Problems on Compton Effect and X-rays (Serway et al. p.94-95)

Answers (R. Egerton)

22. E = hc/(lambda) = (6.63E34)(3.00E8)/(500E9) = 3.97E-19 J = 2.48 eV

momentum = h/(lambda) = (6.63E34)/(500E-9) = 1.33E-27 kg m/s^2

24. (a) wavelength shift = (h/mc) (1- cos(theta))

= (6.63E34)/[(9.11E-31)(3.00E8)] (1-0.866)

= (2.43E-12)(0.134) = 3.25E-13 m

(b) energy of scattered x-ray = hc/(lambda') = hc/(lambda + 3.25E-13)

where lambda = hc/E = (6.63E-34)(3.00E8)/[(300E3)(1.60E-19)] = 4.133E-12m

so scattered-photon energy = (6.63E-34)(3.00E8)/(4.46E-12) = 4.456E-14 J = 278 keV

(c) recoil-electron energy = initial-photon energy - scattered-photon energy

= 300 keV - 278 keV = 22 keV

38. Bragg law is: n(lambda) = 2d sin(theta), where theta is the Bragg angle

The smallest angle corresponds to n = 1, giving sin(theta) = 0.616E-10/(8.00E-10)

= 0.077, so that theta = 4.426 deg and scattering angle = 2(theta) = 8.83 deg

n=2 gives sin(theta) = (2)(0.616E-10)/(8.00E-10) = 0.154, scattering angle = 17.7 deg

n=3 gives sin(theta) = (3)(0.616E-10)/(8.00E-10) = 0.231, scattering angle = 26.6 deg

Positrons and other particles (Serway p. 653)

3. The minimum photon frequency corresponds to the case of zero kinetic energy of the proton and antiproton, in which case: Mc^2 + Mc^2 + 0 = hf + hf

where M = proton rest mass = 1.673E-27 kg, giving hf = Mc^2 = 1.50E-10 J = 938 MeV

minimum frequency = 2.26E23 Hz, wavelength = 1.32E-15 m