Physics 30 Assignment

28 Marks Total ASSIGNMENT 14 Compton, de Broglie, and Wave-Particle Duality This assignment is worth 28 marks. The value of each question is noted in parentheses in the left margin. Note: The answer areas will expand to fit the length of your response. 1. (1) There are similarities and differences between the Photoelectric Effect and Compton Scattering. Complete each of the six partial statements below using the following guide; all you need to provide for an answer is PE, CS, BOTH, or NEITHER. PE if the statement only applies to the Photoelectric Effect CS if the statement only applies to Compton Scattering BOTH if the statement only applies to both the Photoelectric Effect and Compton Scattering NEITHER if the statement applies to Neither the Photoelectric Effect or Compton Scattering Energy is conserved in _____. Answer: Both (1) Photons are observed before and after the interaction in _____. Answer: CS (1) Electrons are observed as the result of the experiment in _____. Answer: both (1) Angles are measured in the experiment in _____. Answer: CS (1) Photons with very low energies such as 5.0 to 10.0 eV is observed in _____. Answer: Pe (1) Ionization occurs in _____. Answer: Both (2) 2. What quantity measured in the Compton effect experiment show the wave-particle duality of light? Answer: The quantity that is measure in Compton effect experiment and shows the wave-particle are energy and momentum, the Compton effect is a result of a high energy photon colliding a target. (5) 3. An X-ray with a frequency of 3.74 × 1020 Hz is incident on a thin piece of metal. The lower frequency X-ray on the other side is observed deflected at 48o. What is the frequency of the deflected X-ray? Answer: lambda1-lambda0 = lambdaC(1-cos?)   lambda1 = lambda0+lambdaC(1-cos?)  lambda0 = c/f0 = 8.0158411E-13 m  If it's a collision with an electron,  lambda1 = 1.6043759E-12 m  f1 = c/lambda1 = 1.8685924E+20 Hz.  1.87 x 1020hz If it's a collision with a proton,  lambda1 = 8.0202133E-13 m  f1 = c/lambda1 = 3.7379612E+20 Hz. 3.74 x 10^20 hz (5) 4. A scientist changes the frequency of an incident X-ray to 4.50 × 1019 Hz and measures the deflected X-ray frequency of 4.32 × 1019 Hz. What was the angle of deflection? Answer: ?? = ?f - ?i ?? = (h/mc)(1-cos?) ? = c/f. ?i = c/f = (3.00 x 10^8 m/s) / (4.50 x 10^19 Hz) = 6.66666 x 10^-12 m ?f = c/f = (3.00 x 10^8 m/s) / (4.32 x 10^19 Hz) = 6.9444 x 10^-12 m ?? = ?f - ?i = (6.9444x 10^-12 m) - (6.666666x 10^-12 m) = 2.77777 x 10^-13 m ?? = (h/mc)(1-cos?) 2.77777 x 10^-13 m = ((6.63 x 10^-34 Js) / (9.11 x 10^-31 kg)(3.00 x 10^8 m/s))(1-cos?) 2.77777 x 10^-13 m = (2.4259056 x 10^-12)(1-cos?) 0.1145044 = 1-cos? cos? = 0.8854955 cos^-1(0.8854955) = 27.687° =27.7° (2) 5. Can the equation E = pc be applied to particles? Why or why not? Answer: No because particles cant travel at the speed of light. (3) 6. A stationary hydrogen atom with a mass of 1.67 × 10-27 kg absorbs a photon of light with 10.2 eV. What is the velocity of the hydrogen atom after absorbing the photon in a perfectly inelastic collision? Answer: Given: M= 1.67 x 10-27 kg E= 10.2 eV or 1.632 x 10^ -18J C= 3.00 x 108 E= pc E= m.v.c V= E/m.c V=1.632 x 10^ -18J / 1.67 × 10-27 kg x 3.00 10^8 m/s V= 3.26 m/s Describe the results of performing Young’s experiment with x-rays and then high speed electrons. Answer: When young's experiment is performed with x-rays, a phenomenon known as Bragg's diffraction occurs according to which electrons are preferentially scattered in certain special directions where as when performed with high speed electrons, termed at the Davisson-Germer experiment which demonstrated the wave nature of the electron, confirming the earlier hypothesis of deBroglie. Putting wave-particle duality on a firm experimental footing, it represented a major step forward in the development of quantum mechanics. The Bragg law for diffraction applied to x-ray diffraction is also applicable to particle waves. Bragg's law states that : When x-rays are scattered from a crystal lattice, peaks of scattered intensity are observed which correspond to the following conditions: The angle of incidence = angle of scattering. The path length difference is equal to an integer number of wavelengths. The condition for maximum intensity contained in Bragg's law above allow us to calculate details about the crystal structure, or if the crystal structure is known, to determine the wavelength of the x-rays incident upon the crystal. n*(wavelength) = 2*d*sin(?) Where d is the spacing between the planes in the atomic lattice, and ? is the angle between the incident ray and the scattering planes. (2) 8. How do the results of performing Young’s experiment with x-rays and then high speed electrons support the wave-particle model? Answer: The experiment showed that an electron traveling at high speeds are capable of making interference patters, which is a characteristic of waves. (1) 9. All of the following quantities can be measured or calculated for light waves and subatomic particles except _____. momentum velocity frequency energy Answer: C When you have completed all of the questions in this assignment, submit your work.