General Physics (PHY 2140) Lecture 14

Transcription

1 General Physics (PHY 2140) Lecture 14 Modern Physics 1. Relativity Einstein s General Relativity 2. Quantum Physics Blackbody Radiation Photoelectric Effect X-Rays Diffraction by Crystals The Compton Effect Chapter 26 Chapter 27

2 Lightning Review Last lecture: Einstein s s Postulates: For all inertial frames of reference: Velocity of light in vacuo,, c, is the same (~ m/s). The form of the laws of physics (E/M, Mechanics, thermodynamics, etc.) are identical. Relativity consequences Time dilation, length contraction Relativistic energy, momentum Δt p Δ t = Relativistic addition of velocities v c L = L 1 v c p 2 2 v ab = v ad 1+ + v v v c db ad db 2 p mv =γ mv v c KE = γmc2 mc2

3 Problem: relativistic proton A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV.. Determine (a) the momentum and (b) the speed of the proton.

4 A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV. Determine (a) the momentum and (b) the speed of the proton. Given: E = 50.0 GeV mc 2 = MeV Find: p =? v =? Recall that E 2 = p 2 c 2 + (mc 2 ) 2. This can be used to solve for p: Thus, ( ) ( ) ( ) E mc mc + KE mc p = = c c ( KE) 2 + 2( mc 2 )( KE) p = = 50.9GeV c c Similarly with velocity: 2 1 E = γ ( mc ) γ = = v c E mc E v= 1 1 = c 2 mc 2 2

5 Problem: relativistic pion The average lifetime of a π meson in its own frame of reference (i.e., the proper lifetime) is s. If the meson moves with a speed of 0.98c,, what is (a) its mean lifetime as measured by an observer on Earth and (b) the average distance it travels before decaying as measured by an a observer on Earth? (c) What distance would it travel if time dilation did not occur?

6 The average lifetime of a π meson in its own frame of reference (i.e., the proper lifetime) is s. If the meson moves with a speed of 0.98c,, what is (a) its mean lifetime as measured by an observer on Earth and (b) the average distance it travels before decaying as measured by an observer on Earth? (c) What distance would it travel if time dilation did not occur? Given: v = 0.98 c τ p = s Find: τ =? d =? d n =? Recall that the time measured by observer on Earth will be longer then the proper time. Thus for the lifetime τ p τ = = v c Thus, at this speed it will travel ( )( )( ) d = vτ = m s s = 38m If special relativity were wrong, it would only fly about 8 8 ( )( )( ) d = vτ = m s s = 7.6m p s

7 Problem: more spaceships A spaceship travels at 0.750c relative to Earth. If the spaceship fires a small rocket in the forward direction, how fast (relative to the ship) must it be fired for it to travel at 0.950c relative to Earth?

8 A spaceship travels at 0.750c relative to Earth. If the spaceship fires a small rocket in the forward direction, how fast (relative to the ship) must it be fired for it to travel at 0.950c relative to Earth? Given: Since v ES = -v SE = velocity of Earth relative to ship, the relativistic velocity addition equation gives v SE = c v RE = c Find: v RS =? v RS = vre v 1+ + v v 2 c RE ES ES c+ ( c) ( 0.950c)( 0.750c) = = c 1+ c 2

9 Mass Inertial vs. Gravitational Mass has a gravitational attraction for other masses F = g G Mass has an inertial property that resists acceleration F i = m i a The value of G was chosen to make the values of m g and m i equal m g r m 2 ' g

10 Einstein s s Reasoning Concerning Mass That m g and m i were directly proportional was evidence for a basic connection between them No mechanical experiment could distinguish between the two masses He extended the idea: no experiment of any type can distinguish the two masses

11 Postulates of General Relativity All laws of nature must have the same form for observers in any frame of reference, whether accelerated or not In the vicinity of any given point, a gravitational field is equivalent to an accelerated frame of reference without a gravitational field This is the principle of equivalence

12 Implications of General Relativity Gravitational mass and inertial mass are not just proportional, but completely equivalent A clock in the presence of gravity runs more slowly than one where gravity is negligible The frequencies of radiation emitted by atoms in a strong gravitational field are shifted to lower frequencies This has been detected in the spectral lines emitted by atoms in massive stars

13 QUICK QUIZ 26.5 Two identical clocks are in the same house, one upstairs in a bedroom and the other downstairs in the kitchen. Which statement is correct? (a) The clock in the kitchen runs more slowly than the clock in the bedroom. (b) The clock in the bedroom runs more slowly than the clock in the kitchen. (c) Both clocks keep the same time.

14 QUICK QUIZ 26.5 ANSWER (a). The downstairs clock runs more slowly because it is closer to the Earth and hence experiences a stronger gravitational field than the upstairs clock does. The time difference between the two clocks is about 4x10-16 s for each second that passes on the clock. Too small to observe without expensive equipment!

