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BIO A 70.0-kg person experiences a whole-body exposure to \(\alpha\) radiation with energy 4.77 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 \(^{226} \mathrm{Ra}\) (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 × 10^9 Bq, (d) 1.48 hours.

Step by step solution

01

Calculate absorbed energy

First, we need to calculate the total absorbed energy in joules. We know that each α particle has energy 4.77 MeV and that there are \(6.25 \times 10^{12}\) particles. Convert MeV to joules using the conversion \(1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}\). Thus, the total energy absorbed is:\[6.25 \times 10^{12} \times 4.77 \times 1.602 \times 10^{-13} \text{ J} = 4.77 \times 1.602 \times 10^{-13} \times 6.25 \times 10^{12} \approx 4.78 \text{ J}\]
02

Calculate absorbed dose in rad

The formula for absorbed dose in rad is given by \( ext{Absorbed Dose} = \frac{\text{Energy absorbed (J)}}{\text{mass (kg)}} \times 100 \). Using the total energy from Step 1 (\(4.78\) J) and the body mass (70.0 kg), we get:\[\text{Absorbed Dose} = \frac{4.78}{70.0} \times 100 = 6.83 \text{ rad}\]
03

Calculate equivalent dose in rem

The equivalent dose in rem can be calculated using the formula \( ext{Equivalent Dose (rem)} = ext{Absorbed Dose (rad)} \times ext{Quality Factor} \). For α particles, the Quality Factor is 20. Thus:\[\text{Equivalent Dose} = 6.83 \times 20 = 136.6 \text{ rem}\]
04

Calculate activity of the source

Activity is calculated using the formula \( A = \frac{N \ln(2)}{t_{1/2}} \), where \(N\) is the number of atoms and \(t_{1/2}\) is the half-life in seconds. First, convert 0.0320 g of \(^ {226} Ra\) to moles, using the molar mass 226 g/mol, and then to number of atoms using Avogadro's number \( 6.022 \times 10^{23} \).Number of moles: \( \text{moles} = \frac{0.0320 \text{ g}}{226 \text{ g/mol}} \approx 1.42 \times 10^{-4} \text{ moles}\)Number of atoms: \(N = 1.42 \times 10^{-4} \times 6.022 \times 10^{23} \approx 8.55 \times 10^{19} \text{ atoms}\)Convert half-life to seconds: \(1600 \, \text{years} \times 365 \times 24 \times 3600 = 5.05 \times 10^{10} \text{ s}\)Activity: \[A = \frac{8.55 \times 10^{19} \times 0.693}{5.05 \times 10^{10}} \approx 1.17 \times 10^{9} \text{ decays/s, or Bq}\]
05

Calculate time for dose to be delivered

To find how long it would take for this dose to be delivered, calculate the time required for \(6.25 \times 10^{12}\) α particles to decay. Using the activity \(A = 1.17 \times 10^{9} \text{ Bq}\) from Step 4:\[\text{Time} = \frac{6.25 \times 10^{12}}{1.17 \times 10^{9}} \approx 5.34 \times 10^{3} \text{ s} \approx 1.48 \text{ hours}\]

