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Chapter 42: Problem 67

An organic sample of mass \(4.00 \mathrm{~kg}\) absorbs \(2.00 \mathrm{~mJ}\) via slow neutron radiation \((\mathrm{RBE}=5)\). What is the dose equivalent (mSv)?

Short Answer

Expert verified
The dose equivalent is 2.5 mSv.

Step by step solution

01

Understand the Given Data

We are given the following:- Mass of sample: \(4.00\, \text{kg}\)- Absorbed energy: \(2.00\, \text{mJ}\)- Relative Biological Effectiveness (RBE): \(5\)
02

Calculate the Absorbed Dose in Grays

The absorbed dose \(D\) is calculated using the formula:\[ D = \frac{E}{m} \]where \(E\) is the absorbed energy in joules and \(m\) is the mass in kilograms. First, convert \(2.00\, \text{mJ}\) to joules:\[ 2.00\, \text{mJ} = 2.00 \times 10^{-3}\, \text{J} \]Thus, the absorbed dose is:\[ D = \frac{2.00 \times 10^{-3}}{4.00} = 0.5 \times 10^{-3}\, \text{Gy} \]
03

Convert the Absorbed Dose to Dose Equivalent

The dose equivalent \(H\) is calculated by multiplying the absorbed dose \(D\) by the RBE:\[ H = D \times \text{RBE} \]Substituting the given values:\[ H = (0.5 \times 10^{-3}\, \text{Gy}) \times 5 = 0.0025\, \text{Sv} \]
04

Convert the Dose Equivalent to Millisieverts

Finally, convert the dose equivalent from sieverts to millisieverts by multiplying by 1000:\[ 0.0025\, \text{Sv} = 2.5\, \text{mSv} \]

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Key Concepts

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

Understanding Absorbed Dose
To fully grasp the concept of absorbed dose, let's break it down. The absorbed dose is a measure of the amount of energy deposited by radiation in a sample. It is expressed in grays (Gy), where one gray corresponds to one joule of energy absorbed per kilogram of material.

To find the absorbed dose, use the formula \( D = \frac{E}{m} \) where \( E \) is the energy absorbed in joules and \( m \) is the mass of the sample in kilograms. In the given exercise, converting the absorbed energy from millijoules to joules allows calculation in the right units. A small absorbed dose might seem negligible, but it is crucial in determining the biological impact of the radiation, making it a foundational concept in radiological science.
  • Unit: Gray (Gy) = Energy (in joules) / Mass (in kilograms)
  • Significance: Determines the biological impact when combined with other factors like RBE.
Exploring Relative Biological Effectiveness (RBE)
Relative Biological Effectiveness (RBE) is a crucial factor when evaluating the impact of radiation on biological tissues. It accounts for different types of radiation causing varying degrees of biological damage, even if the absorbed doses are the same.

RBE is introduced into our dose calculations to provide a better understanding of the potential biological risks associated with different radiation types. It acts as a weighting factor that, when multiplied by the absorbed dose, gives the dose equivalent, presented in sieverts (Sv). In the exercise, RBE is given as 5 for slow neutron radiation, reflecting its relatively higher potential to cause damage compared to other forms of radiation such as x-rays or gamma rays.
  • Purpose: Adjusts absorbed dose to reflect biological harm.
  • Usage: Multiplies absorbed dose to determine the dose equivalent.
  • Example: Neutron radiation might have a higher RBE than x-rays.
Understanding Neutron Radiation
Neutron radiation, unlike electromagnetic radiation like x-rays, consists of free neutrons. These particles are uncharged, allowing them to penetrate materials more deeply without interacting extensively with electrons.

Neutron radiation is particularly significant in contexts such as nuclear reactors and cosmic radiation. Because it doesn't ionize directly upon entry, it transfers energy to nuclei in the absorbing material, making it effective in causing nuclear reactions.

