Problem 31 - College Physics

Question

An X-ray tube is operated at \(50 \mathrm{kV}\). The shortest wavelength photons from this tube are used in a Compton scattering experiment. One of these photons strikes a free electron and is scattered directly back at an angle of \(180^{\circ} .\) (a) What is the wavelength of the incident photon that comes from the X-ray tube? (b) What is the wavelength of the scattered photon?

Step-by-Step Solution

Answer

(a) The wavelength of the incident photon is \(2.48 \times 10^{-11} \text{ m}\). (b) The scattered photon wavelength is \(2.966 \times 10^{-11} \text{ m}\).

Step 1: Determine Incident Photon Energy

The energy of the incident photon is calculated using the voltage of the X-ray tube. This is given by: \[ E_i = eV \] where \( e = 1.6 \times 10^{-19} \) Coulombs (elementary charge) and \( V = 50 \) kV = \( 50 \times 10^3 \) volts. Thus, the energy becomes:\[ E_i = 1.6 \times 10^{-19} \times 50 \times 10^3 = 8.0 \times 10^{-15} \text{ J} \].

Step 2: Convert Energy to Wavelength of Incident Photon

The wavelength \( \lambda_i \) of the incident photon is found using the energy-wavelength relationship: \[ E = \frac{hc}{\lambda} \] which rearranges to: \[ \lambda_i = \frac{hc}{E_i} \] where \( h = 6.626 \times 10^{-34} \text{ Js} \) (Planck's constant) and \( c = 3 \times 10^8 \ \text{m/s} \) (speed of light). Substituting in the values, we get:\[ \lambda_i = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8.0 \times 10^{-15}} = 2.48 \times 10^{-11} \text{ m} \].

Step 3: Apply Compton Wavelength Shift Equation

The Compton wavelength shift equation for a photon scattered at an angle \( \theta \) is: \[ \Delta \lambda = \frac{h}{m_ec} (1 - \cos{\theta}) \] where \( m_e = 9.11 \times 10^{-31} \) kg. For \( \theta = 180^{\circ} \), \( \cos{\theta} = -1 \), so:\[ \Delta \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^8} (1 - (-1)) \approx 4.86 \times 10^{-12}\text{ m} \].

Step 4: Calculate Wavelength of Scattered Photon

The wavelength of the scattered photon \( \lambda_s \) is given by the sum of the incident wavelength and the shift:\[ \lambda_s = \lambda_i + \Delta \lambda \] Substitute the values we have:\[ \lambda_s = 2.48 \times 10^{-11} + 4.86 \times 10^{-12} = 2.966 \times 10^{-11} \text{ m} \].

Key Concepts

X-ray tube operation

X-ray tubes are fascinating devices used to produce X-rays, a type of high-energy electromagnetic radiation. The basic principle behind their operation involves accelerating electrons at high speeds and striking them against a metal target, usually tungsten. When these high-speed electrons interact with the target, they are decelerated rapidly, causing the emission of X-ray photons. This process is known as Bremsstrahlung or "braking radiation." In practical terms, the energy of the X-ray photons produced depends on the voltage across the tube. For example, in an X-ray tube operating at 50 kV, the photons generated will have maximum energy corresponding to the voltage applied. Similar mechanisms apply when using X-rays in medical imaging, material analysis, and in scientific research such as Compton scattering experiments.

Photon energy calculation

The energy of a photon, especially in the context of X-rays, is a crucial concept in physics. It helps us understand the characteristics of the radiation produced. The energy of a photon is directly related to the electrical potential used in the X-ray tube. The formula used to calculate this energy is given by: \[ E_i = eV \] Here, \( e \) is the elementary charge, equivalent to \( 1.6 \times 10^{-19} \) Coulombs, and \( V \) is the voltage in volts applied across the X-ray tube. A higher voltage results in higher energy photons. For instance, using a 50 kV voltage, the photon energy becomes \( 8.0 \times 10^{-15} \) Joules. Understanding this relationship is essential for controlling the intensity and penetration power of the X-rays produced.

Wavelength shift equation

The Compton Effect, explained by the Compton wavelength shift equation, describes the change in wavelength when a photon collides with a particle, such as an electron. The equation is: \[ \Delta \lambda = \frac{h}{m_ec} (1 - \cos{\theta}) \] Where \( h \) is Planck's constant, \( m_e \) is the electron mass, \( c \) is the speed of light, and \( \theta \) is the scattering angle. In cases like a backscatter at an angle of \( 180^\circ \), the equation simplifies because \( \cos{180^\circ} = -1 \). Thus, the shift in wavelength becomes approximately \( 4.86 \times 10^{-12} \) meters. This equation is vital for understanding photon-electron interactions, often resulting in the scattering and altering the path and energy distribution of photons in phenomena such as medical imaging or astrophysical observations.

Planck's constant

Planck's constant, denoted as \( h \), is a fundamental constant in quantum mechanics, valued at approximately \( 6.626 \times 10^{-34} \) Joule seconds. It plays a key role in various equations involving particles and radiation, most notably in the energy-wavelength relationship of photons: \[ E = \frac{hc}{\lambda} \] This relationship links a photon's energy to its wavelength, using the speed of light \( c \). In the context of X-ray photons, Planck's constant allows us to calculate their wavelengths based on their energy, as determined by the X-ray tube's operational parameters. The energy quantified by this constant explains the wave-particle duality observed in light and matter, forming the cornerstone of modern physics theories and technologies.