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Physics 200 Final Exam November 16, 2005 Name______________________________ Part I Multiple Choice Choose the one alternative that best completes the statement or answers the question below. [4pts x 6 = 24 points] 1. A light beam in air is incident on a surface of glass as shown in the figure below. Which of the marked arrows is the actual path of the beam in the glass? Answer: _____A______ (A) J (B) K (C) H air glass H J K 2. A light beam traveling in glass is incident on a glass-air interface as shown in the figure below. If the angle of incidence is smaller than the critical angle for glass, which of the choices given below describes what will happen to the beam? Answer: _____D______ (A) The light beam will be totally reflected along L. L (B) Some of the light will be reflected along L, some of it will pass undeflected along K. glass (C) Some of the light will be refracted along J, some of it will air H be reflected along L. (D) Some of the light will be refracted along H, some of it K J will be reflected along L. 3. In a double-slit experiment with two slits named S1 and S2, the path difference (S1P – S2P) to a point P on the screen that will result in P being the first off-center dark fringe is equal to (A) λ/4 (B) λ/2 (C) λ (D) 2 λ Answer: _____B______ 4. The pattern below shows a double slit interference pattern. The bright fringes are shown in white, and dark fringes are in black. Which of the patterns below will result if the distance between the slits is increased? Answer: ____A_______ (A) (B) (C) Final Exam Fall 2005 (D) Page 1 of 9 5. A positively charged particle moves in a magnetic field as shown in the drawing. (x) means the magnetic field is going into the paper. What is the direction of the force on the particle? (A) Out of the paper (toward you) (B) Into the paper (away from you) (C) To the left ← (D) to the right → Answer: _____C______ 6. Suppose that the distance between two successive nodes in a standing wave pattern on a string is 10 centimeters and the speed of the wave in the string is 40 meters per second. What is the frequency of the wave? (A) 20 Hz (B) 40 Hz (C) 80 Hz (D) 200 Hz Answer: _____D______ (E) 400 Hz Part II Short Answer: These questions can be answered in one, two or three short sentences. [6pts x 4 = 24 points] 1. Explain how fluorescence works. One example is the glow in the dark stickers I showed in class. Another is the minerals that glowed under UV light. The process is – it absorbs a photon of one energy then emits several lower energy photons. When an atom or molecule absorbs a photon it may skip over several levels as it goes from a low energy state to a high energy state. As it deexcites and emits photons, it may deexcite through these smaller steps. That is, it absorbs one high energy photon and gets excited. It then emits more than one lower energy photon as it deexcites. With real numbers, a hydrogen atom may absorb a lot of energy and go from n = 1 to n = 3 (the energy of the photon absorbed is 12.089eV, which is very far ultraviolet). It may then deexcite going from n = 3 to n = 2, and emit a lower energy photon which we can see (the energy of the photon emitted is 1.89 eV which we can see). Then it may deexcite going from n = 2 to n = 1, emitting another lower energy photon (10.2 eV which is ultraviolet, but not as far ultraviolet). The process is – it absorbs a photon of one energy then emits several lower energy photons. 2. State the three main findings that led Bohr to his planetary model for the hydrogen atom. Thomson’s discovery of the electron. Rutherford’s discovery of a small, dense atomic nucleus. Balmer’s formula describing the wavelengths of light coming out of a hydrogen atom. Final Exam Fall 2005 Page 2 of 9 3. Two experimenters study the double slit pattern for photons. Observer A detects the photons with both slits open. Observer B first records the pattern with the top slit open and the bottom closed and then adds the results with the bottom slit open and the top closed. What differences, if any do the two observers notice? Observer A sees a two slit interference pattern, with interference fringe maxima following dsin = m. Observer B does not see the two slit interference. Instead she sees bright areas directly behind each of the slits. Really, she sees single slit diffraction patterns behind each slit, which for the most part look like a single bright area behind each slit. 4. Briefly say deBroglie's hypothesis on the electrons orbiting an atom and what