Problem 68 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(a) 6.83 rad, (b) 136.6 rem, (c) 1.17 × 10^9 Bq, (d) 1.48 hours.

Step 1: 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:\[6.25 \times 10^{12} \times 4.77 \times 1.602 \times 10^{-13} \text{ J} = 4.77 \times 1.602 \times 10^{-13} \times 6.25 \times 10^{12} \approx 4.78 \text{ J}\]

Step 2: 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:\[\text{Absorbed Dose} = \frac{4.78}{70.0} \times 100 = 6.83 \text{ rad}\]

Step 3: 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:\[\text{Equivalent Dose} = 6.83 \times 20 = 136.6 \text{ rem}\]

Step 4: 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: \[A = \frac{8.55 \times 10^{19} \times 0.693}{5.05 \times 10^{10}} \approx 1.17 \times 10^{9} \text{ decays/s, or Bq}\]

Step 5: 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:\[\text{Time} = \frac{6.25 \times 10^{12}}{1.17 \times 10^{9}} \approx 5.34 \times 10^{3} \text{ s} \approx 1.48 \text{ hours}\]

Key Concepts

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.