Question

A \(2.0-\mathrm{MeV}\) X-ray photon is scattered from a free electron at rest into an angle of \(53^{\circ} .\) What is the wavelength of the scattered photon?

Short answer

Answer: The wavelength of the scattered photon is \(6.50\times10^{-12}\) meters.

Step-by-step solution

Finding the Initial Wavelength

First, we need to find the initial wavelength of the X-ray photon. We can convert its energy (given in MeV) to wavelength using the Planck's formula: \(E = h\cdot c / \lambda\) Here, E is the energy, h is the Planck's constant (\(6.63\times10^{-34} \mathrm{J\cdot s}\)), c is the speed of light (\(3\times10^{8} \mathrm{m/s}\)), and λ is the wavelength. First, convert the energy from MeV to J (Joules): $$ 2.0 \mathrm{MeV}\times\frac{1.6\times10^{-13} \mathrm{ J}}{1 \mathrm{MeV}} = 3.2\times10^{-13}\mathrm{J} $$ Now, we can find the initial wavelength: $$ \lambda_0 = \frac{h\cdot c}{E} = \frac{6.63\times10^{-34} \mathrm{J\cdot s}\times 3\times10^{8} \mathrm{m/s}}{3.2\times10^{-13}\mathrm{J}} = 6.20\times10^{-12}\mathrm{m} $$

Applying the Compton Scattering Formula

Now, we will apply the Compton scattering formula that relates the initial wavelength (\(\lambda_0\)), the scattering angle (\(\theta\)), and the final wavelength (\(\lambda\)): \(\Delta\lambda = \lambda - \lambda_0 = \frac{h}{m_e c}(1 - \cos\theta)\) Here, \(m_e\) is the mass of the electron (\(9.11\times10^{-31} \mathrm{kg}\)). Substitute the values given in the problem and the initial wavelength calculated in Step 1: $$ \Delta\lambda = \frac{6.63\times10^{-34} \mathrm{J\cdot s}}{9.11\times10^{-31}\mathrm{kg}\cdot 3\times10^{8} \mathrm{m/s}}(1 - \cos53^{\circ}) = 3.00\times10^{-13}\mathrm{m} $$

Finding the Final Wavelength

Since we have the initial wavelength (\(\lambda_0\)) and the wavelength change (\(\Delta\lambda\)), we can find the final wavelength (\(\lambda\)) of the scattered photon: \(\lambda = \lambda_0 + \Delta\lambda\) Substitute the values we calculated earlier: $$ \lambda = 6.20\times10^{-12}\mathrm{m} + 3.00\times10^{-13}\mathrm{m} = 6.50\times10^{-12}\mathrm{m} $$ So the wavelength of the scattered photon is \(6.50\times10^{-12}\) meters.

Key concepts

X-ray Photon Energy

Understanding the energy of X-ray photons is crucial in physics, especially when dealing with phenomena such as Compton scattering. X-ray photons are a form of electromagnetic radiation with very high energy, characterized by their short wavelengths. The energy (\(E\)) of an X-ray photon can be determined by its frequency or by its wavelength. When examining X-ray photon energy, we frequently consider it in terms of electronvolts (eV), which is a common unit of energy in the context of atoms and subatomic particles. For example, a 2.0 MeV X-ray photon has a very high energy, which corresponds to millions of electronvolts. This significant energy level allows X-ray photons to penetrate various materials and makes them useful in medical imaging and other applications.

Planck's Formula

Planck's formula serves as the cornerstone for understanding the relationship between the energy of a photon and its wavelength. This relationship is expressed by the equation \( E = h \cdot c / \lambda \), where \(E\) represents the energy of the photon, \(h\) is Planck's constant (approximated as 6.63 x 10^(-34) Joule seconds), \(c\) is the speed of light (approximately 3 x 10^8 meters per second), and \(\lambda\) denotes the wavelength.

By using Planck's formula, one can convert back and forth between the energy and wavelength of a photon if the other value is known. This is pivotal, as it provides a means of calculating the key attributes of a photon based solely on its energy or wavelength. For instance, the initial wavelength of a given 2.0 MeV X-ray photon can be precisely determined through this formula, which is an essential step before exploring its behavior during interactions such as Compton scattering.

Compton Wavelength Shift

The Compton wavelength shift illustrates a fundamental aspect of the behavior of photons when they interact with particles such as electrons, leading to the phenomenon known as Compton scattering. This interaction results in a change in the photon's wavelength, which is given by the formula \( \Delta\lambda = \lambda - \lambda_0 = \frac{h}{m_e c}(1 - \cos\theta) \), where \(\lambda_0\) is the initial wavelength, \(\lambda\) is the wavelength after scattering, \(h\) is Planck's constant, \(m_e\) is the mass of the electron, \(c\) is the speed of light, and \(\theta\) denotes the scattering angle.

The Compton wavelength shift is particularly interesting because it embodies the dual particle-wave nature of light; it presents a measurable effect stemming from quantum mechanics. During an interaction, such as the scattering of an X-ray photon off an electron, the energy transferred to the electron results in a longer wavelength for the photon post-collision. Understanding this shift is critical in diagnosing the properties of the interacting particles and the incident photons.