this implies for electrons, neutrons, etc. Prince Louis de Broglie hypothesized that electrons orbiting the nucleus at certain distances acted like standing waves. If we remember standing waves on a string held by a wall at both ends, there are only certain wavelengths that are allowed for that length of string. For length L the wavelengths could be 2L, 2L/2, 2L/3, 2L/4 (that is: 2L, L, 2L/3, L/2, etc.) For a circle of radius r, the circumference at distance r is 2 r and so the standing waves of electrons that are allowed must fit an integral number of times around the circle. The important implication are: This quantizes the angular momentum that is allowed for that electrons, since mvr = nh/2 Also, this says that electrons may behave like waves. This leads to the further implication that neutrons, protons, atoms, baseballs, elephants may all behave like waves. Final Exam Fall 2005 Page 3 of 9 5. Extra Credit (6 points): Why do some things appear white, like this paper you're looking at now or a pile of sugar? We learned about this and had a demonstration in class in which I smashed a glass. Feel free to draw, but you must write descriptive sentences. The obvious answer is that is reflects all wavelengths of light, but then so do mirrors. The special thing about objects appearing white is that they reflect light off of many different layers of material, and scatter the light in many different directions. For example, for light coming straight in, glass in air reflects only 4% of the incoming light off of any surface. If you smash the glass into a powder you now have many glass/air interfaces to reflect the light off of and most of the light is reflected. If these crystals of powder have random directions, and random shapes, the light reflected no longer can form an image like off of a smooth piece of glass. The object appears white. Many materials have a higher index of refraction than glass, like titanium dioxide (n=2.4). More light is reflected off of each interface between titanium oxide and air, so fewer layers are needed to reflect back most of the light. That means a thin layer of titanium dioxide paint can make a wall look white. Water vapor on the other hand has a lower index of refraction, and so a thicker layer of fog is needed to appear white than you would need of titanium dioxide. 6. Another Extra Credit (3 points): Holograms don't record an image of the thing you're "taking a picture of". What do they record? Holograms record an interference pattern of the light from the object and the light from a reference beam. Part III Numerical [Do any 5 of the following 7 problems] 1. We observe two objects. Object A has a mass of 10 kilograms and moves at 2 m/s. Object B has a mass of 5 kilograms and moves at 4 m/s. (a) Which object has more kinetic energy? (show all work) [5 points] A: E = ½ m v2 = ½ (10 kg) ( 2m/s)2 = ½ (10 kg)( 4 m2/s2) = 20 kg m2 / s2 = 20 J B: E = ½ m v2 = ½ (5 kg) ( 4m/s)2 = ½ (5 kg)( 16 m2/s2) = 40 kg m2 / s2 = 40 J Object B has more kinetic energy (b) Which object has more linear momentum? (show all work) [5 points] A; p = mv = (10 kg)(2 m/s) = 20 kg m/s B; p = mv = (5 kg)(4 m/s) = 20 kg m/s Object A and B have the same amount of linear momentum Final Exam Fall 2005 Page 4 of 9 2. A double slit experiment is set up as shown in the figure below. S1 and S2 are the two slits (whose size is shown highly exaggerated for clarity), and point A is midway between S1 and S2, so that any point on AB is equidistant from S1 and S2. P If a laser of wavelength 500 nanometers is used in the S2 experiment, how far away from B will the second offcenter bright fringe on the screen be located? The distance A B between slits S1 and S2 is 0.25 millimeters, and the screen is at a distance of 5 meters from the slits. [10 points] S1 Use the formula dy/L = m The wavelength = 500 x 10 -9 meters. The length L is the distance from the slits to the screen: L = 5 meters. The distance between the slits, d = 0.25 mm = 0.25 x 10 -3 meters. We are looking for the second off center bright fringe, that means the variable m = 2. y is the distance from the center of the interference pattern to the specific bright fringe (the second bright fringe in this problem). That is, it’s the distance between point B and point P. Solve for y: y = m L/d = [(2)( 500 x 10 -9 meters)(5 meters)] / (0.25 x 10 -3 meters) = 2 x 10-2 meters = 2 centimeters. 