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3-de Broglie Wavelength, phenomena and its related concept..pptx
- 1. Do Photons HaveMomentum ? What is momentum ? Photons have energy and a finite velocity so there must be some momentum associated with photons ! Just like Energy, TOTAL MOMENTUM IS ALWAYS CONSERVED De Broglie Waves
- 6. Soal #1:Berapa panjang gelombang sebuah elektron (massa = 9,11 x 10-31 kg) yang bergerak dengan kecepatan 5,31 x 106 m/s? Soal #2: Berapa panjang gelombang dalam meter dari sebuah proton yang bergerak dengan kecepatan 255.000.000 m/s (yang merupakan 85% kecepatan cahaya)? (Asumsikan massa proton adalah 1,673 x 10-27 kg.) Soal #3: Hitung panjang gelombang (dalam nanometer) atom H (massa = 1,674 x 10-27 kg) yang bergerak dengan kecepatan 698 cm/s Soal #4: Berapa panjang gelombang bola bisbol seberat 5,00 ons yang melaju dengan kecepatan 100,0 mil per jam? (5,00 oz = 0,14175 kg dan 100 mph = 44,70 m/s) Soal #5: Sebuah atom helium memiliki panjang gelombang de Broglie 4,30 x 10-12 meter. Berapa kecepatan dan energi kinetiknya? massa= 6.646632348 x 10-27 kg
- 7. Soal #6:Hitung kecepatan elektron (massa = 9,10939 x 10-31 kg) yang memiliki panjang gelombang de Broglie 269,7 pm Soal #7: Hitung kecepatan neutron dengan panjang gelombang 65 pm. Massa neutron = 1,67493 x 10-27 kg Soal #8: Hitung panjang gelombang de Broglie dari sebuah neutron (massa = 1,67493 x 10-27 kg) yang bergerak dengan kecepatan seperlima ratus kecepatan cahaya (c/500). Soal #9: Hitung panjang gelombang suatu benda yang beratnya 100,0 kg dan bergerak dengan kecepatan 160 km/jam. Soal #10: Berapa panjang gelombang de Broglie (dalam nm) dari molekul buckminsterfullerene (C60), yang bergerak dengan kecepatan 100,0 m/s?
- 8. The Compton Effect In1924, A. H. Compton performed an experiment where X-rays impinged on matter, and he measured the scattered radiation. Problem: According to the wave picture of light, the incident X-ray should give up some of its energy to the electron, and emerge with a lower energy (i.e., the amplitude is lower), but should have . Incident X-ray wavelength Scattered X-ray wavelength e Electron comes flying out It was found that the scattered X-ray did not have the same wavelength ! e
- 9. • The scatteringof photons from charged particles is called Compton scattering after Arthur Compton who was the first to measure photon-electron scattering in 1922. • When the incoming photon gives part of its energy to the electron, then the scattered photon has lower energy and according to the Planck relationship has lower frequency and longer wavelength. • The wavelength change in such scattering depends only upon the angle of scattering for a given target particle. • The constant in the Compton formula above can be written: The quantity h/mec is known as the Compton wavelength of the electron; it is equal to 2.43×10−12 m. The wavelength shift λ′ − λ is at least zero (for θ = 0) and at most twice the Compton wavelength of the electron (for θ = 180).
- 10. Phys. Rev. 21,483 – Published 1 May, 1923. DOI: https://doi.org/10.1103/PhysRev.21.483.
- 11. Q1. A high-energy X-rayphoton is used in a laboratory experiment to investigate photon-electron interactions. The photon is incident on a stationary free electron. If the X-ray is scattered at an angle of 90°, calculate the change in the wavelength of the photon. What is the physical significance of this result? Q2. Gamma rays emitted by a radioactive source are scattered by electrons in a target material. You observe a scattered photon at 135°. Given the incident photon has an energy of 511 keV, determine the wavelength of the scattered photon and the kinetic energy transferred to the electron.
- 12. Q3. In a studyof high-energy astrophysics, cosmic photons collide with electrons in the upper atmosphere, producing Compton scattering effects. For an incoming photon of energy 1 MeV scattered at 180°, calculate (a) the scattered photon energy, and (b) the maximum possible kinetic energy of the recoiled electron. Q4. During an experiment in medical imaging, a 100 keV photon is scattered by an electron at an angle of 60°. Determine the Compton shift in wavelength and the energy of the photon after scattering. Explain briefly how this affects image resolution in medical imaging.
- 13. Q5. A researcher wantsto confirm the quantum nature of light using Compton scattering. She uses a beam of photons with a known wavelength of 0.025 nm. If the photons are scattered at 150°, what is the final wavelength of the scattered photons? What does this imply about the particle-like behavior of light? Q6. In a physics class demonstration, a laser with a photon energy of 2.0 eV is directed at electrons at rest. The students are asked to observe any significant Compton scattering effects. Explain whether significant scattering occurs and justify your reasoning using the Compton formula.
- 14. Q7. An unknown photonsource emits radiation that, when scattered at 90°, results in a wavelength increase of 0.0015 nm. What is the Compton wavelength of the particle involved? Based on your result, suggest whether the scattering occurred with an electron, proton, or neutron.
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- 21. Quantum Picture tothe Rescue Incident X-ray e Electron initially at rest (almost) Scattered X-Ray e Ee e Compton found that if you treat the photons as if they were particles of zero mass, with energy and momentum . The collision behaves just as if it were two billiard balls colliding ! Photon behaves like a particle with energy & momentum as given above!
- 22. Photon Momentum IN FREESPACE: IN OPTICAL MATERIALS:
- 23. Manifestation of thePhoton Momentum Conservation of linear momentum implies that an atom recoils when it undergoes spontaneous emission. The direction of photon emission (and atomic recoil) is not predictable. A well-collimated atomic beam of excited atoms will spread laterally because of the recoil associated with spontaneous emission. A source emitting a spherical wave cannot recoil, because the spherical symmetry of the wave prevents it from carrying any linear momentum from the source. SOURCE EMITTING A PHOTON SOURCE EMITTING AN EM WAVE excited atom de -excited atom photon source of excited atoms collimating diaphragms beam spreads laterally because of spontaneous emission
- 24. SOLAR SAIL at rest INCOMINGPHOTONS 1000 W/m2 every second photons with momentum + (1000 J/m2 )/c impact the sail SOLAR SAIL moves REFLECTED PHOTONS 1000 W/m2 every second photons with momentum - (1000 J/m2 )/c leave the sail with momentum + (2000 J/m2 )/c … and gets that much more momentum every second … Pressure acting on the sail = (2000 J/m2 ) /c /second = 6.7 Newtons/km2 Photon Momentum - Moves Solar Sails Photon Momentum - Moves Solar Sails Image by D. Kassing http://en.wikipedia.org/wiki/File:SolarSail- DLR-ESA.jpg on Wikipedia Image in the Public Domain
- 25. Classical Picture QuantumPicture Energy of EM wave ~ (Amplitude)2 Energy per photon photon momentum Photons Electrons knocked loose by photons - - - EM wave Electrons shaken loose by an EM wave - - -
Editor's Notes
- #21 The relationship that p=E/c is obtained from the Energy-momentum formula which Einstein derived (Special Relativity). He showed that for particles which are moving close to the speed of light, they have a total energy given by: E2 = sqrt( p2c2 + m2c4), Where p = momentum, c=speed of light, and m=mass of the particle. For m=0 (like a photon), we get E=pc, or, by rearranging, p=E/c. This explanation is beyond what I expect of you, but I provided it just in case you’re interested in where this relation p=E/c came from.