5 A beam of high energy electrons is incident on a iridium (Z=77) target. One of the the electrons inthe beam ejects a K-shell (n=1) electron from an iridium atom. An M-shell (n= 3) electron then makesa downward transition to fill the result-ing hole, and an X-ray is emitted. After its emission the X-rayphoton encounters a free electron and is Compton scattered by it through an angle of 60°. Find thewavelength of the scattered X-ray photon. This is two-step problem. First work a characteristic X-rayproblem to find the wavelength of the X-ray photon emitted. Then work a Compton scattering problemto find the scattered wavelength of the photon. Round your answer to 3 significant figures and box it. *Hints: (1) Zeff is the number of protons in the nucleus minus the number of shielding electrons.(2) The number of electrons in a full shell is 2n2, where n is the principal quantum number.

Given: The electron makes a transition from n=3 to n=1. The angle between the free electron and Compton scattered photon is 60∘. Introduction: An X-ray is a penetrating form of high-energy electromagnetic radiation. Most X-rays have a wavelength ranging from 10 picometers to 10 nanometers, corresponding to frequencies in the range of 30 petahertz to 30 exahertz.Calculation: Write the expression of the energy of the x-ray photon emitted. E=-Z-121n12-1n22E0 Here, n1 is the initial shell level of the electron, n2 is the final shell level and E0 is the ground state energy of the hydrogen atom. Substitute 77 for Z, 3 for n1, 1 for n2 and 13.6 eV for E0 in the above expression. E=-77-12132-112(13.6 eV)E=-762(-12.089 eV)E=69.83 keV Write the expression of the x-ray photon. λ=hcE Here, h is the Planck's constant, c is the speed of light in the vacuum and E is the energy of the photon. Substitute 4.14×10-15 eVs for h, 3×108 ms-1 for c and 69.83 keV for E in the above expression. λ=4.14×10-15 eVs3×108 ms-169.83×103 eVλ=0.18×10-10 m Write the expression of the scattered wavelength. λ'-λ=hmec1-cosθ Substitute 0.180×10-10 m for λ, 6.63×10-34 Js for h, 9.1×10-31 kg for me, 3×108 ms-1 for c and 60∘ for θ in the above expression. λ'=0.180×10-10 m+6.6×10-34 Js9.1×10-31 kg3×108 ms-11-cos60∘λ'=0.180×10-10 m+0.2421-12×10-11 mλ'=0.180×10-10 m+0.121×10-11 mλ'=0.192×10-10 m Thus, the wavelength of the scattered photon is 0.192×10-10 m.

Answered: 5 A beam of high energy electrons is…