Modern Physics Part II : Introduction to Quantum physics Ch.# 40

1. Modern Physics Part II : Introduction to Quantum physics Ch.# 40 2 weeks (8 lectures) مدرس المادة الدكتور: 1431 – 1430 1

2. Lecture # 1 Ch. 40 1- Planck’s Quantum Hypothesis; Blackbody Radiation Any object in thermal equilibrium at any temperature is constantly emitting and absorbing radiation. However, not all materials are equally capable of absorbing and emitting radiation in different parts of the spectrum (i.e. in all frequency 0 to infinity). According to Kirchhoff’s law, any object which is a good absorber of radiation of a particular wavelength is also a good emitter of radiation of the same wavelength. A body which absorbs radiation of ALL wavelengths is called BLACKBODY. 2

3. What is a blackbody? An object that absorbs all incident radiation, i.e. no reflection. At thermal equilibrium with its surroundings radiates as much energy as it absorbs, i.e. blackbody is perfect emitter as well as absorber 3

4. The radiation is absorbed in the walls of the cavity • This causes a heating of the cavity walls • Atoms in the walls of the cavity will vibrate at frequencies characteristic of the temperature of the walls • These atoms then re-radiate the energy at this new characteristic frequency The spectrum of blackbody radiation has been measured; it is found that the frequency of peak intensity increases linearly with temperature. 4

5. This figure shows blackbody radiation curves for three different temperatures. Note that frequency increases to the left. Plots characteristics: 1. The distribution of frequencies is a function of the temperature of the blackbody. 2. The total amount of radiation emitted increases with increasing temperature. 3. The position of the peak maximum shifts toward higher frequencies with increasing equilibrium temperature 4. The wavelength at the peak of the curve is inversely proportional to the absolute temperature . Wien’s displacement law 5

6. lpeak vs Temperature 2.9 x 10-3 m T(Kelvin) T l peak = 3100K (body temp) 2.9 x 10-3 m 3100 =9x10-6m infrared light 58000K (Sun’s surface) visible light 2.9 x 10-3 m 58000 =0.5x10-6m 6

7. The birth of the quantum theory • This spectrum could not be reproduced using 19th-century physics. A solution was proposed by Max Planck in 1900:The energy of atomic oscillations within atoms cannot have an arbitrary value; it is related to the frequency: The constant h is now called Planck’s constant. Planck found the value of his constant by fitting blackbody curves: Planck’s proposal was that the energy of an oscillation had to be an integral multiple of hf. This is called the quantization of energy. 7

8. Basic Laws of Radiation • All objects emit radiant energy. 8

9. Basic Laws of Radiation • All objects emit radiant energy. • Hotter objects emit more energy than colder objects. 9

10. Basic Laws of Radiation • All objects emit radiant energy. • Hotter objects emit more energy than colder objects. The amount of energy radiated is proportional to the temperature of the object. 10

11. Basic Laws of Radiation • All objects emit radiant energy. • Hotter objects emit more energy than colder objects. The amount of energy radiated is proportional to the temperature of the object raised to the fourth power. •  This is the Stefan Boltzmann Law • F =  T4 • F = flux of energy (W/m2) • T = temperature (K) •  = 5.67 x 10-8 W/m2K4 (a constant) All objects emit radiation whose total intensity is proportional to the fourth power of their temperature. This is called thermal radiation; a blackbody is one that emits thermal radiation only. 11

12. Basic Laws of Radiation • All objects emit radiant energy. • Hotter objects emit more energy than colder objects (per unit area). The amount of energy radiated is proportional to the temperature of the object. • The hotter the object, the shorter the wavelength () of emitted energy. 12

13. Basic Laws of Radiation • All objects emit radiant energy. • Hotter objects emit more energy than colder objects (per unit area). The amount of energy radiated is proportional to the temperature of the object. • The hotter the object, the shorter the wavelength () of emitted energy. • This is Wien’s Law 13