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Key Concepts

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

Absorbed dose
When studying radiation, it's crucial to understand the concept of absorbed dose. The absorbed dose refers to the amount of radiation energy deposited in a material, usually tissue. It is crucial for evaluating potential health effects of radiation exposure. Measured in rads, the formula to calculate this is \( \text{Absorbed Dose (rad)} = \frac{\text{Energy absorbed (J)}}{\text{mass (kg)}} \times 100 \). This calculation gives insight into the overall radiation exposure of a body.
Let's say we have 4.78 joules of absorbed energy in a 70 kg person. Using the formula, we divide the energy by the mass and multiply by 100, resulting in an absorbed dose of 6.83 rad. This measure helps indicate the risk associated with the radiation exposure.
Equivalent dose
To comprehend the potential biological impact of different types of radiation, we use the equivalent dose. It considers not just the absorbed dose, but also the type of radiation, thereby translating physical measurements into health risks.
This dose is measured in rems and can be calculated by multiplying the absorbed dose by a quality factor: \( \text{Equivalent Dose (rem)} = \text{Absorbed Dose (rad)} \times \text{Quality Factor} \). The quality factor for alpha particles is 20, reflecting their potential for causing biological damage.
For example, with an absorbed dose of 6.83 rad from alpha radiation, the equivalent dose is \( 6.83 \times 20 = 136.6 \) rem. This calculation better reflects the comparative biological harm of different radiation types.
Radioactive decay
Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This decay is random for individual atoms but predictable across a large quantity of atoms.
The activity of a radioactive sample, which signifies how many decays occur per second, is measured in becquerels (Bq). It's determined by the number of radioactive atoms and their half-life, a fixed period in which half of the radioactive nuclei decay. For instance, calculate the activity \( A \) of radium-226 with a half-life of 1600 years in a 0.0320 g sample. First, convert mass to moles and then to atoms. Use Avogadro's number, then apply \( A = \frac{N \ln(2)}{t_{1/2}} \). The calculated activity turns out to be approximately \( 1.17 \times 10^{9} \) Bq, reflecting the number of alpha particles emitted per second.
Alpha particles
Alpha particles are a type of ionizing radiation made up of 2 protons and 2 neutrons, essentially a helium nucleus. They are emitted by certain radioactive materials, like radium, during radioactive decay.
Due to their relatively large mass and charge, alpha particles have a low penetration ability. They can be stopped by just a piece of paper or the outer dead layer of human skin. However, if ingested or inhaled, they can be very harmful as they tend to deposit their energy in localized tissues.
  • Alpha particles travel only short distances.
  • They have high linear energy transfer (LET), meaning they deposit a lot of energy in a small area.
  • Useful in smoke detectors and medical treatments but dangerous with internal exposure.
This inability to penetrate effectively makes them less harmful externally but a significant threat internally.

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Most popular questions from this chapter

Consider the nuclear reaction $$ _{14}^{28} \mathrm{Si}+\gamma \rightarrow_{12}^{24} \mathrm{Mg}+\mathrm{X} $$ where \(X\) is a nuclide. (a) What are \(Z\) and \(A\) for the nuclide \(X\) ? (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a \(_{14}^{28}\) Si atom is 27.976927 \( \mathrm{u},\) and the mass of a \(^{24}_{12} \mathrm{Mg}\) atom is 23.985042 \(\mathrm{u}\)

BIO (a) If a chest x ray delivers 0.25 \(\mathrm{mSv}\) to 5.0 \(\mathrm{kg}\) of tissue, how many total joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about 0.10 \(\mathrm{mSv}\) per year at sea level. Assuming an \(\mathrm{RBE}\) of \(1,\) how many rem and rads is this dose, and how many joules of energy does a 75 -kg person receive in a year? (c) How many chest xays like the one in part (a) would it take to deliver the same total amount of energy to a 75 -kg person as she receives from natural radiation in a year at sea level, as described in part (b)?

The common isotope of uranium, \(^{238} \mathrm{U},\) has a half-life of \(4.47 \times 10^{9}\) years, decaying to \(^{234} \mathrm{Th}\) by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by 10.0 g of uranium?

\(\mathrm{A}^{60} \mathrm{Co}\) source with activity \(2.6 \times 10^{-4} \mathrm{Ci}\) is embedded in a tumor that has mass 0.200 \(\mathrm{kg} .\) The source emits \(\gamma\) photons with average energy 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 RBE for these \(\gamma\) rays is 0.70\(?\) (d) What exposure time is required for an equivalent dose of 200 rem?

BIO To Scan or Not to Scan? It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans) using x rays, just to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 \(\mathrm{mSv}\) , applied to the whole body. By contrast, a chest \(x\) ray typically administers 0.20 mSv to only 5.0 kg of tissue. How many chest \(x\) rays would deliver the same total amount of energy to the body of a 75 -kg person as one whole-body scan?

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