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X-rays with a wavelength of , =9.587pm are used in a Compton scattering experiment (i.e. the photons are scattered by essentially free electrons). What is the energy, E , of the incident photons in eV? а. b. When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0° , what is the wavelength of the Compton scattered X-ray? When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 = 145.0°, what is the change in energy for the X-ray photon in eV (or the difference in energy between the scattered photon and the incident photon)? С.

X-rays with a wavelength of , =9.587pm are used in a Compton scattering experiment (i.e. the photons are scattered by essentially free electrons). What is the energy, E , of the incident photons in eV? а. b. When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0° , what is the wavelength of the Compton scattered X-ray? When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 = 145.0°, what is the change in energy for the X-ray photon in eV (or the difference in energy between the scattered photon and the incident photon)? С.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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11. X-rays with a wavelength of 1, = 9.587 pm are used in a Compton scattering experiment (i.e. the photons are scattered by essentially free electrons). What is the energy, E , of the incident photons in eV? а. b. When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0°, what is the wavelength of the Compton scattered X-ray? When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0°, what is the change in energy for the X-ray photon in eV (or the difference in energy between the scattered photon and the incident photon)? d. с. When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0°,what is the change in kinetic energy for the scattered electron in eV?
Transcribed Image Text:11. X-rays with a wavelength of 1, = 9.587 pm are used in a Compton scattering experiment (i.e. the photons are scattered by essentially free electrons). What is the energy, E , of the incident photons in eV? а. b. When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0°, what is the wavelength of the Compton scattered X-ray? When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0°, what is the change in energy for the X-ray photon in eV (or the difference in energy between the scattered photon and the incident photon)? d. с. When the Compton scattered X-ray scatters to an angle (measured from the direction of the incident X-ray) of 0 =145.0°,what is the change in kinetic energy for the scattered electron in eV?
Waves in general: FORMULA PAGE 1 a y 1-dimensional wave equation: 1 a'y ; here v is the speed of the wave v? ôt? Solution: f(x- vt) or f(x+vt) Harmonic or sinusoidal waves: y(x,t)= Asin(kx- ot) 2л k 2n = 27f; v=- T v = f2 General Constants: -34 h = 6.626×10*J.s = 4.13567×10¬eV ·s ; (with recent revisions to the SI system of units Planck's Constant is defined to have an exact value: h= 6.62607015×10¯“J·s) –34 -19 hc = 1240 eV · nm; hc=1239.84eV · nm (for more accuracy); leV =1.6022×10-J = 299,792, 458 m /s (exact); -31 electron mass: m, =9.1094×10' kg proton mass: m, =1.6726×10-27 kg Photons: E = hf hc ; Protons: m,c² = 938.3MEV , Electrons: m.c² = 511.0keV %3| h = 1.0546x10 34J•s = 6.5821×10-1eV ·s Chapter 36. Diffraction Single slit diffraction: Minima: a sin 0, = ma, m=1,2,3,...where a is the slit width, note: there is a maximum at 0 = 0 sin(a) па Intensity: I(0) = ,, a = -sin(0) m a Circular aperture: First minimum: sin 0 = 1.22- Rayleigh's criterion ( 1 <d ): a =1.22- d Double slit experiment with slit separation d and slit width a: sin a Intensity: I(0) = I„(cos? B)| where B = -sin 0 , a = па -sin O Grating equation (normal incidence): d sin 0 = m order in which the grating is being used, d is the line or groove spacing m = 0,1, 2,3,... (maxima), where m is the
Transcribed Image Text:Waves in general: FORMULA PAGE 1 a y 1-dimensional wave equation: 1 a'y ; here v is the speed of the wave v? ôt? Solution: f(x- vt) or f(x+vt) Harmonic or sinusoidal waves: y(x,t)= Asin(kx- ot) 2л k 2n = 27f; v=- T v = f2 General Constants: -34 h = 6.626×10*J.s = 4.13567×10¬eV ·s ; (with recent revisions to the SI system of units Planck's Constant is defined to have an exact value: h= 6.62607015×10¯“J·s) –34 -19 hc = 1240 eV · nm; hc=1239.84eV · nm (for more accuracy); leV =1.6022×10-J = 299,792, 458 m /s (exact); -31 electron mass: m, =9.1094×10' kg proton mass: m, =1.6726×10-27 kg Photons: E = hf hc ; Protons: m,c² = 938.3MEV , Electrons: m.c² = 511.0keV %3| h = 1.0546x10 34J•s = 6.5821×10-1eV ·s Chapter 36. Diffraction Single slit diffraction: Minima: a sin 0, = ma, m=1,2,3,...where a is the slit width, note: there is a maximum at 0 = 0 sin(a) па Intensity: I(0) = ,, a = -sin(0) m a Circular aperture: First minimum: sin 0 = 1.22- Rayleigh's criterion ( 1 <d ): a =1.22- d Double slit experiment with slit separation d and slit width a: sin a Intensity: I(0) = I„(cos? B)| where B = -sin 0 , a = па -sin O Grating equation (normal incidence): d sin 0 = m order in which the grating is being used, d is the line or groove spacing m = 0,1, 2,3,... (maxima), where m is the
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