14.  Stefan Boltzmann Law. F =  T4 F = flux of energy (W/m2) T = temperature (K)  = 5.67 x 10-8 W/m2K4 (a constant)  Wien’s Law 14

15. We can use these equations to calculate properties of energy radiating from the Sun and the Earth. 6,000 K 300 K 15

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18. 1000 100 10 1 0.1 0.01 Electromagnetic Spectrum visible light infrared ultraviolet x-rays microwaves Low Energy High Energy  (m) 18

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20. Blue light from the Sun is removed from the beam • by Rayleigh scattering, so the Sun appears yellow • when viewed from Earth’s surface even though its • radiation peaks in the green 20

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22.  Stefan Boltzman Law. F =  T4 F = flux of energy (W/m2) T = temperature (K)  = 5.67 x 10-8 W/m2K4 (a constant) 22

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24. 1. The human eye is most sensitive to 560-nm light. What is the temperature of a black body that would radiate most intensely at this wavelength? 24

25. 2. (a) Lightning produces a maximum air temperature on the order of 104 K, whereas (b) a nuclear explosion produces a temperature on the order of 107 K. Use Wien’s displacement law to find the order of magnitude of the wavelength of the thermally produced photons radiated with greatest intensity by each of these sources. Name the part of the electromagnetic spectrum where you would expect each to radiate most strongly. (a) (b) 25

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27. Lecture # 2 Ch. 40 3- The Photoelectric Effect What Is The Photoelectric Effect? when light illuminates a piece of metal, the light will kick off electrons from the metal’s surface and these electrons can be detected as a change in the electric charge of the metal or as an electric current, emitted electrons are called PHOTOELCTRONs. Hence the name: photo for light and electric for the current. The explanation behind this simple phenomenon opened the door to revolutionary modern physics concepts regarding the composition of light, quantum mechanics, and what is now referred to as the “wave-particle duality” of nature. 27

28. What is the Photoelectric Effect? • A photon with energy hf strikes an electron and ejects it from the metal. E=hf = K.E. + w :- • w = work to remove electron from metal • K.E. = kinetic energy of ejected electron. 28

29. The remarkable fact that the ejection energy was independent of the total energy of illumination showed that the interaction must be like that of a particle which gave all of its energy to the electron! This fit in well with Planck's hypothesis that: light in the blackbody radiation experiment could exist only in discrete bundles with energy Most commonly observed phenomena with light can be explained by waves. But the photoelectric effect suggested a particle nature for light. 29

30. We can obtain the following results experimentally 1- if v is kept constant, the photoelectric current increases with increasing intensity I of the incident radiation. 2- Photoelectrons are emitted within less than 10-9 sec after the surface is illuminated by light, i.e. NO TIME LAG BETWEEN ILLUMINATION AND EMISSION. 3- The emission of photoelectrons takes place ONLY if the frequency of the incident radiation is equal to or greater than a certain minimum frequency vs called the THRESHOLD FREQUENCY . vs is different for different surfaces. 30

31. 4-The maximum kinetic energy, Kmax, of photoelectrons is independent of the intensity I of the incident light. The stopping potential is the same for light of three different intensities but of the same frequency I3˃ I2 ˃ I1 31

32. 5- The maximum kinetic energy, Kmax, of photoelectrons depend on the frequency of the incident radiation. The stopping potential is different for different v even though I is the same vs1˃ vs2 ˃ vs3 32

33. 6- There is a linear relation between Kmax, and v. 33

34. Explanation of Classical “Problems” • The effect is not observed below a certain cutoff frequency since the photon energy must be greater than or equal to the work function • Without this, electrons are not emitted, regardless of the intensity of the light • The maximum Kmax depends only on the frequency and the work function, not on the intensity 34

