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43.68. A \(70.0-\mathrm{kg}\) person experiences a whole-body exposure to \(\alpha\) radiation with energy 4.77 \(\mathrm{MeV}\) . A total of \(6.25 \times 10^{12} \alpha\) particles are absorbed. (a) What is the absorbed dose in rad? (b) What is the equivalent dose in rem?(c) If the source is 0.0320 \(\mathrm{g}\) of 26 \(\mathrm{Ra}\) (half- life 1600 \(\mathrm{y}\) ) somewhere in the body, what is the activity of this source? (d) If all the alpha particles produced are absorbed, what time is required for this dose to be delivered?

Short Answer

Expert verified
(a) 6.83 rad. (b) 136.6 rem. (c) 1.17 billion Bq. (d) 89 minutes.

Step by step solution

01

Calculate Energy Absorbed in Joules

First, convert the energy per alpha particle from MeV to Joules. Use the conversion factor: 1 MeV = 1.602 x 10^{-13} J. So, for one alpha particle: \( 4.77 \times 1.602 \times 10^{-13} = 7.64 \times 10^{-13} \) J. Multiply this by the total number of particles (\( 6.25 \times 10^{12} \)): \( 7.64 \times 10^{-13} \times 6.25 \times 10^{12} = 4.78 \) Joules.
02

Calculate Absorbed Dose in rad

The absorbed dose in rad is calculated by dividing the total energy absorbed (in Joules) by the mass (in kg), and then converting from Joules/kg to rad. 1 rad = 0.01 J/kg, so: \( \frac{4.78}{70} = 0.06829 \) J/kg = 6.829 rad.
03

Calculate Equivalent Dose in rem

The dose equivalent in rem is calculated by multiplying the absorbed dose in rad by the radiation weighting factor for alpha particles, which is 20: \( 6.829 \times 20 = 136.58 \) rem.
04

Calculate Activity of the Radium Source

Use the decay constant \( \lambda = \frac{0.693}{T_{1/2}} \). For \( ^{226}\text{Ra} \), \( T_{1/2} = 1600 \) years. Convert to seconds: \( \lambda = \frac{0.693}{1600 \times 365.25 \times 24 \times 3600} = 1.37 \times 10^{-11} \text{s}^{-1} \). Compute the number of atoms \( N = \frac{0.032}{226} \times 6.022 \times 10^{23} = 8.53 \times 10^{19} \). Activity \( A = \lambda N = 1.17 \times 10^9 \) Bq.
05

Calculate Time for Dose Delivery

The dose rate is equal to activity times the energy per decay converted to Joules \( (4.77 \times 1.602 \times 10^{-13}) \) and also accounting for the total number of particles that needs to be absorbed within the given time span. Set \( A \times t = \text{Total Number of Particles} \). From \( 1.17 \times 10^9 \times t = 6.25 \times 10^{12} \), solve for \( t \): \( t = \frac{6.25 \times 10^{12}}{1.17 \times 10^9} = 5.34 \times 10^3 \) seconds or approximately 89 minutes.

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Key Concepts

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

Absorbed Dose
Absorbed dose refers to the amount of energy that radiation deposits in a unit mass of tissue. It describes how much energy from ionizing radiation is absorbed per kilogram of human tissue or other materials. This is important for understanding how radiation can affect the body on a cellular level.
The unit of absorbed dose is the gray (Gy), but in older literature, you might see the rad, where 1 Gy equals 100 rad. The formula used here to calculate absorbed dose in rads was \[\text{Absorbed Dose (D)} = \frac{\text{Total Energy Absorbed (in J)}}{\text{Mass (in kg)}} \times 100\]This shows the conversion factor thanks to which absorbed dose is often expressed in rads alongside modern units.
  • For example, if a person absorbs 4.78 Joules of energy, and their body mass is 70 kg, you calculate the absorbed dose as 6.829 rad.

Understanding this concept helps us predict biological effects, making it a foundational notion in radiation dosimetry.
Equivalent Dose
The equivalent dose takes the absorbed dose and modifies it with a factor that reflects the varying levels of biological impact of different types of radiation. This weighting factor acknowledges that some radiation types, like alpha particles, can be more damaging to biological tissues than others like X-rays or beta particles.
The equivalent dose uses the sievert (Sv) as its unit, but for historical notation, it can also be expressed in rems; where 1 Sv equals 100 rem.
To find the equivalent dose in rem, you multiply the absorbed dose by the radiation weighting factor for alpha particles, which is 20 according to the standard radiation protection guides:\[\text{Equivalent Dose (H)} = \text{Absorbed Dose} \times \text{Radiation Weighting Factor}\]
In our scenario, an absorbed dose of 6.829 rad is multiplied by 20 to get 136.58 rem, suggesting higher potential biological damage due to the nature of alpha particles' radiation.
  • This concept helps us understand the potential impact on human health and aids in formulating safety standards.
Radioactive Decay
Radioactive decay is the process by which unstable atomic nuclei lose energy by emitting radiation. A fundamental aspect of this process is half-life, the time taken for half of the radioactive nuclei in a sample to decay.
For radium-226 (^{226}Ra), which was used in our exercise, the half-life is 1600 years. This long half-life indicates that the radioactive substance remains active for a considerable time, continuing to emit radiation as it decays.
  • The decay constant (\lambda), which helps calculate the decay rate, is determined by the formula: \[\lambda = \frac{0.693}{T_{1/2}}\]
  • For ^{226}Ra, this was calculated as approximately 1.37 \times 10^{-11} \text{s}^{-1}.

This constant helps determine how many decays occur per second. It's crucial for calculating the activity of a radioactive source, which becomes evident in the next section.
Activity Calculation
The activity of a radioactive source is a measure of the number of decays occurring in a given time frame, expressed in becquerels (Bq), where 1 Bq equals 1 decay per second.
In the case of radium-226 with a mass given in the exercise:
  • First, you need to determine the number of radioactive atoms (N), which is calculated using the formula \[N = \frac{\text{mass} \times N_A}{\text{molar mass}}\]
  • Where N_A is Avogadro's number (6.022 \times 10^{23} ext{ mole}^{-1}), resulting in 8.53 \times 10^{19} atoms for the exercise.
  • The activity (A)is then computed using \[A = \lambda \times N\]
  • This produces an activity level of 1.17 \times 10^9 ext{ Bq} for the radium source.

Understanding activity is vital in assessing radiation safety, as it tells how radioactive materials might interact with the environment and human health 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?

43.18. What particle \((\alpha \text { particle, electron, or positron) is emitted in }\) \(_{14}^{27} \mathrm{Si} \rightarrow\) \(_{13}^{27} \mathrm{Al} ;(\mathrm{b})\) \((b)^{2 a g} U \rightarrow\) \(\frac{234}{90} \mathrm{Th}\) \((\mathrm{c})_{33}^{74} \mathrm{As} \rightarrow\) \(\sqrt[74]{5} \mathrm{se}\)

43.8. Calculate (a) the total binding energy and (b) the binding energy per nucleon of \(^{12} \mathrm{C}\) (c) What percent of the rest mass of this nucleus is its total binding energy?

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.)

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?

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