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Chapter 12: Problem 1

What is the short-wavelength limit of the continuum produced by an X-ray tube having a tungsten target and operated at \(50 \mathrm{kV} ?\)

Short Answer

Expert verified
The short-wavelength limit is approximately 0.0248 nm.

Step by step solution

01

Understand the Problem

We need to determine the short-wavelength limit of the X-ray continuum spectrum for an X-ray tube with a tungsten target operated at a potential difference of 50 kV.
02

Use the Formula for Short Wavelength Limit

The short-wavelength limit (\( \lambda_{min} \)) is found using the equation:\[\lambda_{min} = \frac{hc}{eV}\]where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \; \text{Js} \)), \( c \) is the speed of light in a vacuum (\( 3.00 \times 10^{8} \; \text{m/s} \)), \( e \) is the elementary charge (\( 1.602 \times 10^{-19} \; \text{C} \)), and \( V \) is the accelerating voltage (50 kV or 50,000 V).
03

Plug in the Values

Substitute the given values into the formula:\[\lambda_{min} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{1.602 \times 10^{-19} \times 50,000}\]
04

Calculate the Result

Perform the calculation:\[\lambda_{min} = \frac{1.9878 \times 10^{-25}}{8.01 \times 10^{-15}} \approx 2.48 \times 10^{-11} \; \text{m}\]
05

Convert to Nanometers

Since X-ray wavelengths are often expressed in nanometers, convert the result:\[\lambda_{min} = 2.48 \times 10^{-11} \; \text{m} = 0.0248 \; \text{nm}\]

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Key Concepts

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

X-ray tube
An X-ray tube is essential equipment used in various applications like medical imaging and material analysis. It operates by generating X-rays through a process that involves accelerating electrons towards a target material. When these high-speed electrons strike the target, X-rays are produced.

The basic components of an X-ray tube include:
  • A cathode, which emits electrons when heated.
  • An anode, typically made of a metal such as tungsten, where the electrons collide.
  • A vacuum tube, which ensures that electrons travel from the cathode to the anode without interference from air molecules.
The operation relies on applying a high voltage between the cathode and the anode. This high voltage accelerates the electrons, giving them enough energy to produce X-rays when they strike the anode. The choice of target material and the voltage used impact the characteristics of the X-rays emitted, such as their wavelength and intensity.
tungsten target
In an X-ray tube, the target material is crucial as it significantly influences the properties of the generated X-rays. Tungsten is commonly used for this purpose due to several advantageous properties.

Key reasons for using tungsten as a target material include:
  • High atomic number: Tungsten, with its high atomic number, is efficient at producing X-rays because it can decelerate the fast-moving electrons more effectively, leading to the emission of higher energy radiation.
  • High melting point: As the electron collision generates significant heat, tungsten's high melting point (around 3422掳C) makes it ideal for withstanding the thermal stress in an X-ray tube.
  • Good thermal conductivity: It helps to dissipate the heat quickly to avoid damage to the target.
These characteristics make tungsten an ideal candidate for use in X-ray tubes, ensuring efficient generation of X-rays while maintaining the durability of the target material.
accelerating voltage
Accelerating voltage in an X-ray tube plays a critical role in determining the energy and wavelength of the produced X-rays. It refers to the potential difference applied between the cathode and anode.

When a high voltage is applied:
  • The electrons emitted from the cathode accelerate towards the anode.
  • The speed of the electrons increases with higher voltages, resulting in more energetic collisions with the target.
  • These collisions produce higher energy X-rays, capable of shorter wavelengths.
In the context of our problem, an accelerating voltage of 50 kV means that the electrons achieve correspondingly high energies, leading to the production of the X-rays with specific wavelengths calculated using the formula for the short-wavelength limit. The equation \( \lambda_{min} = \frac{hc}{eV} \) illustrates that as the accelerating voltage \( V \) increases, the minimum wavelength \( \lambda_{min} \) produced by the X-ray tube decreases.
continuum spectrum
The continuum spectrum, in the context of X-ray production, refers to the broad range of X-ray wavelengths that can be generated by an X-ray tube. This spectrum arises because the energy of the electrons, and consequently the energy of the emitted X-rays, varies due to the random processes involved in electron deceleration when they strike the target.

Characteristics of the X-ray continuum spectrum include:
  • A broad wavelength range: The X-ray tube produces X-rays across a wide spectrum rather than a single, distinct wavelength.
  • Short-wavelength limit: This represents the smallest wavelength within the spectrum, determined by the maximum energy of the electrons, which depends on the accelerating voltage.
  • Dependence on target and voltage: The specifics of the continuum spectrum are influenced by the target material and the voltage applied, affecting both the intensity and the wavelength range of the X-rays.
Understanding the continuum spectrum is essential for applications requiring precise knowledge of the X-ray characteristics, such as diagnostic imaging and detailed material studies.

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Most popular questions from this chapter

The mass absorption coefficient for Ni, measured with the Cu K \(\alpha\) line, is \(49.2 \mathrm{cm}^{2} / \mathrm{g}\). Calculate the thickness of a nickel foil that was found to transmit \(35.2 \%\) of the incident power of a beam of Cu K \(\alpha\) radiation. The density of Ni is \(8.90 \mathrm{g} / \mathrm{cm}^{3} .\)

A solution of \(\mathbf{I}_{2}\) in ethanol had a density of 0.794 \(\mathrm{g} / \mathrm{cm}^{3} .\) A \(1.50-\mathrm{cm}\) layer was found to transmit \(33.2 \%\) of the radiation from a Mo Ka source. Mass absorption coefficients for I, \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) are \(39.2,0.70,0.00,\) and 1.50 respectively. (a) Calculate the percentage of I\(_{2}\) present, neglecting absorption by the alcohol. (b) Correct the results in part (a) for the presence of alcohol.

Manganese was determined in samples of geological interest via XRF using barium as an internal standard. The fluorescence intensity of isolated lines for each element gave the following data: $$\begin{array}{clc}\text { Wt. \% Mn } & \text { Ba } & \text { Mn } \\\\\hline 0.00 & 156 & 80 \\\0.10 & 160 & 106 \\\0.20 & 159 & 129 \\\0.30 & 160 & 154 \\\0.40 & 151 & 167\end{array}$$ What is the weight percentage manganese in a sample that had a Mn-to-Ba count ratio of 0.936?

Aluminum is to be used as windows for a cell for X-ray absorption measurements with the Ag Ka line. The mass absorption coefficient for aluminum at this wavelength is \(2.74 ;\) its density is \(2.70 \mathrm{g} / \mathrm{cm}^{3} .\) What maximum thickness of aluminum foil could be used to fabricate the windows if no more than \(2.8 \%\) of the radiation is to be absorbed by them?

For Mo K \(\alpha\) radiation (0.711 A) the mass absorption coefficients for K, I, H, and O are \(16.7,39.2,0.0,\) and \(1.50 \mathrm{cm}^{2} / \mathrm{g},\) respectively. (a) Calculate the mass absorption coefficient for a solution prepared by mixing \(11.00 \mathrm{g}\) of \(\mathrm{KI}\) with \(89.00 \mathrm{g}\) of water. (b) The density of the solution described in (a) is \(1.086 \mathrm{g} / \mathrm{cm}^{3} .\) What fraction of the radiation from a Mo Ka source would be transmitted by a 0.40-cm layer of the solution?
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