Physics 6C Final Practice Solutions
Use the following information for problems 1 and. A beam of white light with frequency between 4.00 x 10 14 Hz and 7.90 x 10 14 Hz is incident on a sodium surface, which has a work function of.8 ev. 1) What is the range of frequencies in this beam of light for which electrons are ejected from the sodium surface? An easy way to do this one is to find the frequency of a photon that has an energy of.8 ev. E photon h f.8ev (4.14 10 f 5.5 10 14 Hz 15 ev s) (f) Now we just have to realize that any photon with a higher frequency will have more energy, and will also eject electrons. So the answer is (d).
Use the following information for problems 1 and. A beam of white light with frequency between 4.00 x 10 14 Hz and 7.90 x 10 14 Hz is incident on a sodium surface, which has a work function of.8 ev. ) What is the maximum kinetic energy of the photoelectrons that are ejected from this surface? We use the most energetic photon available the one with the highest frequency. Its energy will be higher than the work function, and the difference will be the leftover KE. KE KE KE max max max h f (4.14 10 0.99eV 15 ev s) (7.9 10 14 Hz).8eV Convert to Joules: KE max (0.99eV) (1.6 10 19 J ev ) 1.6 10 19 J Answer (b)
3) When light of a particular wavelength and intensity shines on the surface of a given metal, no electrons are ejected. Which of the following changes will most likely cause the emission of photoelectrons? a) Decreasing the frequency of the light b) Increasing the intensity of the light c) Decreasing the wavelength of the light d) Increasing the wavelength of the light In order for electrons to be ejected, the photons have to have more energy. To do this we need to either increase the frequency or decrease the wavelength. Changing the intensity will not change the energy of individual photons. Answer (c).
4) What is the De Broglie wavelength for a Helium atom (atomic mass=4.00603 u) traveling at a speed of 10 3 m/s? Use the conversion factor 1 u = 1.66054 x 10-7 kg. This one is just a formula: h m v 1.0 10 34 6.63 10 J s (4.00603u) (1.66054 10 10 m 7 kg u ) (10 3 m s ) Answer (b). You may have to round up carefully on this one.
5) The neutral pion (π 0 ) is an unstable particle produced in high-energy particle collisions. Its mass is 64 times that of the electron, and it exists for an average lifetime of 8.4x10-17 s before decaying into gamma-ray photons. Find the uncertainty in the mass of the pion, expressed as a fraction of its mass. Δm/m is approximately: We use the uncertainty principle for this one: E E E h t (8.4 10 17 1.56 10 s) 18 J 6.63 10 34 J s At this point use E=mc : 1.56 10 18 m 1.4 10 J 35 m c kg Divide by the mass of the pion to get the ratio we need: m m pion 35 1.4 10 kg 8 5.8 10 31 64 9.1 10 kg Answer (c).
6) You and a friend travel through space in identical spaceships. Your friend informs you that he has made some length measurements and that his ship is 150m long and yours is 10m long. What is the relative speed of the two ships? Use the length contraction formula: L L 10m (150m) 0.8 1 0.64 1 v c 0 0.36 1 v 0.6c Answer (c). v c v c v c 1 v c
7) A laser is used to move the electron in a hydrogen atom from the n= to the n=4 state. What is the wavelength of the light used to excite this electron? The energy of the absorbed photon is the difference in the energies of the excited states. E E n 4 13.6eV n 13.6eV E 4 13.6eV.55eV Now find the wavelength of a photon with this energy: h c Ephoton 15 (4.14 10 ev s) (3 10.55eV 7 4.87 10 m 490nm 8 m s ) Answer (b).
8) Assume the excited hydrogen atoms in problem 7 will emit photons until they fall back to the ground state. What is the minimum wavelength of the emitted photons? Since we want the minimum wavelength, we want the photon with the maximum energy. That will be the one that drops all the way to the ground state (i.e. from n=4 to n=1). 13.6eV E E4 13.6eV 4 1 1.75eV Now find the wavelength of a photon with this energy: h c Ephoton 15 (4.14 10 ev s) (3 10 1.75eV 8 9.7 10 m 97nm 8 m s ) Answer (b).
