Calculate the dose in rem/y for the lungs of a weapons plant employee who inhales and retains an activity of \(1.00 \mu \mathrm{Ci}^{239} \mathrm{Pu}\) in an accident. The mass of affected lung tissue is \(2.00 \mathrm{kg}\) and the plutonium decays by emission of a 5.23-MeV \(\alpha\) particle. Assume a RBE value of 20.

Short Answer

Expert verified
The dose in rem/y for the lungs of the weapons plant employee who inhales and retains an activity of \(1.00 \mu \mathrm{Ci}^{239} \mathrm{Pu}\) is approximately 9.78 rem/y.

Step by step solution

01

Convert activity to decays per second (Bq)

To convert the activity from microcuries to Bq (decays per second), we will use the conversion factor 1 Ci = \(3.7 \times 10^{10}\) Bq: Activity (Bq) = Activity (\(\mu\)Ci) × \(3.7 \times 10^{10}\) Bq/Ci × \(10^{-6}$$\mu\)Ci/Ci Activity (Bq) = \(1.00 \times 3.7 \times 10^{10} \times 10^{-6}\) Bq Activity (Bq) = 37,000 Bq
02

Calculate energy deposited per second in the lung tissue

Energy of emitted alpha particles = 5.23 MeV To convert MeV to Joules, we will use the conversion factor 1 eV = \(1.602 \times 10^{-19}\) J: Energy of emitted alpha particles (J) = 5.23 MeV × \(1.602 \times 10^{-13}\) J/MeV Energy of emitted alpha particles (J) = \(8.384 \times 10^{-13}\) J Now, we can calculate the energy deposited in the lung tissue per second (Power): Power (J/s) = Activity (Bq) × Energy of emitted alpha particles (J) Power (J/s) = 37,000 Bq × \(8.384 \times 10^{-13}\) J Power (J/s) = \(3.102 \times 10^{-8}\) J/s
03

Convert the energy to dose in Gy (Gray)

To find the dose in Gy, we need to divide the power by the mass of the affected lung tissue: Dose (Gy/s) = Power (J/s) / Mass (kg) Dose (Gy/s) = \(3.102 \times 10^{-8}\) J/s / 2.00 kg Dose (Gy/s) = \(1.551 \times 10^{-8}\) Gy/s
04

Convert the dose from Gy to rem, considering the RBE value

Now, we can convert the Gy dose into rem using the given RBE of 20: Dose (rem/s) = Dose (Gy/s) × RBE Dose (rem/s) = \(1.551 \times 10^{-8}\) Gy/s × 20 Dose (rem/s) = \(3.102 \times 10^{-7}\) rem/s
05

Convert the dose from rem/s to rem/y

Lastly, we need to convert the dose from rem/s to rem/y: Dose (rem/y) = Dose (rem/s) × Number of seconds in a year Dose (rem/y) = \(3.102 \times 10^{-7}\) rem/s × \(3.1536 \times 10^{7}\) s/y Dose (rem/y) ≈ 9.78 rem/y Thus, the dose in rem/y for the lungs of the weapons plant employee who inhales and retains an activity of \(1.00 \mu \mathrm{Ci}^{239} \mathrm{Pu}\) is approximately 9.78 rem/y.

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Key Concepts

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

Activity Calculation
When we refer to 'activity' in radiation physics, we are talking about how many nuclear decays occur in a material over time. This rate of decay is crucial for determining how much radiation is released by a radioactive source. To accurately calculate activity, which is measured in becquerels (Bq), one must consider the type and amount of radioactive material.

For example, in the exercise provided, the activity of plutonium-239 is converted from microcuries to becquerels using the conversion factor of 1 curie (Ci) to 3.7 x 1010 Bq. Understanding this conversion is essential, as different units are used depending on the context. Health physicists, engineers, and technicians rely on these calculations to ensure safety and compliance with regulatory limits.
Energy Conversion
Dealing with radioactive materials often requires conversions between different units of energy. Energy released in radioactive decay is typically given in megaelectronvolts (MeV), a unit common in nuclear physics. To relate this energy to other forms such as heat or work, it is often necessary to convert MeV to joules (J), the SI unit of energy and work.

In the provided example, each decay releases 5.23 MeV of energy, which is then converted to joules using the factor that 1 eV equals 1.602 x 10-19 joules. This step is vital for further calculations, such as determining the absorbed dose, because it places the radioactive decay energy in a context understandable in terms of the effects on human tissue or other materials.
Absorbed Dose
The absorbed dose is a measure of the radiation energy absorbed per unit mass of tissue and is expressed in the unit gray (Gy). It is a critical concept for understanding the potential for biological damage from exposure to radiation. One gray is equivalent to the absorption of one joule of radiation energy by one kilogram of matter.

In practical terms, as seen in our example, it's important to divide the total energy deposited in the tissue (in joules) by the mass of the tissue (in kilograms) to obtain the absorbed dose in grays. However, to fully assess the potential biological effects of the radiation, one must also consider the type of radiation involved, which is where the concept of relative biological effectiveness comes into play.
Relative Biological Effectiveness (RBE)
The concept of Relative Biological Effectiveness (RBE) is used to compare the biological damage caused by different types of ionizing radiation. Not all radiation has the same effect on living tissue; for instance, alpha particles are much more damaging than beta particles or gamma rays when they all deposit the same energy in a tissue.

In our scenario, the RBE for alpha particles is given as 20 times more damaging than the standard reference of 1 for x-rays. Thus, the absorbed dose in gray must be multiplied by the RBE to convert it to an equivalent dose in rems (or sieverts in SI units), which better represents the biological risk. The practice of considering RBE is essential for ensuring that safety standards and radiological protections reflect the true potential for harm from different types of radiation exposure.

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Most popular questions from this chapter

A radioactive sample initially contains \(2.40 \times 10^{-2}\) mol of a radioactive material whose half-life is 6.00 h. How many moles of the radioactive material remain after \(6.00 \mathrm{h}\) ? After 12.0 h? After 36.0 h?

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An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than \(1 / 1000\) the normal amount of \(^{14} \mathrm{C}\). Estimate the minimum age of the charcoal, noting that \(2^{10}=1024\).

Verify that the total number of nucleons, and total charge are conserved for each of the following fusion reactions in the proton-proton chain. (i) \(^{1} \mathrm{H}+^{1} \mathrm{H} \rightarrow^{2} \mathrm{H}+e^{+}+v_{\mathrm{e}},\) (ii) \(^{1} \mathrm{H}+^{2} \mathrm{H} \rightarrow^{3} \mathrm{He}+\gamma,\) and (iii) \(^{3} \mathrm{He}+^{3} \mathrm{He} \rightarrow^{4} \mathrm{He}+^{1} \mathrm{H}+^{1} \mathrm{H}\). (List the value of each of the conserved quantities before and after each of the reactions.)

The detail that you can observe using a probe is limited by its wavelength. Calculate the energy of a particle that has a wavelength of \(1 \times 10^{-16} \mathrm{m},\) small enough to detect details about one-tenth the size of a nucleon.

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