Question

An X-ray photon with an energy of 50.0 keV strikes an electron that is initially at rest inside a metal. The photon is scattered at an angle of \(45^{\circ}\), What are the kinetic energy and the momentum (magnitude and direction) of the electron after the collision? You may use the non relativistic relationship between the kinetic energy and the momentum of the electron.

Short answer

In this problem, we determined the kinetic energy and momentum of an electron after being struck by an X-ray photon using Compton scattering. The kinetic energy of the electron is found to be \(0.54 \times10^{-15}\, J\), while the magnitude of the electron's momentum is \(3.45\times10^{-23}\, kg\,m/s\). The direction of the electron's momentum is opposite to the final momentum of the photon.

Step-by-step solution

Calculate the initial and final wavelengths of the photon

The initial energy of the photon is given as 50.0 keV. We need to convert it to its corresponding wavelength using Planck's relation, \(E = h c / \lambda\), where \(E\) is the energy of the photon, \(h\) is the Planck's constant (6.626 x \(10^{-34}\) J*s), \(c\) is the speed of light (3 x \(10^8\) m/s) and \(\lambda\) is the wavelength of the photon. The energy unit needs to be in joule: 50.0 keV \(\times\) 1.602 x \(10^{-16}\) J/eV = 8.01 x \(10^{-15}\) J. Now, we can find the initial wavelength of the photon: \(\lambda_i = \frac{h c}{E_i}~=~\frac{(6.626 \times10^{-34} \,J*s) (3\times10^8\, m/s)}{8.01\times10^{-15}\,J}$$\approx~ 2.48\times10^{-11}\,m\). Next, we'll use the Compton Scattering formula to find the final wavelength of the photon, \(\lambda_f\): \(\lambda_f−\lambda_i~=~\left(\frac{h}{m_ec}\right)(1−cos\theta)=~\frac{(6.626 \times10^{-34}\, J*s)(1-(\cos{45^{\circ}}))}{(9.11\times10^{-31}\,kg)(3\times10^8\, m/s)}\approx 1.76\times10^{-12}\,m \). Now, we can calculate the final wavelength \(\lambda_f = \lambda_i + (\lambda_f - \lambda_i) = 2.48\times10^{-11}\,m + 1.76\times10^{-12}\,m = 2.66\times10^{-11}\,m\).

Calculate the energy transferred to the electron

Now we will calculate the final energy \(E_f\) and energy transferred \(\Delta E\) to the electron using Planck's relation. The final energy of the photon is: \(E_f = \frac{h c}{\lambda_f} = \frac{(6.626 \times10^{-34}\,J*s)(3\times10^8\,m/s)}{2.66\times10^{-11}\,m}\approx7.47\times10^{-15}\,J\). Now we can calculate the energy trasferred to the electron, \(\Delta E = E_i - E_f = (8.01 - 7.47) \times10^{-15}\,J =\ 0.54 \times10^{-15}\,J\). Consequently, this energy will be equal to the kinetic energy of the electron after the collision.

Calculate the momentum of the electron

Since we are allowed to use the non-relativistic relationship between kinetic energy and momentum, we will use \(KE = \frac{p^2}{2m_e}\), where \(KE\) is the kinetic energy and \(p\) is the momentum of the electron. Rearranging the equation to solve for momentum: \(p = \sqrt{2m_e KE} = \sqrt{2(9.11\times10^{-31}\,kg)(0.54\times10^{-15}\,J)} \approx 3.45\times10^{-23}\, kg\,m/s\).

Calculate the direction of the electron's momentum

We have the momentum magnitude; now, we need to calculate the direction of the momentum. For this, we'll use the law of cosines for the momentum triangle: \(p_e^2 = p_i^2 + p_f^2 - 2p_ip_f \cos\theta\) Since the electron is initially at rest, \(p_i = 0\), and we simplify the equation: \(p_e^2 = p_f^2 - 2p_f(0)\cos\theta = p_f^2\) Comparing this to the magnitude of the electron's momentum calculated earlier \((3.45\times10^{-23}\,kg\,m/s)\), we notice they have the same value, so the electron's momentum is in the opposite direction to the final momentum of the photon. #Conclusion# The kinetic energy of the electron after the collision is \(0.54 \times10^{-15}\, J\). The magnitude of the electron's momentum is \(3.45\times10^{-23}\, kg\,m/s\), and the direction is opposite to the final momentum of the photon.