35. Einstein’s “quantum theory” Explanation • A tiny packet of light energy, called a photon, would be emitted when a quantized oscillator jumped from one energy level to the next lower one • Extended Planck’s idea of quantization to electromagnetic radiation • The photon’s energy would be E = hƒ • Each photon can give all its energy to an electron in the metal • The maximum kinetic energy of the liberated photoelectron is Kmax = hƒ – Φ • Φ is called the work function of the metal 35

36. Photoelectric effect (Albert Einstein) • The photoelectric effect is a phenomenon where electrons are ejected from the surface of a metal (or some other material) when light shines on it • shine light rays --> current • increase intensity(total energy) --> current goes up • That makes sense. • The strange thing about this phenomenon, is that a certain minimum frequency of light is required before any current is detected. 36

37. Verification of Einstein’s Theory • Experimental observations of a linear relationship between KE and frequency confirm Einstein’s theory • The x-intercept is the cutoff frequency 37

38. Cutoff Wavelength • The cutoff wavelength is related to the work function • Wavelengths greater than lC incident on a material with a work function f don’t result in the emission of photoelectrons 38

39. Example: Light of wavelength λ=5893 A° (589.3 nm)is incident on a potassium surface. The stopping potential for the emitted electrons (i.e. the maximum energy of the photoelectron) is 0.36 volt. Calculate: A) the work function, and B) the threshold frequency (or cutoff wavelength). A) The work function is , B) The cutoff wavelength is given by the relation 39

40. 4- Application of the Photoelectric Effect The most obvious example is probably solar energy, which is produced by photovoltaic cells. These are made of semi-conducting material which produce electricity when exposed to sunlight. An everyday example is a solar powered calculator and a more exotic application would be solar power satellites that orbit around the earth. - The photo-cell The photo-cell is the most ranging of applications of the photoelectric effect. It is most commonly found in solar panels. it works on the basic principle of light striking the cathode which causes the emission of electrons, which in turn produces a current. 40

41. 5- Compton effect Arthur H. Compton observed the scattering of x-rays from electrons in a carbon target and found scattered x-rays with a longer wavelength than those incident upon the target. The shift of the wavelength increased with scattering angle according to the Compton formula (Compton Shift Equation): 41

42. Compton explained and modeled the data by assuming a particle (photon) nature for light and applying conservation of energy and conservation of momentum to the collision between the photon and the electron. The scattered photon has lower energy and therefore a longer wavelength according to the Planck relationship. Q: In the Compton shift equation, show that the maximum shift is given as , where is the Compton wavelength 42

43. Derivation of the Compton shift equation: Energy conservation: Momentum conservation: Three equations, three unknowns (l', v and f)  eliminating v and f  relation between l', l0 and q : 43

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45. 7. Calculate the energy, in electron volts, of a photon whose frequency is 3.10 GHz, and determine the corresponding wavelength for this photon and state the classification of the electromagnetic spectrum 45

46. 22. X-rays having an energy of 300 keV undergo Compton scattering from a target. The scattered rays are detected at 37.0° relative to the incident rays. Find (a) the Compton shift at this angle, (b) the energy of the scattered x-ray, and (c) the energy of the recoiling electron. : a: : b: c: 46

47. An X-ray photon with =60 nm is scattered over 1500 by a target electron. • Find the change of its wavelength. • Find the angle between the directions of motion of the recoil electron and the incident photon. • Find the energy of the recoil electron. (a) 47

48. momentum conservation: (b) along y axis: along x axis: (c) 48

49. An X-ray photon with wavelength 0.8nm is scattered by an electron at rest. After the scattering the electron recoils with a speed equal to 1.4106m/s (non-relativistic case). NOTE: IN THE FOLLOWING, DO THE CALCULATIONS TO AT LEAST 4-DECIMAL PLACE ACCURACY (a) Calculate the energy of the scattered photon in units eV (use conservation of energy). (b) Calculate the Compton shift in the photon’s wavelength, in meters. (c) Calculate the angle through which the photon was scattered. (a) 49

50. (b) (c) 50