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\(\mathrm{A}^{60} \mathrm{Co}\) source with activity \(15.0 \mathrm{Ci}\) is embedded in a tumor that has a mass of \(0.500 \mathrm{~kg}\). The Co source emits gamma-ray photons with average energy of \(1.25 \mathrm{MeV}\). Half the photons are absorbed in the tumor, and half escape. (a) What energy is delivered to the tumor per second? (b) What absorbed dose (in rad) is delivered per second? (c) What equivalent dose (in rem) is delivered per second if the \(\mathrm{RBE}\) for these gamma rays is \(0.70 ?\) (d) What exposure time is required for an equivalent dose of 200 rem?

Short Answer

Expert verified
Energy delivered to the tumor per second is 1.11 Watts, absorbed dose is 222 rad/s, equivalent dose is 155 rem/s, and exposure time is approximately 1.29 seconds.

Step by step solution

01

Convert Activity from Curie to Disintegrations per Second

First, we need to convert the activity from curies to disintegrations per second (becquerels). 1 Ci is equal to \( 3.7 \times 10^{10} \) disintegrations per second. Therefore, the activity in becquerels is \( 15.0 \times 3.7 \times 10^{10} \).
02

Calculate Energy Emission per Second

Next, we calculate the total energy emitted per second. Since each disintegration releases a photon with an energy of 1.25 MeV, we convert this energy into joules (\(1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J} \)). Then we find the energy emitted per second as \( E = 15.0 \times 3.7 \times 10^{10} \times 1.25 \times 1.602 \times 10^{-13} \).
03

Calculate Energy Absorbed by the Tumor

The exercise states that only half of the photons are absorbed by the tumor. Therefore, the energy absorbed per second is half of the total energy emitted calculated in Step 2. Thus, \( E_{\text{absorbed}} = \frac{1}{2} \times E \).
04

Calculate Absorbed Dose in rad

The absorbed dose (in rad) is calculated as the energy absorbed per unit mass of the tumor. Since 1 rad is equal to 0.01 J/kg, we convert the absorbed energy from joules to rads for the mass of the tumor (0.5 kg). \( \text{Dose} = \frac{E_{\text{absorbed}}}{0.5} \times 100 \).
05

Calculate Equivalent Dose in rem

The equivalent dose is calculated by multiplying the absorbed dose by the relative biological effectiveness (RBE). Thus, the equivalent dose in rem is \( \text{Equivalent Dose} = \text{Dose} \times \text{RBE} = \text{Dose} \times 0.70 \).
06

Calculate Exposure Time for Equivalent Dose of 200 rem

To find the exposure time required for a dose of 200 rem, we use the rate at which the equivalent dose is being delivered. The time \( t = \frac{200}{\text{Equivalent Dose}} \).

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Key Concepts

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

Absorbed Dose
In radiation dosimetry, the absorbed dose is a measure of how much radiation energy is deposited in a given mass of tissue. It is expressed in units called rads, where 1 rad is equivalent to the absorption of 0.01 joules of radiation energy per kilogram of tissue. Calculating the absorbed dose helps in understanding the potential physical effects on the tissue.
For example, when calculating the absorbed dose from a \( ^{60} \mathrm{Co} \) source, as in the exercise, it is important to first determine the energy absorbed by the tissue. If the activity of the source and energy of the emitted gamma rays are known, the energy absorbed by the tumor can be calculated by focusing on how much energy is deposited per second. This is particularly crucial in medical contexts like radiation therapy, where precise dosages are necessary for effective treatment.
Understanding absorbed dose is fundamental for anyone working in environments where radiation is present, as it directly relates to the safety and efficacy of radiation applications.
Equivalent Dose
The concept of equivalent dose is vital when considering the biological effects of different types of radiation on human tissue. While absorbed dose tells us how much energy is deposited, it does not account for the type of radiation. This is where equivalent dose comes into play.
Equivalent dose is measured in sieverts (Sv) but often converted to rems (where 1 Sv = 100 rems), and it is calculated by multiplying the absorbed dose by a factor that describes the biological damage caused by the type of radiation, known as the relative biological effectiveness (RBE) or quality factor.
In the case of gamma rays released from a \( ^{60} \mathrm{Co} \) source, with an RBE of 0.70, the equivalent dose would be the absorbed dose multiplied by this RBE. This evaluation allows healthcare professionals to compare the biological effects of radiation from different sources, ensuring the safe and effective delivery of treatment dosages.
Gamma Rays
Gamma rays are a form of electromagnetic radiation with very high energy and the ability to penetrate deep into tissues. They are commonly used in medical imaging and therapy due to their penetrating power.
The gamma rays emitted from radioactive sources, like \( ^{60} \mathrm{Co} \), are critical in various medical treatments, especially in the treatment of cancer via radiation therapy. These rays can target and destroy cancer cells, although they must be used carefully to minimize damage to surrounding healthy tissues.
Understanding gamma rays' characteristics, including their energy and penetration ability, is essential for their safe use in medical settings. This involves knowing how to calculate their absorption and ways to shield or mitigate their effects in environments where their use is prevalent.
Radiation Therapy
Radiation therapy is a method used to treat cancer and other conditions by using high doses of radiation to kill or damage cancer cells. It can shrink tumors and is often used in conjunction with other treatments.
The precise calculation of radiation exposure, as per the absorbed and equivalent doses, is crucial in radiation therapy to ensure maximum efficiency with minimal damage to healthy tissues. The use of sources like \( ^{60} \mathrm{Co} \) in radiation therapy takes advantage of the intense energy of gamma rays to target specific areas in the body.
While highly effective, radiation therapy requires careful planning and dosimetry calculations to optimize the treatment protocol, ensuring that the prescribed dose is administrated safely and effectively.

