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Chapter 42: Problem 21

(II) A 1.0-cm-thick lead target reduces a beam of gamma rays to \(25 \%\) of its original intensity. What thickness of lead will allow only one \(\gamma\) in \(10^{6}\) to penetrate (see Problem 20)?

Short Answer

Expert verified
The required thickness of lead is approximately 7.98 cm.

Step by step solution

01

Define the relationship between intensity and thickness

When a gamma ray beam passes through a material, its intensity is reduced according to the exponential attenuation law. This is expressed by the formula: \[ I = I_0 e^{-\mu x} \] where \( I \) is the final intensity, \( I_0 \) is the initial intensity, \( \mu \) is the linear attenuation coefficient of the material, and \( x \) is the thickness of the material.
02

Determine the linear attenuation coefficient (\(\mu\))

We know that a 1.0-cm thick lead target reduces the gamma rays intensity to \(25\%\). This translates to: \[ \frac{I}{I_0} = 0.25 = e^{-\mu \times 1.0} \]. Taking the natural logarithm of both sides, we have: \[ \ln(0.25) = -\mu \times 1.0 \]. This gives us: \[ \mu = -\ln(0.25) \].
03

Use the attenuation formula to find required thickness

We need to find the thickness \( x \) that reduces the intensity to \( \frac{1}{10^6} \) of its original value, given by: \[ \frac{1}{10^6} = e^{-\mu x} \]. Taking the natural logarithm of both sides results in:\[ \ln\left(\frac{1}{10^6}\right) = -\mu x \]. Solve for \( x \) as follows:\[ x = -\frac{\ln\left(\frac{1}{10^6}\right)}{\mu} \].
04

Calculate the thickness (\(x\))

First, substitute the value of \( \mu \) from Step 2 into the expression for \( x \): \[ x = -\frac{\ln(10^{-6})}{-\ln(0.25)} \]. This simplifies to: \[ x = \frac{6 \cdot \ln(10)}{\ln(4)} \].Calculating this expression gives: \[ x \approx 7.98\, \text{cm} \].

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Key Concepts

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

Linear Attenuation Coefficient
The linear attenuation coefficient, denoted by \( \mu \), is a critical parameter in understanding how materials interact with gamma rays. It represents the probability per unit length of a gamma ray interacting with a material and being absorbed or scattered. When gamma rays pass through a medium, they are progressively weakened due to these interactions.
  • \( \mu \) is specific to the material and type of radiation.
  • The larger the \( \mu \), the more effective the material is at attenuating gamma rays.
  • It is usually measured in units of cm\(^{-1}\).
In the exercise, the linear attenuation coefficient is derived from observing how much the gamma ray intensity diminishes after passing through a 1.0 cm thick lead target. By measuring this reduction, it allows us to calculate \( \mu \) through the exponential attenuation law.
Exponential Attenuation Law
The exponential attenuation law describes how the intensity of a gamma ray beam decreases as it travels through a material. This relationship is expressed through the formula: \[ I = I_0 \cdot e^{-\mu x} \]Where:
  • \( I \) is the final intensity of the gamma rays.
  • \( I_0 \) is the initial intensity before entering the medium.
  • \( \mu \) is the linear attenuation coefficient, representing how the material reduces the gamma ray intensity per unit length.
  • \( x \) is the thickness of the material the gamma rays travel through.
This law illustrates the exponential decay of intensity, signifying a consistent fractional reduction per unit thickness. It is essential for calculating how different materials shield against radiation by adjusting the thickness \( x \) until a desired level of intensity reduction is achieved.
Gamma Ray Intensity Reduction
Gamma ray intensity reduction is a practical application of both the linear attenuation coefficient and the exponential attenuation law. It is the percentage by which the original intensity of gamma rays is reduced after passing through a medium. The exercise outlines this concept by showing how materials can attenuate gamma rays to specific levels:
  • The problem demonstrates how a 1.0 cm thick lead sheet decreases the intensity to 25% of its original.
  • Solving further involved the calculation of lead thickness to reduce gamma ray intensity to one in a million (\(\frac{1}{10^6}\)).
These calculations are vital to designing radiation shielding for safety and protection in various industries, such as medical imaging and nuclear power, where reducing gamma ray exposure is crucial. By understanding how to control the thickness and type of material, one can effectively manage the levels of exposure.

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Most popular questions from this chapter

A \(1.2-\mu \mathrm{Ci}^{137} \mathrm{Cs}\) source is used for 1.6 hours by a \(65-\mathrm{kg}\) worker. Radioactive 135 \(\mathrm{Cs}\) decays by \(\beta^{-}\) decay with a half-life of 30 yr. The average energy of the emitted betas is about 190 keV per decay. The \(\beta\) decay is quickly followed by a \(\gamma\) with an energy of 660 \(\mathrm{keV} .\) Assuming the person absorbs all emitted energy, what effective dose (in rem) is received?

(I) A dose of 4.0Sv of \(\gamma\) rays in a short period would be lethal to about half the people subjected to it. How many grays is this?

The average yearly background radiation in a certain town consists of 29 mrad of X-rays and \(\gamma\) rays plus 3.6 mrad of particles having a QF of \(10 .\) How many rem will a person receive per year on the average?

(I) What is the effective cross section for the collision of two hard spheres of radius \(R_{1}\) and \(R_{2} ?\)

(I) The cross section for the reaction \(\mathrm{n}+{ }_{5}^{10} \mathrm{~B} \rightarrow{ }_{3}^{7} \mathrm{Li}+{ }_{2}^{4} \mathrm{He}\) is about \(40 \mathrm{bn}\) for an incident neutron of low energy (kinetic energy \(\approx 0\) ). The boron is contained in a gas with \(n=1.7 \times 10^{21}\) nuclei \(/ \mathrm{m}^{3}\) and the target has thickness \(\ell=12.0 \mathrm{~cm} .\) What fraction of incident neutrons will be scattered?
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