Problem 38 - Physics for Scientists and Engineers with Modern Physics

Question

(I) If the shortest-wavelength bremsstrahlung X-rays emitted from an X-ray tube have \(\lambda=0.027 \mathrm{nm}\), what is the voltage across the tube?

Step-by-Step Solution

Answer

The voltage across the tube is approximately 46,000 volts.

Step 1: Understand the relationship between wavelength and voltage

Bremsstrahlung X-rays intensity depends on the energy of electrons, which is related to the tube voltage. The energy of the most energetic photons is given by Planck's equation: \( E = hf \), where \( f \) is the frequency. Using the speed of light, \( c = \lambda \cdot f \), the equation for energy becomes \( E = \frac{hc}{\lambda} \).

Step 2: Calculate energy in electron volts

To find the voltage, solve for energy in joules, then convert to electron volts. The Planck's constant \( h \) is \( 6.63 \times 10^{-34} \mathrm{Js} \) and the speed of light \( c \) is \( 3 \times 10^8 \mathrm{m/s} \). Substituting into \( E = \frac{hc}{\lambda} \), where \( \lambda = 0.027 \times 10^{-9} \mathrm{m} \), we have:\[E = \frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{0.027 \times 10^{-9}}\]Calculate \( E \) in joules.

Step 3: Convert energy from joules to electron volts

The energy obtained in joules needs to be converted to electron volts. Since \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \), the energy in electron volts is:\[E_{\text{eV}} = \frac{E_{\text{Joules}}}{1.6 \times 10^{-19}}\]

Step 4: Determine the voltage across the tube

The energy of the electrons accelerated by the tube voltage \( V \) corresponds to the energy of the resulting photons. Thus, the calculated energy \( E_{\text{eV}} \) is equal to \( V \), i.e., \( V = E_{\text{eV}} \). Substitute the calculated \( E_{\text{eV}} \) to find the voltage.

Key Concepts

Tube Voltage

When you hear "tube voltage" in the context of X-ray tubes, it's referring to the electric potential difference applied across the X-ray tube. This voltage accelerates electrons from the cathode towards the anode. The energy that these electrons have when they strike the anode is critical because it determines the energy and the characteristics of the X-rays produced, like their wavelength and intensity.

Tube voltage is crucial because it affects:

• X-ray penetration: Higher tube voltage results in more penetrating X-rays, useful for examining denser tissues or thicker objects in medical imaging.
• X-ray intensity: The intensity of the X-rays is directly proportional to the tube voltage squared, increasing the strength of the X-ray beam.
• Wavelength of X-rays: A higher voltage leads to shorter wavelength X-rays, which are more energetic and potentially more effective for certain types of imaging.

Understanding tube voltage can help predict the behavior of electromagnetic radiation emitted during the X-ray generation process and optimize it for specific applications.

Planck's Equation

Planck's equation is a fundamental relation in the physics of electromagnetic radiation, connecting the energy of photons with their frequency. It is expressed with the formula: \( E = hf \), where \( E \) is the energy, \( h \) is Planck's constant \( (6.63 \times 10^{-34} \mathrm{Js}) \), and \( f \) is the frequency.

This equation is crucial in the context of X-rays because:

• Energy Calculation: By using Planck's equation, we can determine the energy of the photons generated in an X-ray tube, which is directly related to the tube voltage and the speed and impact of electrons.
• Wavelength-Frequency Relationship: Frequency \( (f) \) is linked to the wavelength \( (\lambda) \) through the speed of light \( (c) \), using the relation \( c = \lambda \cdot f \), allowing energy to be expressed as \( E = \frac{hc}{\lambda} \).
• Determining Photon Characteristics: Understanding this equation helps link physical properties of electromagnetic radiation, such as wavelength and frequency, to energy, allowing scientists and engineers to tailor X-ray characteristics for specific uses.

Overall, Planck's equation provides a bridge between wave-like and particle-like descriptions of electromagnetic radiation, crucial for many applications, especially in medicine and materials science.

Electromagnetic Radiation

Electromagnetic radiation encompasses a broad range of wavelengths, from radio waves with very long wavelengths to gamma rays with extremely short ones. X-rays are a form of electromagnetic radiation that falls between ultraviolet light and gamma rays in terms of wavelength.

Key characteristics of electromagnetic radiation include:

• Invisible to the naked eye: While some electromagnetic radiation is visible (like visible light), others, including X-rays, are not detectable by human eyes.
• Wave-particle duality: As with other types of electromagnetic radiation, X-rays exhibit both wave-like and particle-like properties, facilitating various practical applications, like imaging the structure of objects.
• Interaction with matter: Different wavelengths and energies of electromagnetic radiation interact uniquely with matter, allowing for uses in imaging, communication, and even medical treatment.

Understanding these characteristics allows us to harness the specific benefits of electromagnetic radiation, such as X-rays for medical imaging, by optimizing parameters like tube voltage to control their production in X-ray machines. This not only enhances image quality but also reduces potential risks of radiation exposure.