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Most popular questions from this chapter

A photon with a wavelength of \(3.50 \times 10^{-13} \mathrm{~m}\) strikes a deuteron, splitting it into a proton and a neutron. (a) Calculate the kinetic energy released in this interaction. (b) Assuming the two particles share the energy equally, and taking their masses to be \(1.00 \mathrm{u},\) calculate their speeds after the photodisintegration.

Because iodine in the body is preferentially taken up by the thyroid gland, radioactive iodine in small doses is used to image the thyroid, and in large doses it is used to kill thyroid cells to treat some types of cancer or thyroid disease. The iodine isotopes used have relatively short half-lives, so they must be produced in a nuclear reactor or accelerator. One isotope frequently used for imaging is \({ }^{123} \mathrm{I} ;\) it has a half-life of 13.2 hours and emits a \(0.16 \mathrm{MeV}\) gamma ray. One method of producing \({ }^{123} \mathrm{I}\) is in the nuclear reaction \({ }^{123} \mathrm{Te}+\mathrm{p} \rightarrow{ }^{123} \mathrm{I}+\mathrm{n}\). The atomic masses relevant to this reaction are \(^{123} \mathrm{Te}: 122.90427 \mathrm{u} ;{ }^{123} \mathrm{I}: 122.90559 \mathrm{u} ; \mathrm{n}: 1.008665 \mathrm{u} ;{ }^{1} \mathrm{H}: 1.007825 \mathrm{u}\). The iodine isotope commonly used for treatment is \({ }^{131} \mathrm{I}\), which is produced by irradiating \({ }^{130} \mathrm{Te}\) in a nuclear reactor to form \({ }^{131} \mathrm{Te}\). The \({ }^{131}\) Te then decays to \({ }^{131}\) I. \({ }^{131}\) I undergoes \(\beta\) decay with a halflife of 8.04 days, emitting electrons with energies up to \(0.61 \mathrm{MeV}\) and gamma rays with energy \(0.36 \mathrm{MeV}\). A typical thyroid cancer treatment might involve the administration of \(3.7 \mathrm{GBq}\) of \({ }^{131} \mathrm{I}\). Which reaction produces \({ }^{131}\) Te in the nuclear reactor? A. \({ }^{130} \mathrm{Te}+\mathrm{n} \rightarrow{ }^{131} \mathrm{Te}\) B. \({ }^{130} \mathrm{I}+\mathrm{n} \rightarrow{ }^{131} \mathrm{Te}\) C. \({ }^{132} \mathrm{Te}+\mathrm{n} \rightarrow{ }^{131} \mathrm{Te}\) D. \({ }^{132} \mathrm{I}+\mathrm{n} \rightarrow{ }^{131} \mathrm{Te}\)

Calcium- 47 is a \(\beta^{-}\) emitter with a half-life of 4.5 days. If a bone sample contains \(2.24 \mathrm{~g}\) of this isotope, at what rate will it decay?

Nuclear weapons tests in the \(1950 \mathrm{~s}\) and \(1960 \mathrm{~s}\) released significant amounts of radioactive tritium \(\left({ }_{1}^{3} \mathrm{H},\right.\) half-life 12.3 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, \({ }_{2}^{3} \mathrm{He},\) to the remaining tritium in the water. For example, if the ratio of \({ }_{2}^{3} \mathrm{He}\) to \({ }_{1}^{3} \mathrm{H}\) 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 \({ }_{2}^{3} \mathrm{He}\) to \({ }_{1}^{3} \mathrm{H}\) is \(4.3: 1.0 .\) How many years ago did this water sink below the surface?

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