Question

X-rays of wavelength \(\lambda=0.120 \mathrm{nm}\) are scattered from carbon. What is the Compton wavelength shift for photons deflected at a \(90.0^{\circ}\) angle relative to the incident beam?

Short answer

Solution: Using the Compton scattering formula, the Compton wavelength shift for photons deflected at a 90.0° angle relative to the incident beam is 2.431 x 10^{-12} meters.

Step-by-step solution

Write down the known information and constants

Before starting to solve the problem, let's write down all the known information and constants: - Initial wavelength of the X-rays, \(\lambda = 0.120\) nm - Scattering angle of the photon, \(\theta = 90.0^{\circ}\) - Compton wavelength (a constant), \(\lambda_c = 2.426 \times 10^{-12}\) m Note that we will need to convert the initial wavelength from nm to meters to have consistent units.

Convert the initial wavelength to meters

To convert the initial X-ray wavelength from nm to meters, we use the conversion factor \(1\) nm \(= 10^{-9}\) m: \(\lambda = 0.120 \; \mathrm{nm} \times \frac{10^{-9} \; \mathrm{m}}{1 \; \mathrm{nm}} = 1.20 \times 10^{-10} \; \mathrm{m}\).

Use the Compton scattering formula to find the wavelength shift

The Compton scattering formula is given by: \(\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)\), where \(\Delta\lambda\) is the wavelength shift, \(\lambda'\) is the final wavelength, \(h\) is the Planck's constant, \(m_e\) is the electron mass, \(c\) is the speed of light, and \(\theta\) is the scattering angle. Using the values of \(h = 6.626 \times 10^{-34}\) Js, \(m_e = 9.109 \times 10^{-31}\) kg, \(c=3 \times 10^8 \; \mathrm{m/s}\), and \(\theta = 90.0^{\circ}\): \(\Delta\lambda = \frac{(6.626 \times 10^{-34} \; \mathrm{Js})}{(9.109 \times 10^{-31} \; \mathrm{kg})(3 \times 10^8 \; \mathrm{m/s})}(1 - \cos(90.0^{\circ}))\)

Calculate the wavelength shift

Now, we compute the numerical value for the wavelength shift: \(\Delta\lambda = \frac{(6.626 \times 10^{-34} \; \mathrm{Js})}{(9.109 \times 10^{-31} \; \mathrm{kg})(3 \times 10^8 \; \mathrm{m/s})}(1 - 0) = 2.431 \times 10^{-12} \; \mathrm{m}\). So, the Compton wavelength shift for photons deflected at a \(90.0^{\circ}\) angle relative to the incident beam is \(2.431 \times 10^{-12}\) meters.

Key concepts

Compton Wavelength Shift

The Compton wavelength shift is a key phenomenon in understanding the particle-like properties of light. During Compton scattering, when X-rays hit an electron, the X-ray's wavelength changes due to the transfer of energy during the collision.

The magnitude of the shift depends on the scattering angle, \theta, which is the angle at which the photon is deflected relative to its original path. A larger angle typically results in a greater shift. Notably, the Compton formula \[\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)\] articulates this relation, where \(\Delta\lambda\) is the change in the X-ray's wavelength, \(\lambda\) is the initial wavelength, \(\lambda'\) is the final wavelength, \(h\) is Planck's constant, \(m_e\) is the electron's rest mass, and \(c\) is the speed of light.

For a 90-degree scattering angle, as in our exercise, the cosine term becomes zero, simplifying the calculation and maximizing the wavelength shift. This fundamental understanding of Compton scattering unlocks insights into the dual nature of light and helps refine quantum mechanics theories.

X-ray Wavelength Conversion

X-ray wavelength conversion is a practical aspect that plays a pivotal role when working with Compton scattering problems. It involves converting the given wavelength into a consistent unit of measurement. To our advantage, the conversion factors are standardized, and for X-rays, the conversion from nanometers to meters involves multiplying by \(10^{-9}\).

For instance, in our problem, an initial X-ray wavelength of \(\lambda = 0.120\) nm is converted into meters by using \(\lambda = 0.120 \times 10^{-9}\) m. Accurate conversion ensures that the subsequent calculations for measuring the Compton wavelength shift are correct, as having consistent units across all values is essential in physics. It's a simple yet crucial step to avoid discrepancies in the final results and to maintain the integrity of the scientific method.

Planck's Constant

Planck's constant \(h\), is a fundamental concept in quantum physics, evidencing the quantization of energy. It is the proportionality constant between the energy (\(E\)) of a photon and the frequency (\(u\)) of its associated electromagnetic wave, established by the relation \(E = hu\).

With a value of approximately \(6.626 \times 10^{-34}\) joule-seconds (Js), Planck's constant plays a central role in the Compton scattering formula. It allows us to calculate the shift in wavelength of light as a consequence of its particle-like interactions with matter. In our exercise, Planck's constant was crucial to determine the Compton wavelength shift for the scattered X-ray photon. Understanding its impact on energy quantization is essential for students grappling with the wave-particle duality of light and the broader implications for quantum theory.