Physics 2203, Fall 2012 Modern Physics. Friday, Sept. 7th, 2012: Chapter 4: QuanAzaAon of light Start Chapter 5: QuanAzaAon of Atomic Energy Levels

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1 Physics 2203, Fall 2012 Modern Physics. Friday, Sept. 7th, 2012: Chapter 4: QuanAzaAon of light Start Chapter 5: QuanAzaAon of Atomic Energy Levels Announcements: First computer exercise next Wednesday. Home work #3 on Ch. 4 and 5 will be posted today. I will be in Washington, DC, Monday and Wednesday. Professor Zhang will teach this class

2 Isaac Newton ( ) English Physicist, MathemaAcian James Clerk Maxwell ( ) ScoYsh Physicist, MathemaAcian. Physicists might have rewri[en the biblical story of creaaon as follows! In the beginning He created the heaven and the earth F G mm' And He said let there be light Q E ida = E ids = dφ B dt BidA = 0 ε 0 Bids = µ 0 I +ε 0 µ 0 dφ E dt r 2 = ma

3 The first hint that there was something wrong with this picture of the creaaon came from the the intensity of radiaaon coming for a heated cavity (black body) as a funcaon of the wave length. Ultraviolet catastrophe 1000 K QuanAzed Light Max Planck ( ) German, Nobel prize in 1918

4 Max Planck The objecave is to explain the amount of light intensity that is radiated from an object as a funcaon of temperature. We consider what is described as a blackbody. The walls are black with a very small hole, which is the blackbody 1918 Nobel Prize

5 Black Body RadiaBon: Create a cavity with only a very small opening. This pictures shows light entering the cavity but since emission and absorpaon are connected by Kirchhoff s Theorem a perfect absorber is an ideal radiator. Josef Stefan ( ) observed that the total power e total per unit area was proporaonal to T 4 e total = e f df = σt 4 0 In 1893 Wilhelm Wien proposed a general form of the black body radiaaon formula, which gave the correct dependence of the maximium in the distribuaon Wien s displacement law λ max T = 2.898x10 3 mik u( f,t ) = Af 3 β f /T e u is energy density/f

6 The Sun s radius is R s =7.0x10 8 m and the average Earth Sun distance is R=1.5x10 11 m. The power per unit area from the Sun is 1400 W/m 2. What is the temperature of the Sun? We need to connect R s (sun) to R on earth: e total (R s ) = σt 4 σ = 5.67x10 8 W im 2 ik 4 Conservation of energy e total (R s )i4πr s 2 = e total (R)i4πR 2 e total (R s ) = e total (R)iR2 R s 2 T = e total (R)iR2 2 σ R s 1/4 = 5800 K ObservaBon: The peak in the efficiency of the human eye occurs at a wave length of 500 nm. What temperature are you most sensiave to? WHY! T = 2.898x10 3 mik λ max = 2.898x10 3 mik 500x10 9 m = 5800K

7 Very careful measurements in infrared and far infrared regions of the spectrum showed that Wien s law failed in this region. u(λ,t ) = Aλ 3 β /λt e Look at the Wien equaaon for long wave lengths. u(λ,t ) = Aλ 3 e β /λt A λ 1 β 3 λt A λ 3 Experiment by Rubens showed a T dependence!!

8 On a Sunday evening in October of 1900 Planck discovered the famous blackbody formula, which ushered in the quantum theory. earlier in the day he had visited Rubens and found out about the T dependence of u(f,t). Planck s constant h u( f,t ) = 8π f 3 1 h=6.26x10 24 J/K c 3 e hf /kbt 1 h= 4.135x10 15 ev/k Boltzman s constant High frequency limit. k B =1.380x10 34 Js u( f,t ) = 8π f 3 1 8π f 3 c 3 e hf /kbt 1 c 3 Wien's formula Low frequency limit. e hf /k BT u( f,t ) = 8π f 3 1 8π f 3 1 = 8π f 2 c 3 e hf /kbt 1 c 3 1+ hf / k B T + -1 c 3 k B T h Linear

9 Planck believed that blackbody radiaaon was produced by vibraang submicroscopic electric charges called Resonators. In classical Maxwellian theory a Resonators could oscillate at any frequency and could lose any fracaon of its energy. Planck had to assume the Resonators could only have discrete values of the Energy and that the change in energy was quanbzed. E resonator = nhf n=1, 2, Angular frequency ω=f/2π E resonator = n ω n=1, 2, ΔE = hf ΔE = ω

