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- Wisconsin
- University of Wisconsin - Madison
- Physics
- Physics 208
- Carlsmith
- Exam3 Solution fall06

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Name: ____SOLUTIONS__________________________________ Student ID: ___________________________ Section #: _________ Physics 208 Exam 3 Nov. 29, 206 Print your name and section clearly above. If you do not know your section number, write your TA’s name. Your final answer must be placed in the box provided. You must show al your work to receive ful credit. If you only provide your final answer (in the box), and do not show your work, you wil receive very few points. Problems wil be graded on reasoning and intermediate steps as wel as on the final answer. Be sure to include units, and also the direction of vectors. You are alowed one 8½ x 1” shet of notes and no other references. The exam lasts exactly 90 minutes. Problem 1: ____ / 20 Planck’s constant h = 6.626 x 10 -34 m 2 kg/s Problem 2: ____ / 20 Planck’s constant x velocity of light hc = 1240 eV x nm Problem 3: ____ / 20 Bohr radius a o = 0.053 nm Problem 4: ____ / 20 m elctron c 2 = 51,00 eV µ B = Bohr magneton = Problem 5: ____ / 20 5.78382 x 10 -5 eV/T h = h/2π TOTAL: ____ / 10 2 1) [20 pts, 4 pts each] Multiple choice/short answers. i) In Bohr’s model of the Hydrogen atom, which of the folowing clasical ideas is stil assumed to be valid? (circle AL that are corect) a. The electrons circulate about the nucleus in circular orbits. b. Atoms radiate because of the acceleration of the electrons. c. The force of atraction betwen the electrons and the nucleus is the usual Coulomb force. d. Al orbital radi are possible. e. None of the above. i) According to Bohr theory, when a Hydrogen atom akes a transition from an n = 5 to an n=2 state the radial distance of the electron from the nucleus changes by: a. 3a 0 b. 25a 0 c. 21a 0 d. 5a 0 e. a 0 ii) If a photon is emited in the same transition as in i), its energy wil be: a. 0.21 x 13.6 eV b. 21 x 13.6 eV c. 0.04 x 13.6 eV d. 25 x 13.6 eV e. none of the above iv) The n=3, ℓ=1 states of a Hydrogen atom al have the same energy in the absence of a magnetic field. In a 1 T magnetic field, the energy separation betwen the states with n=3 and ℓ=1 is: a. 5.78382 x 10 -5 eV b. 1.157676 x 10 -4 eV c. 0 d. 1.736515 x 10 -4 eV e. None of the above. v) True-false question: Electron microscopes can have much larger magnifying power than optical microscopes because the shorter wavelength of electrons alows beter resolution of images. TRUE: r=n 2 a o , so Δr=(5 2 - 2 ) a o =21a o Photon energy = diference betwen initial and final state energies = ! "13.6eV/5 2 "13.6eV/2 ( ) 13.6eV1/2"1/5 2 ( ) =0.21#13.6eV There are thre such states, all of the same energy in zero magnetic field corresponding to m l =-1, 0 or +1. In a magnetic field, al thre have diferent energy as ! " 13.6 n 2 "µ B m l . The energy diference betwen states is µ B =5.79"10 #5 eV/T ( ) 1T () =5.79"10 #5 eV 3 2) Thin Lenses [20 pts, 4 pts each] An object O (represented by the arow on the left of the lens) has an height of h=1.1 cm and is placed at a distance p = 3 cm from a converging thin lens with a focal length of 10 cm. a) Use principal rays to locate the image graphicaly in the figure below. b) State whether the image is real or virtual and find its distance from the thin lens. Virtual. ! 1 3 + 1 q = 1 10 " 1 q =0.1#0.333=#0.233 q=#4.29cm c) Find the lateral magnification. Lateral mag = -q/p = 4.29/3=1.43 d) Find the height of the image. ! Image height = Magobject height ( ) =1.43"1.1cm () =1.57cm e) Where should we put the real object to obtain a real and inverted image? Chose one of the folowing answers: 1) p = f 2) p > f 3) p < f q = Value Units M = Value Units h’ = Value Units f f p O h 4 Interference 3) [20 pts, 5 pts each] Two narow slits separated by d =1.8 m are iluminated by yelow light of wavelength 589 nm from a sodium lamp. The interference patern is observed on a scren 4 m far away. For this proble, ignore the difraction efects from individual slits. a) Calculate the spacing betwen the bright fringes. Constructive interference: dsinθ=mλ tanθ=y/L, small angle approx: tanθ~sinθ dy/L~mλ -> spacing = Lλ/d =(4m)x(589x10 -9 m)/(1.8x10 -3 m)=1.31 m b) If the scren extends on the two sides of the central maximum of the interference patern by 4 cm on each side (total scren length 8 cm), how many bright fringes in total wil be visible on the scren? On left side, highest order visible is (1.8mm)(40mm)/4000mm=m(589x10 -6 m) m=30.5. So max order visible is m=30. Same on other side. Add central line to get 61 total bright fringes. c) Calculate the phase diference betwen the 2 interfering waves at a point P on the scren a distance y = 2m from the center of the central fringe. Path length dif = dsinθ~yd/L Phase diference = 2π(yd/L)/λ ! =2" 2mm#1.8mm 4000589#10 $6 mm ( ) % & ’ ( * =9.6rad d) If the maximum intensity on the scren is I max = 0.05 W/m 2 , which is the light intensity at point P? I P = I max cos 2 (Φ/2) = 3.8e-4 W/m 2 y = Value Units # fringes= Value Units Φ = Value Units I P = Value Units 5 4) Difraction of waves [a) 10 pts] In the same situation that we considered in problem 3), let us now consider also the difraction efects of the slits that have a width of a = d/10 = 0.18 m. a) How many bright fringes are seen in the central difraction maximum? [Hint: remember that the single slit difraction modulates the overal intensity of the two- slit interference patern.] The first difraction minimum is given by ! asin"=#. The y-position on the scren is given by tan#sin"=y min /L So the first difraction min occurs at ! y min =L"/a4000mm ( ) 589#10 $6 mm ( ) /0.18mm=13.1mm From before, the interference maxima are 1.31 m apart. So the tenth int. max. occurs at the difraction minimum and is not visible. 9 fringes on either side, + 1 in the middle, gives 19 fringes visible. N = Value Units 6 Difraction of electrons [b) and c) 5 pts each] A beam of electrons of momentum p moves horizontaly toward a narrow horizontal square slit of width a. After leaving the slit, the beam spreads verticaly. The half-width θ of this angular spread is ilustrated in the figure, where θ is the angle of the first minimum of difraction. Derive a formula for the angle θ, using two complementary aproaches: b) Find the de Broglie wavelength of the electrons, and treat the problem as difraction of a wave by a single slit. Expres θ in terms of p, a and h. [Hint: Consider θ very smal and find the first minimum in the single slit patern.] The deBroglie wavelength is λ=h/p. The first minimum in the single slit pattern is at θ=λ/a in the small-angle approximation, so we have θ=h/pa. x Electron beam p y a 7 c) Use the uncertainty principle to determine the vertical component p y of the momentum after the electrons pas through the slit, and again expres θ in terms of p, a and h. [To get the same answer as in (b) let the quantity on the right side of the uncertainty principle equation be h rather than h/4π and consider that only the y-component of the momentum varies when elctrons pass through the slit.] The uncertainty principle is ∆y ∆p y =h, where ∆y is the slit width, a. Then ∆p y =h/a. This is the y- component of the momentum after the electron leaves the slit. The x- component is stil p. The deflection angle is thus θ=tan - 1 (p y /p x )=tan - 1 (h/p a). In the smal angle aproximation, tanθ~θ we get θ~ h/p a as before. 5) Photoelectric Efect [20 pts, 5 pts each] The work function for cesium is 1.9 eV. a) Calculate the cut-of wavelength for the photoelectric efect. The cut-of wavelength is the maximum wavelength for which electrons are ejected from the metal. This is when the photon energy is equal to the work function. ! hc " = 1240 eV#nm " =1.9eV$"=652.6nm b) Calculate the maximum kinetic energy of ejcted electrons when the wavelength of the incident light is 30 nm. If the incident light has λ=30nm, it has energy ! hf= hc " . The maximum kinetic energy of an electron is this energy minus the work function ! = 1240 eV"nm 300 #1.9eV=2.23 eV λ c = Value Units K max = Value Units 8 c) What is the wavelength of the ejected electrons that have the maximum kinetic energy? The wavelength is the deBroglie wavelength, ! "=h/p. The momentum is related to the kinetic energy as ! KE=p 2 /m"p=2mKE Then ! "= h 2mKE = hc 2mc 2 KE = 1240 eV#nm 20.511$10 6 eV ( ) 2.23eV ( ) =0.82 nm d) If the intensity of the beam of photons on the cesium surface is 1 W/m 2 , assuming that the wavelength of the photons is 60 nm, calculate the number of photons that strike an area of 1 cm 2 each second. 1 W/m 2 =1 (J/s)/m 2 = 6.242x10 18 (eV/s)/m 2 A 60 nm photon has energy ! E= hc " = 1240 eV#nm 600 =2.067eV The flux is then 6.242x10 18 (eV/s)/m 2 / (2.067 eV/photon) =3.02x10 18 (photons/s)/m 2 Since there are 10 4 cm 2 in 1 m 2 , the flux / cm 2 is 3.02x10 14 (photons/s)/cm 2 λ= Value Units Ν= Value Units Mark Rzchowski Microsoft Word - Phy208Exam3_V4Sol.doc

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