Question

[10%) A photon of initial energy 0.1 MeV undergoes Compton scattering at an angle of 60°. What is the energy of the scattered photon? (The rest energy of electron is mec 0.511 MeV.) A. B. C. D. 8.361x104 eV 8.55×104 eV 8.741x104 eV 9.111x104 eV "V E-pc [10%] It's known that energy levels of hydrogen follows E,--13.6/n2 eV with n 1, 2, 3, . .. Consider a process that an electron in hydrogen makes a transition from the n 2 state to the n 1 state without emitting a photon. Instead, the excess energy is transferred to an outer electron in the n 4 state, which is ejected by the atom. What is the kinetic energy of the ejected electron? A. 2.55 eV B. 9.35 eV C. 10.2 eV D. 11.05 eV 12

Answer 1

Formula E_0 = hc/lambda_0 implies lambda_0 = 6.626 times 10^-34 times 3 times 10^3 /1.602 times 10^-19 times 0.1 times 10^6 lambda_0 = 1.241 times 10^-2 nm wavelength shift due to compton scattering at 60 degree is, Delta lambda = hc /m_e c^2 (1 - cos theta) = 1.240 times 10^3/511 times 10^3 (1 - cos 60) Delta lambda = 1.213 times 10^-3 nm Then lambda' = lambda_0 + Delta lambda implies 1.362 times 10^-2 nm Then E' = hc/lambda' implies 6.626 times 10^-34 times 3 times 10^3 /1.362 times 10^-11 E' = 9.104 times 10^4 ev almostequalto 9.111 times 10^4 ev option (D) is correct.