Problem 18 - University Physics with Modern Physics

Question

X rays are produced in a tube operating at 18.0 \(\mathrm{kV}\) . After emerging 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

(a) Calculate the original wavelength using \(\lambda = \frac{hc}{eV}\). (b) Use the Compton shift formula for the new wavelength. (c) Convert the scattered wavelength back to energy in electron volts.

Step 1: Calculate the Original X-ray Wavelength

The energy of the x rays is related to the potential difference by the formula: \(E = eV\), where \(e\) is the charge of the electron \(1.6 \times 10^{-19} \text{ C}\) and \(V = 18.0 \text{ kV}\). The energy \(E\) is thus \(18.0 \times 10^3 \times 1.6 \times 10^{-19}\). The wavelength \(\lambda\) is related to energy by \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J}\cdot\text{s}\) and \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\). Calculate \(\lambda\) using \(\lambda = \frac{hc}{eV}\).

Step 2: Calculate the Wavelength of Scattered X-rays

The change in wavelength due to Compton scattering is given by the Compton wavelength shift formula: \(\Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos \theta)\), where \(m\) is the electron rest mass \(9.11 \times 10^{-31} \text{ kg}\). For \(\theta = 45.0^\circ\), \(\Delta \lambda = \frac{h}{mc}(1 - \cos 45.0^\circ)\). Compute \(\lambda' = \lambda + \Delta \lambda\).

Step 3: Calculate the Energy of Scattered X-rays

Convert the scattered wavelength \(\lambda'\) to energy using \(E' = \frac{hc}{\lambda'}\) and express it in electron volts. Since \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\), calculate the energy in joules first and then convert to electron volts using \(E'_{\text{eV}} = \frac{E'}{1.6 \times 10^{-19}}\).

Key Concepts

X-ray Wavelength

When dealing with X-rays, one of the important characteristics to determine is their wavelength. The X-ray wavelength gives us insight into the properties and behavior of the radiation. The energy of X-rays can be derived from the potential difference in the apparatus that produces them. This energy is given the formula: \[ E = eV \] where \( e \) symbolizes the charge of the electron \( 1.6 \times 10^{-19} \) coulombs, and \( V \) is the potential difference, often measured in kilovolts (kV). For instance, in our example, with a 18.0 kV potential difference, the energy of the X-rays would be calculated accordingly.
Understanding and determining the wavelength involves using the relationship between energy \( E \), Planck’s constant \( h \), and the speed of light \( c \). The equation \[ E = \frac{hc}{\lambda} \] helps to derive the wavelength \( \lambda \) as: \[ \lambda = \frac{hc}{eV} \] This relationship demonstrates why high-energy X-rays feature short wavelengths, essential for penetrating matter and being used in imaging technologies. With this, we simplify the otherwise complex nature of radian physics into calculations we can perform to ensure safe and effective application.

Photon Energy

Photon energy is crucial in understanding the interaction of X-rays with matter. It represents the measurable capacity to perform work due to its electromagnetic wave nature. Each photon carries a quantized amount of energy, calculated via \[ E = \frac{hc}{\lambda} \] After Compton scattering, the energy of these X-ray photons can change. The initial X-ray photons, with their original wavelength \( \lambda \), have their energy converted into scattered photons with a new wavelength \( \lambda' \).
This shift leads to modifications in energy that can be calculated by transforming the scattered wavelength back into energy terms. This new energy is represented as \( E' \) and found in joules first, before being rendered into a more convenient form for practical use, such as electron volts (eV). This conversion is crucial since equipment precision often demands such a standard unit to ensure meticulous results. - Convert energy in joules using \( E' = \frac{hc}{\lambda'} \).- Transform \( E' \) into electron volts through: \[ E'_{\text{eV}} = \frac{E'}{1.6 \times 10^{-19}} \]
Grasping photon energy alterations enables scientists and medical practitioners to utilize X-ray technologies effectively, harnessing changes due to scattering in fields such as medical diagnostics and material analysis.

Compton Wavelength Shift

The Compton wavelength shift represents a fundamental effect in which X-rays, when interacting with electrons, undergo a change in wavelength. This change provides crucial data about the scattering process, instrumental in establishing the behavior of the X-ray photons involved.
Compton scattering occurs when photons collide with particles like electrons, resulting in a shift in the photon's pathway and energy distribution. The degree of wavelength change is encapsulated in the Compton wavelength shift formula: \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos \theta) \] where:

• \( h \) is Planck’s constant.
• \( m \) is the rest mass of the electron.
• \( \theta \) is the scattering angle.

This formula gives us the extent of change in wavelength, \( \Delta \lambda \), contingent on the angle \( \theta \) at which the photon is scattered. For instance, a scattering angle of \( 45^circ \) produces a particular wavelength variance. Understanding this shift helps in interpreting the effects of scattering on the energy and wavelength of the X-rays, a pivotal insight in both the theoretical and applied physics realms.
The insights gained from Compton wavelength shift allow enhanced design of devices using radiation, improve safety protocols, and help advance applications in scientific research and healthcare.