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Chapter 43: Problem 69

43.69. A "co source with activity \(2.6 \times 10^{-4} \mathrm{Cl}\) is embedded in a tumor that has mass 0.500 \(\mathrm{kg}\) . The source emits \(\gamma\) photons with average energy 1.25 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 RBE for these \(\gamma\) rays is 0.70\(?\) (d) What exposure time is required for an equivalent dose of 200 rem? (a) after 4.0 minutes and

Short Answer

Expert verified
About 9.63 x 10^-7 J/s is delivered; 1.926 x 10^-5 rad/s absorbed dose; 1.3482 x 10^-5 rem/s equivalent dose; 4.0 minutes exposure time for 200 rem.

Step by step solution

01

Understanding the Problem

We have a gamma source with an activity of \(2.6 \times 10^{-4} \mathrm{Ci}\), 50% photon absorption in a tumor of 0.500 kg, and the energy of each photon is 1.25 MeV. We need to calculate energy delivered, absorbed dose, equivalent dose, and exposure time for a specific dose.
02

Calculate Activity in Decays per Second

Convert activity from curies to decays per second: \(1 \, \mathrm{Ci} = 3.7 \times 10^{10} \, \mathrm{decays/sec}\). So the activity is \(2.6 \times 10^{-4} \times 3.7 \times 10^{10} = 9.62 \times 10^6 \, \mathrm{decays/sec}\).
03

Calculate Energy Per Decay

The energy per decay is given as 1.25 MeV. Converting to joules using \(1 \, ext{MeV} = 1.602 imes 10^{-13} \, ext{J}\), it is \(1.25 \times 1.602 \times 10^{-13} = 2.0025 \times 10^{-13} \, ext{J}\).
04

Compute Total Energy Delivered Per Second

Since half the energy is absorbed, the energy absorbed per second is \(\frac{1}{2} \times 9.62 \times 10^6 \times 2.0025 \times 10^{-13} = 9.63 \times 10^{-7} \, ext{J/s}\).
05

Calculate Absorbed Dose in rad

The absorbed dose is energy per unit mass in rads, with 1 rad = 0.01 J/kg. Thus, absorbed dose = \(\frac{9.63 \times 10^{-7}}{0.5} \times 0.01 = 1.926 \times 10^{-5} \, ext{rad/s}\).
06

Calculate Equivalent Dose in rem

Equivalent dose = \( ext{Absorbed dose} \times \text{RBE} \). Using RBE of 0.70, \(1.926 \times 10^{-5} \times 0.70 = 1.3482 \times 10^{-5} \, ext{rem/s}\).
07

Determine Exposure Time for 200 rem

To achieve an equivalent dose of 200 rem, exposure time \(t = \frac{200}{1.3482 \times 10^{-5}} = 1.483 \times 10^7 \, ext{seconds}\), or approximately 4.0 minutes.

