Chapter 18: Problem 7
How many times the half-absorption thickness of a material is required to reduce the intensity of an \(x\) -ray beam to (a) \(1 / 16\), (b) \(1 / 20\), or (c) \(1 / 200\) of the incident intensity?
Short Answer
Expert verified
(a) 4 layers, (b) ~4.32 layers, (c) ~7.64 layers.
Step by step solution
01
Understanding Half-Absorption Thickness
The half-absorption thickness (also known as the half-value layer, HVL) is the thickness of a material needed to reduce the x-ray intensity by half. This reduction can be described using the exponential attenuation formula: \( I = I_0 \times (\frac{1}{2})^n \), where \( I \) is the final intensity, \( I_0 \) is the initial intensity, and \( n \) is the number of half-value layers.
02
Solving for Part (a)
We need to determine \( n \) such that the intensity is \( \frac{1}{16} I_0 \). Set up the equation: \( (\frac{1}{2})^n = \frac{1}{16} \). Recall that \( \frac{1}{16} = (\frac{1}{2})^4 \). Therefore, \( n = 4 \) half-value layers are needed.
03
Solving for Part (b)
For an intensity of \( \frac{1}{20} I_0 \), set the equation: \( (\frac{1}{2})^n = \frac{1}{20} \). We solve \( n \log(\frac{1}{2}) = \log(\frac{1}{20}) \). Calculating, \( n \approx \frac{\log(20)}{\log(2)} \approx 4.32 \). Thus, approximately \( 4.32 \) half-value layers are needed.
04
Solving for Part (c)
We need \( n \) so that the intensity is \( \frac{1}{200} I_0 \). Set up the equation \( (\frac{1}{2})^n = \frac{1}{200} \). Solving \( n \log(\frac{1}{2}) = \log(\frac{1}{200}) \), we find \( n \approx \frac{\log(200)}{\log(2)} \approx 7.64 \). Thus, approximately \( 7.64 \) half-value layers are needed.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
X-Ray Intensity Attenuation
X-ray intensity attenuation describes the reduction of x-ray beam intensity as it passes through a material. This phenomenon is crucial in various applications like medical imaging and material analysis. As an x-ray beam travels through a substance, particles within the material absorb or scatter the x-ray photons, leading to a decrease in intensity.
The degree of attenuation depends on several factors:
The degree of attenuation depends on several factors:
- The type and density of the material: Denser materials typically attenuate x-rays more effectively than less dense ones.
- The thickness of the material: Thicker materials absorb more x-rays, reducing the intensity further.
- The energy of the x-ray photons: Higher energy photons penetrate materials more effectively, resulting in less attenuation.
Exponential Attenuation Formula
The exponential attenuation formula is a mathematical representation of how x-ray intensity decreases as it passes through a material. This relationship is expressed as: \[ I = I_0 \times e^{-\mu x} \]where:
- \( I \) is the final intensity after passing through the material,
- \( I_0 \) is the initial intensity at the start,
- \( \mu \) is the linear attenuation coefficient, a material-specific constant,
- \( x \) is the thickness of the material.
Half-Value Layer (HVL)
The Half-Value Layer (HVL) is a measure used to express how well a material can attenuate x-rays. It is defined as the thickness of the material required to reduce the x-ray intensity by half.
The importance of HVL comes from its ability to standardize the effectiveness of different materials in absorbing x-ray radiation. In practical terms, the HVL tells us how much protection a material provides and helps in calculating the necessary thickness for shielding.Calculating the HVL involves using the exponential formula, focusing on the condition that halves the initial intensity. The formula becomes: \[ I = I_0 \times (\frac{1}{2})^{n} \]where:
The importance of HVL comes from its ability to standardize the effectiveness of different materials in absorbing x-ray radiation. In practical terms, the HVL tells us how much protection a material provides and helps in calculating the necessary thickness for shielding.Calculating the HVL involves using the exponential formula, focusing on the condition that halves the initial intensity. The formula becomes: \[ I = I_0 \times (\frac{1}{2})^{n} \]where:
- \( I \) is the reduced intensity after passing through \( n \) layers of HVL,
- \( n \) is the number of HVL required for a desired reduction.