Problem 36 - University Physics with Modern Physics

Question

\(X\) rays are produced in a tube operating at 18.0 \(\mathrm{kV}\) . After cmerging from the tube, \(x\) rays with the minimum wavelength produced strike a target and are Compton-scattered through an angle of \(45.0^{\circ} .\) (a) What is the original \(x\) -ray wavelength? (b) What is the wavelength of the scattered \(x\) rays? (c) What is the energy of the scattered \(x\) rays (in electron volts)?

Step-by-Step Solution

Answer

The original X-ray wavelength is \(\lambda_{0} = 6.91 \times 10^{-11} \, \text{m}\). The wavelength of scattered X-rays is \(\lambda = 7.22 \times 10^{-11} \, \text{m}\). The energy of scattered X-rays is approximately 17.2 keV.

Step 1: Determine the Minimum Wavelength

The minimum wavelength of X-rays (\[ \lambda_{0} \]) in the tube can be calculated using the equation for the shortest wavelength limit of X-ray tubes: \[\lambda_{0} = \frac{hc}{eV},\]where:- \(h = 6.626 \times 10^{-34} \, \text{Js}\) (Planck's constant),- \(c = 3.00 \times 10^{8} \, \text{m/s}\) (speed of light),- \(e = 1.602 \times 10^{-19} \, \text{C}\) (elementary charge),- \(V = 18.0 \, \text{kV} = 18,000 \, \text{V}\).Insert these values into the equation:\[\lambda_{0} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{(1.602 \times 10^{-19})(18,000)}.\]Calculate \(\lambda_{0} \) to find the original wavelength.

Step 2: Calculate the Wavelength of Scattered X-rays

Compton's scattering equation is given by:\[\lambda - \lambda_{0} = \frac{h}{mc} (1 - \cos \theta),\]where:- \(\lambda\) is the wavelength of the scattered X-rays,- \(mc = 9.109 \times 10^{-31} \, \text{kg} \cdot 3.00 \times 10^{8} \, \text{m/s}\),- \(\theta = 45.0^{\circ}\) is the Compton scattering angle.\(\lambda_{0}\) is the wavelength found in Step 1. Substituting the known values:\[\lambda - \lambda_{0} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 3.00 \times 10^{8}} (1 - \cos 45^{\circ}).\]Solve for \(\lambda\).

Step 3: Calculate Energy of Scattered X-rays

The energy \( E \) of the scattered X-rays can be determined from:\[E = \frac{hc}{\lambda},\]where \( \lambda \) is the wavelength calculated in Step 2. Insert \(h\), \(c\), and \(\lambda\):\[E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{\lambda}\]Convert this energy from joules to electron volts by dividing by the elementary charge \(e = 1.602 \times 10^{-19} \, \text{C}\). Calculate \(E\) to find the energy in electron volts.

Key Concepts

X-ray wavelength

X-ray wavelength is a critical concept in understanding how X-rays interact with matter. It refers to the distance between two consecutive peaks of the electromagnetic wave. In X-ray physics, wavelength is crucial since it determines the energy and the penetrating power of the X-rays.
The relationship between energy (E) and wavelength (λ) is expressed by the equation: \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant and \(c\) is the speed of light. This equation shows that energy is inversely proportional to wavelength: as wavelength decreases, energy increases.
For X-rays produced in a tube, the minimum wavelength is particularly interesting. The minimum wavelength is often produced by the process of stopping high-energy electrons and is determined by the equation: \(\lambda_{0} = \frac{hc}{eV}\), where \(V\) is the accelerating voltage of the X-ray tube.
Understanding the X-ray wavelength helps in a variety of scientific and medical applications, influencing how X-rays are used in imaging and experimentation.

scattered X-ray energy

Scattered X-ray energy arises when X-rays interact with matter, deflecting from their original path. This concept is integral to Compton scattering, a phenomenon where X-rays collide with electrons, causing a change in the wavelength and, subsequently, the energy of the X-rays.
The change in wavelength can be calculated using Compton's equation: \(\lambda - \lambda_{0} = \frac{h}{mc} (1 - \cos \theta)\). Here, \(\lambda_{0}\) is the initial wavelength, \(\lambda\) is the wavelength of the scattered X-ray, \(m\) is the mass of the electron, and \(\theta\) is the scattering angle.
The energy of the scattered X-rays is then determined by substituting the new wavelength into the energy equation \(E = \frac{hc}{\lambda}\). This energy is often converted to electron volts (eV) by using the elementary charge \(e\).
Understanding scattered X-ray energy is vital in fields such as material science and medical diagnostics, where it aids the understanding of material structures and provides crucial information in imaging.

minimum wavelength of X-rays

The minimum wavelength of X-rays is a defining characteristic produced in X-ray tubes and is crucial in determining their utility in various applications. It represents the shortest possible wavelength obtainable, which corresponds to the maximum photon energy according to the formula \(\lambda_{0} = \frac{hc}{eV}\).
This formula arises from the conversion of the kinetic energy of accelerated electrons into electromagnetic radiation. When electrons are accelerated through a potential difference \(V\) and suddenly stopped, they emit X-rays with a minimum wavelength, resulting from the direct conversion of their kinetic energy to X-ray photons.
The minimum wavelength is directly dependent on the voltage of the X-ray tube. Higher voltages produce shorter wavelengths with more energetic X-rays. These energetic X-rays are capable of penetrating further into materials, which is particularly useful in applications ranging from cancer treatment to the internal imaging of objects.

• Higher energy (shorter wavelength) X-rays are more penetrating.
• Knowledge of the minimum wavelength assists in designing X-ray equipment.
• It's essential for ensuring the safety and effectiveness of medical diagnostic tools.

Understanding the minimum wavelength of X-rays ensures optimal usage in both scientific research and clinical settings.