10 Classical Rayleigh Jeans blackbody Ultraviolet catastrophe u(λ,t )dλ = 8π λ 4 k B Tdλ u( f,t )df = 8π f 2 c 3 k B Tdf 1000 K Planck s quantum blackbody u(λ,t )dλ = u( f,t )dλ = λ 5 c 3 8πhcdλ ( e hc/λkbt 1) 8πhf 3 df ( e hf /kbt 1)

11 In 1905 Einstein published a seminal paper on the paracle nature of light Photoelectric effect. Millikan set out to confirm these predicaons. R. A. Millikan: Phys. Rev (1916).

12 The observaaons consist of the measuring magnitude of the photoemi[ed current as a funcaon of: Frequency of the light f. The intensity of the light I. The bias voltage between the cathode and anode. the nature of the metal. Classical InterpretaBon: The energy in a light wave is proporaonal to the Intensity I, so The emi[ed electron should have more energy with larger Intensity. If the Intensity I is large enough there will always be electrons emi[ed, independent of f. There should be Ame dependence in the signal.

13 K max = 1 2 m v 2 e max = ev s ObservaAon #1: K max does not depend upon intensity. ObservaAon #2: K max is a linear funcaon of frequency. ObservaAon #3: There is a threshold frequency f 0. ObservaAon #4: There is no Ame lag between the start of illuminaaon and the start of the photocurrent. h[p://phet.colorado.edu/en/simulaaon/photoelectric

14 Why do the I V curves in the figure rise gradually between V s and 0? Why isn t there an abrupt raise in the photocurrent at V s? What does the slope of this curve tell you about the distribuaon of electrons inside the metal?

15 Einstein proposed that light comes in quanta, called a Photon. These Photons behave like paracles and can only be produced and absorbed as compete units. Energy of a Photon: E=hf= ω: h is Planck's constant. Momentum of a photon: p= E c = hf c = h λ A photon is absorbed by exciang an electron in the solid. As we will learn the maximum energy electrons are at the Fermi energy. The excited electron now has a kineac energy inside given by E=hf+E F, but it has to escape for the solid. It losses kineac energy equal to the work funcaon φ. 2 mv max 2 = hf φ h[p://phet.colorado.edu/en/simulaaon/photoelectric

16 ev 0 (cut off)=0 hf 0 = φ f 0 = φ h ω 0 = φ

17 Planck viewed this theory with skepacism and Millikan set out to prove him wrong. Millikan set out to measure the cut off as a funcaon of f. The slope should be Planck s constant and the intercept the work funcaon. In 1916 he published data proving Einstein right The measured slope was 4.1x10 15 ev. s, in good agreement with the value of h. The intercept is about 2 ev which is OK for an alkali metal. h[p://phet.colorado.edu/en/simulaaon/photoelectric

18 Suppose that light of total intensity 1.0µW/cm 2 falls on a clean iron sample 1.0 cm 2 in area. Assume that the sample reflects 96% of the light only 3% of the incident light lies in the violet region of the spectrum above the threshold frequency. a) What intensity is actually available for the photoelectric effect? I = (0.03)(0.040)I 0 = (0.03)(0.040)(1.0µW / cm 2 ) = 1.2nW / cm 2 b) Assuming that all the photons in the violet region have an effecave wavelength of 250 nm, how many electrons will be emi[ed per second? N(electrons / sec) = 1.2x10 9 W hf N = = λ ( ) 1.2x10 9 hc ( 250x10 9 m) ( 1.2x10 9 J / s) ( 6.6x10 34 Jis) ( 3x10 8 m / s) = 2 1.5x109

19 Suppose that light of total intensity 1.0µW/cm 2 falls on a clean iron sample 1.0 cm 2 in area. Assume that the sample reflects 96% of the light only 3% of the incident light lies in the violet region of the spectrum above the threshold frequency. c) Calculate the current in the phototube in amperes. N = 1.5x10 9 I = N(1.6x10 19 C) = 2.4x10 10 A d) If the work funcaon is 4.5 ev, what is f 0? f 0 = φ h = 4.5eV 4.14x10 15 evis = 1.1x1015 Hz e) Find the stopping voltage V s if the light is 250 nm. ( )( 3.0x10 8 m / s) ev s = hf φ = hc λ φ = 4.14x10 15 evis 250x10 9 m 4.5eV = 0.46V

