Question

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

Short answer

Answer: The wavelength of the scattered photon is approximately \(7.92 \times 10^{-12} \text{ m}\).

Step-by-step solution

Convert the given energy into initial photon wavelength

First, we need to convert the given energy (2.0 MeV) into the initial photon wavelength using the formula: \(E = \frac{h c}{\lambda}\) Where \(E\) is the energy, \(h\) is the Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. We are given \(E = 2.0 \times 10^6 \text{ eV}\), so we can rewrite the formula as: \(\lambda = \frac{h c}{E}\) Remember to convert the energy from eV to Joules: \(E_\text{J} = E \times e = 2.0 \times 10^6 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.2 \times 10^{-13} \text{ J}\) Then, plug in the values for \(h\) and \(c\): \(\lambda_i = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.0 \times 10^8 \text{ m/s}}{3.2 \times 10^{-13} \text{ J}} = 6.20 \times 10^{-12} \text{ m}\)

Use the Compton scattering formula to find the change in wavelength

The Compton scattering formula relates the initial photon wavelength (\(\lambda_i\)), final photon wavelength (\(\lambda_f\)), electron's rest mass, and the scattering angle (\(\theta\)). It is given by: \(\lambda_f - \lambda_i = \frac{h}{m_e c} (1 - \cos{\theta})\) Where \(\frac{h}{m_e c}\) is the Compton wavelength and is approximately equal to \(2.43 \times 10^{-12} \text{ m}\). The scattering angle is given in degrees, so we need to convert it to radians: \(\theta_\text{rad} = 53^{\circ} \times \frac{\pi}{180} \approx 0.925 \text{ rad}\) Now, plug in the initial wavelength, Compton wavelength, and the scattering angle: \(\Delta\lambda = \lambda_f - \lambda_i = 2.43 \times 10^{-12} \text{ m} \times (1 - \cos{0.925}) = 1.72 \times 10^{-12} \text{ m}\)

Find the final photon wavelength

Now that we have the change in wavelength, we can find the final photon wavelength: \(\lambda_f = \lambda_i + \Delta\lambda = 6.20 \times 10^{-12} \text{ m} + 1.72 \times 10^{-12} \text{ m} = 7.92 \times 10^{-12} \text{ m}\) So, the wavelength of the scattered photon is approximately \(7.92 \times 10^{-12} \text{ m}\).