In the context of the exercise, slow neutron radiation has a high RBE, meaning it can inflict considerable biological damage. Understanding neutron radiation is crucial for fields that involve exposure, such as nuclear engineering, space travel, and medical therapies using neutron beams for cancer treatment.
  • Nature of Particles: Neurons have no charge.
  • Penetration Ability: High, due to low interaction with electrons.
  • Applications: Used in medicine, research, and nuclear energy.

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Most popular questions from this chapter

Plutonium isotope \({ }^{239}\) Pu decays by alpha decay with a halflife of \(24100 \mathrm{y}\). How many milligrams of helium are produced by an initially pure \(12.0 \mathrm{~g}\) sample of \({ }^{239} \mathrm{Pu}\) at the end of \(20000 \mathrm{y}\) ? (Consider only the helium produced directly by the plutonium and not by any by-products of the decay process.)

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays $$ { }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C} \quad \text { and } \quad{ }^{223} \mathrm{Ra} \rightarrow{ }^{219} \mathrm{Rn}+{ }^{4} \mathrm{He} $$ Calculate the \(Q\) value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is \(30.0 \mathrm{MeV}\). What is the barrier height for \({ }^{14} \mathrm{C}\) emission? (Be careful about the nuclear radii.) The needed atomic masses are \(\begin{array}{ll}{ }^{223} \mathrm{Ra} & 223.01850 \mathrm{u} \\ { }^{209} \mathrm{~Pb} & 208.98107 \mathrm{u}\end{array}\) $$ { }^{14} \mathrm{C} \quad 14.00324 \mathrm{u} $$ \(14.00324 \mathrm{u}\) \(4.00260 \mathrm{u}\) \({ }^{4} \mathrm{He}\) $$ { }^{219} \mathrm{Rn} \quad 219.00948 \mathrm{u} $$

What is the mass excess \(\Delta_{1}\) of \({ }^{1} \mathrm{H}\) (actual mass is \(\left.1.007825 \mathrm{u}\right)\) in (a) atomic mass units and (b) \(\mathrm{MeV} / \mathrm{c}^{2}\) ? What is the mass excess \(\Delta_{\mathrm{n}}\) of a neutron (actual mass is \(1.008665 \mathrm{u}\) ) in (c) atomic mass units and (d) \(\mathrm{MeV} / \mathrm{c}^{2} ?\) What is the mass excess \(\Delta_{120}\) of \({ }^{120} \mathrm{Sn}\) (actual mass is \(119.902197 \mathrm{u})\) in (e) atomic mass units and (f) \(\mathrm{MeV} / \mathrm{c}^{2}\) ?

SSM After long effort, in 1902 Marie and Pierre Curie succeeded in separating from uranium ore the first substantial quantity of radium, one decigram of pure \(\mathrm{RaCl}_{2}\). The radium was the radioactive isotope \({ }^{226} \mathrm{Ra}\), which has a half-life of \(1600 \mathrm{y}\). (a) How many radium nuclei had the Curies isolated? (b) What was the decay rate of their sample, in disintegrations per second?

After a brief neutron irradiation of silver, two isotopes are present: \({ }^{108} \mathrm{Ag}\left(T_{1 / 2}=2.42 \mathrm{~min}\right)\) with an initial decay rate of \(3.1 \times\) \(10^{5} / \mathrm{s}\), and \({ }^{110} \mathrm{Ag}\left(T_{1 / 2}=24.6 \mathrm{~s}\right)\) with an initial decay rate of \(4.1 \times\) \(10^{6} / \mathrm{s}\). Make a semilog plot similar to Fig. 42-9 showing the total combined decay rate of the two isotopes as a function of time from \(t=0\) until \(t=10 \mathrm{~min}\). We used Fig. 42-9 to illustrate the extraction of the half-life for simple (one isotope) decays. Given only your plot of total decay rate for the two-isotope system here, suggest a way to analyze it in order to find the half-lives of both isotopes.
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