3. A photoelectric-effect experiment with cesium yields a stopping potential of 0.95 volts when the incident light has a wavelength of 435.8 nm. (a) What is the work function for cesium? [5 points] We know: eVstop = hf – W Which we can rewrite: W = hf – eVstop = hc/ – eVstop = 1240eVnm/435.8nm – e(0.95V) = 2.85eV – 0.95eV = 1.9 eV (b) What is the threshold wavelength for cesium? [5 points] The threshold frequency is the frequency of light needed to overcome the work function of the material We can rewrite: W = hfthreshold = hc/ threshold as: threshold = hc/ W = 1240eV nm / 1.9 eV = 653 nm Final Exam Fall 2005 Page 5 of 9 4. X-rays with a wavelength of 0.02000 nm are scattered through 90 degrees by an electron. The wavelength of the scattered x-rays is measured to be 0.02243 nm. What is the kinetic energy of the electron after the scattering (in eV)? [10 points] The total initial energy is equal to the total final energy E electron initial + E photon initial = E electron final + E photon final The electron is initially at rest so: E electron initial = 0 0 + hc/ initial = E electron final + hc/ final E electron final=hc/ initial - hc/ final=1240eVnm/0.02000nm – 1240eVnm/0.02243nm = 62000 eV – 55283 eV = 6717 eV = energy given to the electron from the photon 5. For a hydrogen atom, find the wavelength of the photon that is emitted when the atom jumps from the n = 4 state to the n = 1 state using the Ritz-Rydberg formula. [10 points] The Ritz-Rydberg formula is 1/ = R(1/m2 - 1/n2 ), where R = 1.097x 10-2 / nm (note: when you see it written nm-1, that just means divide by nm) 1/ = (1.097x 10-2 / nm) (1/m2 - 1/n2 ) = (1.097x 10-2 / nm)(1/ 12 - 1/ 42) = 1/ = (1.097x 10-2 / nm)(1/ 1 - 1/ 16) = (1.097x 10-2 / nm)(0.9375) = 0.001028 /nm So = 1 / (0.001028/nm) = 97.2 nm Final Exam Fall 2005 Page 6 of 9 6. The index of refraction for diamond is 2.42. The diamond is in air. (a) A ray of light falls on a diamond at an angle of incidence equal to 40 degrees. Find the angle of refraction. [5 points] nair sinair = ndiamond sindiamond sindiamond = nair sinair / ndiamond =(1)( sin 40o)/ (2.42) = 0.643/2.42 = 0.266 air = sin-1 (0.266) = 15 degrees (b) Find the critical angle for diamond. [5 points] At the critical angle for a light ray in diamond, the angle of the light ray in air is 90 degrees from the normal. That is, light is just moving parallel with the surface and can't get out. Any angle greater than or equal to the critical angle will transmit NO light out of the diamond into the air, it will all be internal reflection. sindiamond = nair sinair / ndiamond with air = 90o, sinair = 1 sindiamond-critical = nair / ndiamond = 1 / 2.42 = 0.413 diamond-critical = sin-1 (0.413) = 24 degrees 7. Suppose that light with a frequency of 1 x 1014 Hz is traveling in glass (n=1.5). (a) What would the wavelength be in air? [3 points] v = f/so = v/f. For air, v = c = c/f = (3 x 10 8 m/s) / (1 x 10 14 /s) = 3 x 10 -6 m (b) What is the speed of this light inside the glass? [3 points] The index of refraction (n) tells us the ratio of the speed of light in a vacuum to the speed of light in c c that material: n , so v v n 8 v =(3 x 10 m/s) / (1.5) = 2 x 10 8 m/s (c) What is the wavelength of this light inside the glass? [4 points] v c 3x108 m / s 2 x10 6 m 14 f nf (1.5)(1x10 / s) Final Exam Fall 2005 Page 7 of 9 Data and useful information: speed of light: c = 3 x 108 m/s Planck’s Constant: h = 6.626 x10-34 joule•seconds x x1 average velocity vave 2 t 2 t1 Snell's law: n1 sin1 = n2 sin2 2 Sine Wave: y A sin cf x Double Slit: d sin m y m Maxima [Also good for Diffraction Grating] L Produces constructive interference Produces destructive interference d Path Difference = λ Path Difference = / 2 F qE F qvB if v is perpendicular to B plus a direction given by the right hand rule! For Blackbody Radiation watts m2 K 4 -3 maxT constant 2.9 x10 meter Kelvin or 2.9 x106 nanometer Kelvin R T4 5.67 x 108 Definitions: work, energy, potential energy, kinetic energy Kinetic energy is proportional to v 2 Momentum is proportional to v Useful notes and data Final Exam Fall 2005 Page 8 of 9 e 1.6 x 10-19 C me 9.1 x 10-31 kg 1 electron - volt 1 eV 1.6 x 10-19 joules h 6.63 x 10-34 joule s or h 4.14 x 10-15 eV s c 3 x 108 m / s A convenient combihnation : hc 1240 electron - volt nanometers eVstopping hf - W or Vstopping h W f e e or R 1.097 x 102 (nm) 1 Energy levels of hydrogen - 13.6 eV n2 1 1 R 2 2 m n 1 photon energy hf hc Prefixes for powers of 10: 1018 exa 1018 atto 1015 peta 1015 femto 1012 tera 1012 pico 109 giga 109 nano 106 mega 106 micro 103 kilo 103 milli 10 2 1 10 hecto deka Final Exam Fall 2005 10 10 2 1 centi deci Page 9 of 9 Vstopping hc W (all in electron volts )