9) An object is 4m away from a wall. You are to use a concave mirror to project an image of the object on the wall, with the image.5 times the size of the object. What should the radius of curvature of the mirror be? a) 1.1m b).m c) 4.4m d) 5.5m 4m d i d o Since the image is real, it will be inverted, so we should have a magnification m=-.5. We can use the magnification formula to find the ratio of the object and image, then use the picture to find a value for d o. Plug in to find focal length. Radius is twice the focal length. m = d i d o =.5 d i =.5d o d i = 4m + d o =.5d o 4m = 1.5d o d o = 3.m 1 f = 1 + 1 = 1 + 1 = 1 d o d i d o.5d o 3.m + 1.5(3.m) 1 f = +0.45 f = +.m R = +4.4m answer c)
10) A 39 Pu (plutonium) nucleus typically decays by emitting an α-particle with an energy of about 5.1 MeV. What is the resulting daughter nucleus? Alpha-decay ejects a helium nucleus from the nucleus of the large atom, leaving it with fewer protons, and 4 fewer nucleons overall. So we need to use the periodic table and find the element that has fewer protons than plutonium. This is uranium. The number of nucleons should be 4 fewer, so we get 35 U. Answer (a).
11) Nucleus X has 10 times as many nucleons as nucleus Y. Compared to Y, X has a) a larger radius and about the same density b) a larger radius and a smaller density c) a smaller radius and about the same density d) a larger radius and a larger density We have a formula for nuclear radius: 15 3 R (1. 10 m) (A 1 ) So nucleus X should have a larger radius since it has a larger value of A. However, nuclear density is about the same for all nucleons (we did this in class). Answer (a).
1) Originally the decay rate of a sample was 16 decays per second. Now it has a rate of decays per second. The half-life is 5 years. How many years have passed? If you notice that the numbers come out perfectly, then you can do this one without much calculating. The decay rate has dropped from 16 to. This is a factor of 8. Since 3 =8, we know the time should be 3 half-lives, or 15 years. Answer (d).
13) An ancient bone has been recovered from an archeological dig. It is determined that the ratio of the isotope carbon-14 to total carbon is only 0% of the ratio currently present in the atmosphere. The half-life of carbon-14 is 5730 years. How old is the bone? We will need to use our exponential decay formula for this one. The decay constant comes from the half-life: N(t) N 0 e t 0. ln() T1 (0%) N e ln(0.) t t 13,300yr 1.1 10 0 N e 0 t 4 (1.1 10 1 yr 4 1 yr ) t Answer (c).
14) 16 7N is an unstable isotope of nitrogen with a half-life of about 7 seconds. When it undergoes - decay an electron, an antineutrino, and one other particle are emitted. What is the other particle? The other particle will be the daughter nucleus, which should have one more proton than the original nitrogen atom. From looking at the periodic table, we find that it must be oxygen. Answer (c).
15) A light ray in water is incident at an angle of.8 on an oil film with thickness 560 nm and index of refraction 1.45. Does the transmitted light ray bend toward or away from the normal? Assume that the index of refraction of water is 1.33. a) Toward b) Away c) The ray does not bend at all. d) The ray is completely reflected. The ray is entering a higher index material, so it should bend toward the normal. Snell s Law will calculate the angle if we need it. n 1 sin θ 1 = n sin θ 1.33 sin.8 = 1.45 sin θ θ = 0.8 Oil n =1.45 Water n 1 =1.33
16) An oil film floats on water. The refractive index of the oil is 1.4 and the index for water is 1.33. The film appears blue because it reflects light with wavelength 400 nm very well. Calculate the minimum thickness of the oil film. Ray 1 reflects from a higher index, but ray reflects from a lower index, so there is a relative phase shift. Our formulas are: m 0 t destructive n (m 1 ) 0 t constructive n Air n=1 Oil n=1.40 Water n=1.33 1 We use the constructive formula, with m=0 to get the thinnest possible film: 400nm 1 (0 ) 1.4 t 71nm min t Answer (a).