15 More Implications of General Relativity A gravitational field may be transformed away at any point if we choose an appropriate accelerated frame of reference a freely falling frame Einstein specified a certain quantity, the curvature of time-space space,, that describes the gravitational effect at every point

16 Curvature of Space-Time There is no such thing as a gravitational field According to Einstein Instead, the presence of a mass causes a curvature of time-space in the vicinity of the mass This curvature dictates the path that all freely moving objects must follow

17 Testing General Relativity General Relativity predicts that a light ray passing near the Sun should be deflected by the curved space-time created by the Sun s s mass The prediction was confirmed by astronomers during a total solar eclipse

18 Black Holes If the concentration of mass becomes great enough, a black hole is believed to be formed In a black hole, the curvature of space- time is so great that, within a certain distance from its center, all light and matter become trapped

19 Quantum Physics Chapter 27 Objectives 27.1 Explain how blackbody radiation implies the quantization of electromagnetic energy Describe how Einstein's photoelectric equation implies the existence of the photon Define bremsstrahlung radiation Explain how X-ray diffraction can be used to determine crystal structure Solve sample problems that demonstrate the particle-like aspects of radiation as predicted by the Compton effect Identify the particle and wave-like aspects of electromagnetic radiation Identify the wave-like aspects of particles, citing examples Define the wave function for particles Summarize the key aspects of the uncertainty principle Describe the operation of the scanning tunneling microscope.

20 Need for Quantum Physics Problems remained from classical mechanics that relativity didn t t explain Blackbody Radiation The electromagnetic radiation emitted by a heated object Photoelectric Effect Emission of electrons by an illuminated metal Spectral Lines Emission of sharp spectral lines by gas atoms in an electric discharge tube

21 Development of Quantum 1900 to 1930 Physics Development of ideas of quantum mechanics Also called wave mechanics Highly successful in explaining the behavior of atoms, molecules, and nuclei Quantum Mechanics reduces to classical mechanics when applied to macroscopic systems Involved a large number of physicists Planck introduced basic ideas Mathematical developments and interpretations involved such people as Einstein, Bohr, Schrödinger, de Broglie, Heisenberg, Born and Dirac

22 Blackbody Radiation An object at any temperature is known to emit electromagnetic radiation Sometimes called thermal radiation Stefan s s Law describes the total power radiated P Stefan s constant = σ AeT The spectrum of the radiation depends on the temperature and properties of the object 4 emissivity

23 Black Body Radiation

24 Blackbody Radiation Graph Experimental data for distribution of energy in blackbody radiation As the temperature increases, the total amount of energy increases Shown by the area under the curve As the temperature increases, the peak of the distribution shifts to shorter wavelengths

25 Wien s s Displacement Law The wavelength of the peak of the blackbody distribution was found to follow Wein s Displacement Law λ max T = x 10-2 m K λ max is the wavelength at the curve s s peak T is the absolute temperature of the object emitting the radiation

26 The Ultraviolet Catastrophe Classical theory did not match the experimental data At long wavelengths, the match is good At short wavelengths, classical theory predicted infinite energy At short wavelengths, experiment showed no energy This contradiction is called the ultraviolet catastrophe

27 Planck s s Resolution Planck hypothesized that the blackbody radiation was produced by resonators Resonators were submicroscopic charged oscillators The resonators could only have discrete energies E n = n h ƒ n is called the quantum number ƒ is the frequency of vibration h is Planck s s constant, x J s Key point is quantized energy states

28 Problem: a lightbulb Assuming that the tungsten filament of a lightbulb is a blackbody, determine its peak wavelength if its temperature is K.

29 Assuming that the tungsten filament of a lightbulb is a blackbody, determine its peak wavelength if its temperature is K. Given: Recall Wein;s law: λ max T = x 10-2 m K. It can be used to solve for λ max : T = 2900 K λ max = m K T Find: λ max =? Thus, λ max 2 ( m K) = = m= 2900K nm This is infrared, so most of the electric energy goes into heat!

30 Photoelectric Effect When light is incident on certain metallic surfaces, electrons are emitted from the surface This is called the photoelectric effect The emitted electrons are called photoelectrons The effect was first discovered by Hertz The successful explanation of the effect was given by Einstein in 1905 Received Nobel Prize in 1921 for paper on electromagnetic radiation, of which the photoelectric effect was a part

31 Photoelectric Effect Schematic When light strikes E, photoelectrons are emitted Electrons collected at C and passing through the ammeter are a current in the circuit C is maintained at a positive potential by the power supply

32 Photoelectric Current/Voltage The current increases with intensity, but reaches a saturation level for large ΔV s No current flows for voltages less than or equal to ΔV s, the stopping potential The stopping potential is independent of the radiation intensity Graph

33 Features Not Explained by Classical Physics/Wave Theory No electrons are emitted if the incident light frequency is below some cutoff frequency that is characteristic of the material being illuminated The maximum kinetic energy of the photoelectrons is independent of the light intensity The maximum kinetic energy of the photoelectrons increases with increasing light frequency Electrons are emitted from the surface almost instantaneously, even at low intensities

34 Einstein s s 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 s idea of quantization to electromagnetic radiation The photon s s energy would be E = hƒh Each photon can give all its energy to an electron in the metal The maximum kinetic energy of the liberated photoelectron is KE = hƒ h Φ Φ is called the work function of the metal