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Key Concepts

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

Gamma Rays
Gamma rays are a form of electromagnetic radiation, much like X-rays, but with higher energy and the ability to penetrate materials more deeply. They originate from the radioactive decay of atomic nuclei. While gamma rays are invisible and cannot be sensed by humans, their energetic photons can impact living tissues, making them important in fields like radiation therapy and nuclear medicine.
Gamma rays possess short wavelengths and high frequencies, allowing them to carry massive amounts of energy. This energy is harnessed in medical treatments to target diseased tissues, such as tumors, but it is also what demands careful control and management due to potential biological damage.
  • Gamma rays have no mass or charge.
  • They travel at the speed of light, as all electromagnetic waves do.
  • Their ability to ionize and break chemical bonds poses risks as well as therapeutic opportunities.
Absorbed Dose
The absorbed dose quantifies the amount of energy deposited by radiation in a specific mass of tissue. It is a key concept in radiation dosimetry, serving as a fundamental measure of the radiation's potential to cause biological effects. The unit for absorbed dose is the gray (Gy), where 1 Gy is equivalent to an energy absorption of 1 joule per kilogram.
In many instances, the absorbed dose is still measured in rads, a traditional unit where 1 rad is equivalent to 0.01 joules per kilogram. Evaluating the absorbed dose helps determine the ionizing radiation's direct impact on tissues, either damaging cellular structures or aiding in therapeutic outcomes, depending on usage.
  • Used broadly in medicine to calculate doses in cancer treatments.
  • It establishes a correlation between exposure and biological effect.
  • Vital in assessing radiation protection standards.
Equivalent Dose
The equivalent dose considers not only the absorbed dose but also the biological impact of the type of radiation received. Different types of radiation have various effectiveness at causing biological damage, known as the relative biological effectiveness (RBE).
The unit for equivalent dose is the sievert (Sv), and it is calculated by multiplying the absorbed dose by the RBE. This conversion allows for a more comprehensive assessment of the potential harm caused by different forms of radiation, offering a standardized measure that helps protect individuals from harmful radiation effects.
  • This measurement is essential in radiation protection, optimizing safe levels of exposure.
  • Different tissues have different sensitivities; hence, equivalent dose helps in risk assessment.
  • Used extensively in occupational health and safety regulations within radiological environments.
Radioactive Decay
Radioactive decay is the process by which unstable atomic nuclei lose energy by emitting radiation in the form of particles or electromagnetic waves like gamma rays. Over time, this decay transforms them into more stable forms.
The rate of radioactive decay is characterized by the half-life, the time required for half of the radioactive atoms in a sample to decay. This natural process is fundamental to a variety of applications, such as dating archaeological finds through carbon dating or powering pacemakers with long-lasting isotopes.
  • Radioactivity is measured in activity, where 1 curie equals 3.7 脳 1010 decays per second.
  • Understanding decay helps in managing and utilizing radioactive sources carefully in medical and industrial sectors.
  • The decay process reduces the radioactivity and changes the elemental composition over time.

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Most popular questions from this chapter

43.70. The nucleus \(\frac{15}{8} \mathrm{O}\) has a half-life of \(122.2 \mathrm{s} ; \mathrm{g} \mathrm{O}\) has a half-life of 26.9 s. If at some time a sample contains eqnal amounts of \(\frac{15}{8} \mathrm{O}\) and \(_{8}^{19} \mathrm{O}\) what is the ratio of \(_{8}^{15} 0\) to \(_{8}^{19} \mathrm{O}\) (b) after 15.0 minutes?

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 \(\mathrm{MHz}\) ?

43.72. 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?

43.24. Radioactive Tracers. Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. 131 \(\mathrm{I}\) , a \(B^{-}\) emitter with a half-life of 8.0 \(\mathrm{d}\) is one such tracer. Suppose a scientist introduces a sample with an activity of 375 \(\mathrm{Bq}\) and watches it spread to the organs. (a) Assuming that the sample all went to the thyroid gland, what will be the decay rate in that gland 24 \(\mathrm{d}\) (about 2\(\frac{1}{2}\) weeks) later? (b) If the decay rate in the thyroid 24 d later is actually measured to be 17.0 \(\mathrm{Bq}\) , what percentage of the tracer went to that gland? (c) What isotope remains after the \(\mathrm{I}-131\) decays?

43.52. Comparison of Energy Released per Gram of Fuel. (a) When gasoline is burned, it releases \(1.3 \times 10^{8} \mathrm{J}\) of energy per gallon \((3.788 \mathrm{L})\) . Given that the density of gasoline is 737 \(\mathrm{kg} / \mathrm{m}^{3}\) , express the quantity of energy released in \(\mathrm{J} / \mathrm{g}\) of fuel. (b) During fission, when a neutron is absorbed by a \(^{235} \mathrm{U}\) nucleus, about 200 \(\mathrm{MeV}\) of energy is released for each nucleus that undergoes fission. Express this quantity in \(\mathrm{J} / \mathrm{g}\) of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one 4 \(\mathrm{He}\) nucleus with two leftover protons and the liberation of 26.7 \(\mathrm{MeV}\) of energy. The fuel is the six protons. Express the energy produced here in units of \(\mathrm{J} / \mathrm{g}\) of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other (d) Our sun produces energy at a measured rate of \(3.86 \times 10^{26} \mathrm{W}\) . If its mass of \(1.99 \times 10^{30} \mathrm{kg}\) were all gasoline, how long could it last before consuming all its fuel? (Historical note: Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)
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