
Absorbed dose is a measure of how much energy from radiation is deposited in a material or tissue. It is expressed in grays (Gy), where 1 gray equals 1 joule of energy absorbed per kilogram of tissue. Calculating the absorbed dose helps in assessing how much radiation energy is impacting a certain mass.
For instance, if a radioactive substance releases energy within the body, the absorbed dose provides insight into how much energy is absorbed per kilogram of tissue. In our exercise, the energy from alpha particles is entirely absorbed by a specific mass of lung tissue. To calculate this, we first determine the total energy released by the radioactive decay in one year. Then, by dividing this energy by the mass of tissue absorbing it (like the 0.50 kg of lung tissue), we determine the absorbed dose. This dose directly influences the biological effects experienced by the tissue.
Equivalent Dose
Equivalent dose is a refined concept used to assess the biological impact of exposure to different types of radiation. Simply put, it factors in the type of radiation and its relative biological effect. The unit for equivalent dose is the sievert (Sv).
The equivalent dose is calculated by multiplying the absorbed dose by a radiation weighting factor. This factor is higher for radiation types that cause more biological damage. For alpha particles, this factor is 20.
So, if a given absorbed dose was calculated as 1 gray, the equivalent dose for alpha particle radiation would be \(H = 20 \times D\). This means that for equivalent doses, it isn't only about how much energy is absorbed but also considers the potential damage from the specific type of radiation. This is crucial for understanding the potential health risks associated with different radiation exposures.
Alpha Decay
Alpha decay is a type of radioactive decay where an unstable atomic nucleus emits an alpha particle ( \(\alpha\) particle). This process decreases the atomic number by two units and the mass number by four units, leading to the transformation of the original element into another element.
For example, when uranium-238 (\(^{238}_{92}U\)) undergoes alpha decay, it becomes thorium-234 (\(^{234}_{90}Th\)). This decay reduces the atomic number due to the loss of two protons and reduces the mass due to the combined mass of the two protons and two neutrons ejected.
Alpha decay is significant because it is a common form of decay in heavy elements and releases a notable amount of energy. This energy, when absorbed by biological tissues, leads to significant radiation damage which must be carefully managed, as illustrated in our exercise's dose calculations.

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Most popular questions from this chapter

43.80. Industrial Radioactivity. Radioisotopes are used in a variety of manufacturing and testing techniques. Wear measurements can be made using the following method. An automobile engine is produced using piston rings with a total mass of 100 \(\mathrm{g}\) , which includes 9.4\(\mu \mathrm{Ci}\) of \(^{59} \mathrm{Fe}\) whose half-life is 45 days. The engine is test-run for 1000 hours, after which the oil is drained and its activity is measured. If the activity of the engine oil is 84 decays/s, how much mass was worn from the piston rings per hour of operation?

43.3. Hydrogen atoms are placed in an external magnetic field. The protons can make transitions between states in which the nuclear spin component is parallel and antiparallel to the field by absorbing or emitting a photon. What magnetic-field magnitude is required for this transition to be induced by photons with frequency 22.7 \(\mathrm{MHz}\) ?

43.22. Radioactive isotopes used in cancer therapy have a "shelf- life," like pharmaccuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of "Co is 5000 Ci. When its activity falls below 3500 \(\mathrm{Ci}\) , it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these "Co sources in your inventory was manufactured on October \(6,2004\) . It is now April \(6,2007\) . Is the source still usable? The half-life of \(^{6}\) Co is 5.271 years.

43.28. The ratio of \(^{14} \mathrm{C}\) to \(^{12} \mathrm{C}\) in living matter is measured to be \(^{14} \mathrm{C} /^{2} \mathrm{C}=1.3 \times 10^{-12}\) at the present time. A \(12.0-\mathrm{g}\) sample of carbon produces 180 decays/min due to the small amount of \(^{14} \mathrm{C}\) in it. From this information, calculate the half-life of "C.

43.74. In the 1986 disaster at the Chernobyl reactor in the Soviet Union (now Ukraine), about \(\frac{1}{8}\) of the \(^{13} \mathrm{Cs}\) present in the reactor was released. The isotope \(^{137} \mathrm{Cs}\) has a half-life for \(\beta\) decay of 30.07 \(\mathrm{y}\) and decays with the emission of a total of 1.17 \(\mathrm{MeV}\) of energy per decay. Of this, 0.51 \(\mathrm{MeV}\) goes to the emitted electron and the remaining 0.66 MeV to a \(\gamma\) ray. The radioactive 137 \(\mathrm{Cs}\) is absorbed by plants, which are eaten by livestock and humans. How many \(^{137} \mathrm{Cs}\) atoms would need to be present in each kilogram of body tissue if an equivalent dose for one week is 3.5 \(\mathrm{Sv}\) ? Assume that all of the energy from the decay is deposited in that 1.0 \(\mathrm{kg}\) of tissue and that the RBE of the electrons is 1.5 .

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