35 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 KE depends only on the frequency and the work function, not on the intensity

36 More Explanations The maximum KE increases with increasing frequency The effect is instantaneous since there is a one-to to-one one interaction between the photon and the electron

37 Verification of Einstein s s Theory Experimental observations of a linear relationship between KE and frequency confirm Einstein s s theory The x-intercept x is the cutoff frequency f c = Φ h

38 Problem: photoelectric effect Electrons are ejected from a metallic surface with speeds ranging up to m/s when light with a wavelength of λ = 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

39 Electrons are ejected from a metallic surface with speeds ranging g up to m/s when light with a wavelength of λ = 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface? Given: v = 4.6x10 5 m/s λ = 625 nm Find: φ =? ν =? Recall that KE max =hf - φ. This can be used to solve for φ. First find the kinetic energy mv 1 ( )( ) max KEmax = = kg ms = J Thus, hc Φ= max = max = λ Cutoff frequency is 19 hf KE KE J f c 19 Φ J = = = h J s Which equals 1.4 ev 14 Hz

40 Photocells Photocells are an application of the photoelectric effect When light of sufficiently high frequency falls on the cell, a current is produced Examples Streetlights, garage door openers, elevators

41 X-Rays Electromagnetic radiation with short wavelengths Wavelengths less than for ultraviolet Wavelengths are typically about 0.1 nm X-rays have the ability to penetrate most materials with relative ease Discovered and named by Roentgen in 1895

42 Production of X-rays, X 1 X-rays are produced when high-speed electrons are suddenly slowed down Can be caused by the electron striking a metal target A current in the filament causes electrons to be emitted These freed electrons are accelerated toward a dense metal target The target is held at a higher potential than the filament

43 Production of X-rays, X 2 An electron passes near a target nucleus The electron is deflected from its path by its attraction to the nucleus This produces an acceleration It will emit electromagnetic radiation when it is accelerated

44 Diffraction of X-rays X by Crystals For diffraction to occur, the spacing between the lines must be approximately equal to the wavelength of the radiation to be measured For X-rays, X the regular array of atoms in a crystal can act as a three-dimensional grating for diffracting X-raysX

45 Schematic for X-ray X Diffraction A continuous beam of X-X rays is incident on the crystal The diffracted radiation is very intense in certain directions These directions correspond to constructive interference from waves reflected from the layers of the crystal The diffraction pattern is detected by photographic film

46 Photo of X-ray X Diffraction Pattern The array of spots is called a Laue pattern The crystal structure is determined by analyzing the positions and intensities of the various spots This is for NaCl

47 Bragg s s Law The beam reflected from the lower surface travels farther than the one reflected from the upper surface If the path difference equals some integral multiple of the wavelength, constructive interference occurs Bragg s s Law gives the conditions for constructive interference 2 d sin θ = m λ,, m = 1, 2, 3 3

48 The Compton Effect Compton directed a beam of x-rays x toward a block of graphite He found that the scattered x-rays x had a slightly longer wavelength that the incident x-raysx This means they also had less energy The amount of energy reduction depended on the angle at which the x-rays x were scattered The change in wavelength is called the Compton shift

49 Compton Scattering Compton assumed the photons acted like other particles in collisions Energy and momentum were conserved The shift in wavelength is h Δ λ = λ λo = (1 cos θ) mc e

50 Compton Scattering, final The quantity h/m e c is called the Compton wavelength Compton wavelength = nm Very small compared to visible light The Compton shift depends on the scattering angle and not on the wavelength Experiments confirm the results of Compton scattering and strongly support the photon concept

51 QUICK QUIZ 27.1 An x-ray x photon is scattered by an electron. The frequency of the scattered photon relative to that of the incident photon (a) increases, (b) decreases, or (c) remains the same.

52 QUICK QUIZ 27.1 ANSWER (b). Some energy is transferred to the electron in the scattering process. Therefore, the scattered photon must have less energy (and hence, lower frequency) than the incident photon.

53 QUICK QUIZ 27.2 A photon of energy E 0 strikes a free electron, with the scattered photon of energy E moving in the direction opposite that of the incident photon. In this Compton effect interaction, the resulting kinetic energy of the electron is (a) E 0, (b) E, (c) E 0 E, (d) E 0 + E, (e) none of the above.

54 QUICK QUIZ 27.2 ANSWER (c). Conservation of energy requires the kinetic energy given to the electron be equal to the difference between the energy of the incident photon and that of the scattered photon.

55 QUICK QUIZ 27.3 A photon of energy E 0 strikes a free electron with the scattered photon of energy E moving in the direction opposite that of the incident photon. In this Compton effect interaction, the resulting momentum of the electron is: (a) E 0 /c (b) < E 0 /c (c) > E 0 /c (d) (E( 0 E)/ )/c (e) (E( E o )/c Ephoton = pc

56 QUICK QUIZ 27.3 ANSWER (c). Conservation of momentum requires the momentum of the incident photon equal the vector sum of the momenta of the electron and the scattered photon. Since the scattered photon moves in the direction opposite that of the electron, the magnitude of the electron s momentum must exceed that of the incident photon. p electron E+ E = c 0