20 Wilhelm Röntgen ( Nobel Prize in recogniaon of the extraordinary serves he has rendered by the discovery of the remarkable rays subsequently named awer him. Bremsstrahlung breaking radiaaon CharacterisAc RadiaAon

21 CharacterisAc RadiaAon Bremsstrahlung λ min Voltage on tube V 0 Maximum energy of photon is ev 0 ev 0 = E max = hf max = hc λ min = ω Duane Hunt Rule λ min = 1.24x10 6 V m V 0

22 Intensity (a.u.) Ru 4s Ru 4p Sr 4s Sr 4p O 2p-Ru 4d Ru 3d 3/2 Sr 3 Ru 2 O 7 Sr 3p 1/2 Ru 3d 5/2 Sr 3p 3/2 Sr 3d 5/2 Sr 3d 3/2 Intensity (a.u.) Femi Edge!E~0.15eV Binding Energy (ev) Binding Energy (ev)

23 Arthur Compton ( ) This is a very beauaful sca[ering experiment. An incident xray or gamma ray is sca[ered through an angle θ by a collision with an electron a[ached to a nucleus. The electron is sca[ered through an angle φ. Α calculaaon of conservaaon of momentum and energy is needed to determine the change in the wave length of the sca[ered photon. Nobel Prize 1927 for his discovery of the effect named a8er him, Compton effect The momentum of the photon is p p = E c = h λ The momentum of the electron is p e = E 2 ( mc 2 ) 2 c

24 Conservation of E: E+m e c 2 = E '+ E e Conservation of momentum: p=p'cosθ+p e cosφ p'sinθ=p e sinφ Eliminate φ p e 2 = p' 2 + p 2 2 pp'cosθ E e = hf hf '+ m e c 2 A lot of algebra! λ ' λ 0 = h m e c ( 1 cosθ)

25 Arthur Compton ( ) Nobel Prize 1927 for his discovery of the effect names a8er him, Compton effect λ 2 λ 1 = h mc ( 1 cosθ) = λ C ( 1 cosθ) λ C = h mc = hc mc = 1.24x103 ev nm x10 5 ev = nm hc = 1.24x10 3 evinm

26 X rays of wavelength λ=0.200 nm are aimed at a block of carbon. The sca[ered xrays are observed at an angle of 45 o the the incident beam. Calculate the increased wavelength of the sca[ered x rays at this angle. λ ' λ 0 = h m e c ( 1 cosθ) = hc m e c 2 ( 1 cosθ ) = λ 1 1 C 2 Δλ = ( nm) ( 0.293) = nm λ ' = Δλ + λ 0 = nm

27 (a) Why are x ray photons used in the Compton experiment, rather than visiblelight photons? To answer this let us find Δλ for three different incident photon wavelengths. (1) very high energy γ rays with λ= Å, (2) x rays with λ=0.712 Å, and (3) green light from Hg lamp, with λ=5461 Å. Note 10 Å=1 nm Δλ = ( Å) ( 1) = Å For all incident wavelengths (b) The electrons in C have a 4 ev binding energy. Why can this be ignored in the Compton equaaon for x rays? Look at λ=0.712 Å. Energy E=hf= hc λ E = 12400eViÅ Now calculate Δλ λ 0 γ ray Δλ λ 0 = Å Å = 2.29 x ray Δλ λ 0 = Å 0.712Å = green Δλ λ 0 = Å 5461ÅÅ = 4.45x Å = 17,400 ev Resolving power λ Δλ

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29 Experimentalist had measured very carefully the emission and absorpaon spectra of H. They found an empirical formula to describe their observaaons. In 1885 Balmer described the visible spectrum: Balmer Series 1 λ = R : n= 3, 4, 5,... 2 n 2 R= nm In general 1 λ = R 1 n' 1 2 n 2 Lyman is n =1 Paschen is n =3 E = hc λ ( ) = hcr 1 E n,n' n' 1 2 n 2

30 E = hf = hc λ ( ) = hcr 1 ΔE n,n' n' 2 1 n 2 Rydberg Energy E R = hcr = 13.6eV Ionization potential n'=1 to n= I = = 13.6ev Define E n = E R n 2 h[p://web.phys.ksu.edu/vqm/sowware/online/vqm/html/h2spec.html

31 Classical picture H spectra, noace that most of the lines are not in the visible! Visible

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