Fysik 1 Kompendium: Del 2 Standard Modellen samt Ljus

Transcription

1 Fysik 1 Kompendium: Del 2 Standard Modellen samt Ljus Klass: Na2 Lärare: VT14

2 Relativitet: Heureka Kapitel 13 (s ) samt Ergo Formel Skriv ner här alla formel som du kommer i kontakt med under din läsning. Se till att även skriva ner vad varje variabel betyder och de konstanter som behövs. 2

3 Relativitet: Heureka Kapitel 13 (s ) samt Ergo Definitioner Fyll i några korta meningar som beskriver betydelse av varje term. De fyra krafterna: Gravitation: Elektromagnetiska Kraften Starka Kraften Svaga Kraften nukleon/nuklid 3

4 Relativitet: Heureka Kapitel 13 (s ) samt Ergo ljusets dubbelnatur Kvantmekanik LHC/ATLAS Standardmodellen kvark 4

5 Relativitet: Heureka Kapitel 13 (s ) samt Ergo Higgs-boson Frågor 1. Vilka typer av forskning gör dem på Cern? 2. Hur skiljer sig den standardmodellen jämfört med Bohrs eller Rutherfords? 5

6 Relativitet: Heureka Kapitel 13 (s ) samt Ergo Vad menas med tomrum? Är det korrekt att säga att en atom mest innehåller luft? 4. Varför kallas det den starka kraften? Beskriv Newtons teoretiska kanon? 5. Vad kom Einstein fram till om ljusets energi? 6. Vad är Unification Teori och varför är det så eftersökt? 7. Ge exempel på ljusets dubbelnatur. 6

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8 8 Standardmodellen

9 * Kvarker påverkas av den starkakraften * Leptons påverkas INTE av starkakraften 9

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14 Rekommenderade Övningar: Holt-Physics-Problem-Workbook samt Heurek Kap 13 14

15 NAME DATE CLASS Holt Physics Problem 23A QUANTUM ENERGY P R O B L E M Free-electron lasers can be used to produce a beam of light with variable wavelength. Because the laser can produce light with wavelengths as long as infrared waves or as short as X rays, its potential applications are much greater than for a laser that can produce light of only one wavelength. If such a laser produces photons of energies ranging from ev to ev, what are the minimum and the maximum wavelengths corresponding to these photons? S O L U T I O N Given: E1 = ev = (1.034 ev) e JV = J E 2 = ev = (620.6 ev) e JV = J h = J s c = m/s Unknown: l min =? l max =? Use the equation for the energy of a quantum of light. Use the relationship between the frequency and wavelength of electromagnetic waves. E = hf f = l c Substitute for f in the first equation, and rearrange to solve for wavelength. E = h c l l = h c E Substitute values into the equation. l max = l max = m l max = l min = ( J s)( m/s) ( J) 1210 nm ( J s)( m/s) ( J) l min = m Copyright Holt, Rinehart and Winston. All rights reserved. l min = 2.00 nm 186 Holt Physics Problem Workbook 15

16 NAME DATE CLASS ADDITIONAL PRACTICE 1. In 1974, IBM researchers announced that X rays with energies of J had been guided through a light pipe similar to optic fibers used for visible and near-infrared light. Calculate the wavelength of one of these X-ray photons. 2. Some strains of Mycoplasma are the smallest living organisms. The wavelength of a photon with J of energy is equal to the length of one Mycoplasma. What is that wavelength? 3. Of the various types of light emitted by objects in space, the radio signals emitted by cold hydrogen atoms in regions of space that are located between stars are among the most common and important. These signals occur when the spin angular momentum of an electron in a hydrogen atom changes orientation with respect to the spin angular momentum of the atom s proton. The energy of this transition is equal to a fraction of an electron-volt, and the photon emitted has a very low frequency. Given that the energy of these radio signals is ev, calculate the wavelength of the photons. 4. The camera with the fastest shutter speed in the world was built for research with high-power lasers and can expose individual frames of film with extremely high frequency. If the frequency is the same as that of a photon with J of energy, calculate its magnitude. 5. Wireless cable television transmits images using radio-band photons with energies of around J. Find the frequency of these photons. Copyright Holt, Rinehart and Winston. All rights reserved. 6. In physics, the basic units of measurement are based on fundamental physical phenomena. For example, one second is defined by a certain transition in a cesium atom that has a frequency of exactly s 1. Find the energy in electron-volts of a photon that has this frequency. Use the unrounded values for Planck s constant (h = J s) and for the conversion factor between joules and electron volts (1 ev = J). 7. Consider an electromagnetic wave that has a wavelength equal to 92 cm, a length that corresponds to the longest ear of corn grown to date. What is the frequency corresponding to this wavelength? What is its photon energy? Express the answer in joules and in electron-volts. 8. The slowest machine in the world, built for testing stress corrosion, can be controlled to operate at speeds as low as m/s. Find the distance traveled at this speed in 1.00 year. Calculate the energy of the photon with a wavelength equal to this distance. 16 Problem 23A 187

17 Atomic Physics Chapter 23 Additional Practice 23A Givens 1. E = J C = m/s h = J s Solutions l = h c = E ( J s)( m/s) J l = m = nm 2. E = J C = m/s h = J s 3. E = ev C = m/s h = J s l = h c = E l = m l = h c = E l = m ( J s)( m/s) J ( J s)( m/s) ( ev)( J/eV) II 4. E = J h = J s E = hf f = E h J f = J s f = Hz Copyright Holt, Rinehart and Winston. All rights reserved. 5. E = J h = J s 6. f = s 1 h = J s 1 ev = J 7.l= 92 cm = m c = m/s h = J s h = ev s f = E 23 h = J 34 = Hz J/ s E = hf E = ( J s)( s 1 ) J/eV E = ev f = l c f = m/s m f = Hz = 330 MHz E = hf Section Two Problem Workbook Solutions II Ch

18 Givens Solutions E = ( J s)( Hz) E = J E = ( ev s)( Hz) E = ev II 8. v = m/s t = 1.00 year l = x c = m/s h = J s x = v t x = ( m/s)(1.00 year) days 24 h 1 year 1 day s 1 h x = E = hf = h c l E = m hc = x ( J s)( m/s) m E = J Additional Practice 23B 1. hf t = 4.5 ev KE max = 3.8 ev h = ev s f = [KE max + hf t ] [3.8 ev ev] = h 4 15 = Hz ev s 2. hf t = 4.3 ev KE max = 3.2 ev h = ev s 3. hf t,cs = 2.14 ev hf t,se = 5.9 ev h = ev s c = m/s KE max = 0.0 ev for both cases KE max = hf hf t f = KE max + hf t h 3. 2 ev ev f = ev s f = Hz a. KE max = hf hf t = 0.0 ev = h c hf t l hc l = h ft hc ( ev s)( m/s) l Cs = = hf t, Cs 2.14 ev l Cs = m = nm hc ( ev s)( m/s) b.l Se = = hf t, Se 5.9 ev Copyright Holt, Rinehart and Winston. All rights reserved. l Se = m = nm II Ch Holt Physics Solution Manual 18