17) A single-slit experiment is performed with light of wavelength of 500 nm, and the first dark fringe appears at a distance of 4cm from the central bright maximum. If this experiment is now performed with the same setup, but underwater, how far from center will the first dark fringe be? Assume the index of refraction of water is 1.33. This is a single-slit experiment, so we can use the formula for the position of the dark fringes. Use m=1 to get the first dark fringe from center. d sin( ) m This will get us the angle. Now use a right triangle to get distance on the screen. y R tan() We can combine these formulas if we realize that for small angles, tan(θ)~sin(θ). y mr d Before we plug in the numbers, realize that the only difference between the experiments will be the wavelength of the light. Underwater, the wavelength will be shortened. So we can set up a ratio to solve for the distance. y y y water air water y n water air air 4cm 1.33 3cm Answer (b).
18) A double-concave lens is made from glass with index of refraction n=1.6. The radius of curvature is 90cm for each side. Find the focal length of this lens when used in air. a) 50cm b) -90cm c) -75cm d) -50cm Assuming the light comes from the left, as shown in the diagram, we will get a negative value for R 1 and a positive for R. The convention is that the incoming light side is negative, and the outgoing side is positive. Just plug into the formula: 1 f = n 1 1 1 = 1.6 1 R 1 R 1 90cm 1 90cm = (0.6) 1 45cm f = 75cm
19) The sun produces energy by nuclear fusion reactions, in which matter is converted to energy. The rate of energy production is 3.8 x 10 6 Watts. How many kilograms of mass does the sun convert to energy each second? Recall that a Watt is a Joule/Second. Convert energy to mass using Einstein s formula: E m c 3.8 10 m 6 4. 10 J (m) (3 10 9 kg 8 m s ) Answer (a).
0) Calculate the mass defect for the decay of 8 4Be into two alpha particles. The mass of 8 4Be is 8.005305 u, and the mass of an alpha particle is 4.00148 u. Use the conversion factor 1 u = 931.5 MeV We need to find the difference between the mass of the original Be nucleus (don t forget to account for the electrons!), and the mass of the alpha particles that result from the decay: m m Beryllium nucleus 8.005305u 4(0.000549u) (4.00148u) (0.000149u) (931.5 MeV u ) 0.14MeV 0.000149u Answer (c).
1) Polarizer 1 has a vertical transmission axis. Polarizer has a transmission axis tilted 45 from the vertical, and polarizer 3 has a horizontal transmission axis. The light incident on polarizer 1 is polarized vertically. If polarizer is removed, what will be the effect on the intensity of light at point P? a) the intensity will be zero (no light will emerge from polarizer 3) b) the intensity will increase by a factor of c) the intensity will decrease, but will be greater than zero d) the intensity will be unchanged When polarizer # is removed, it leaves #1 and #3 next to each other. Their transmission axes are perpendicular to each other, so no light gets through. Answer (a).
) At a point in space an electromagnetic wave is propagating in the negative y direction. At a certain instant, the magnetic field of the wave at this point is in the positive x direction. At this point and this instant, the electric field of the wave is a) in the positive z direction b) in the positive x direction c) in the negative y direction d) in the negative z direction The right-hand-rule for a plane EM wave is the following: E, B and v are mutually perpendicular. Start with your fingers along E, bend them toward B, and your thumb is along the velocity (propagation direction). In our problem we know that the E-field must be in the z-direction (since B and v are already in x or y). Try it with E-field in the +z direction you get velocity in the +y direction. It must be the other way since velocity comes out in the y direction. So the answer is E-field in z direction
3) An astronaut traveling with a speed v relative to Earth takes her pulse and measures 70 beats per minute. Mission control on Earth, which monitors her heart, observes a rate of 43 beats per minute. What is the astronaut s speed relative to Earth? a) 0.5c b) 0.7c c) 0.8c d) 0.9c We will use the time dilation formula for this one. It may help to convert the given measurements into time intervals first, although this is not necessary. 1 min 70 beats 60 sec 1 min = 0.86 sec beat 1 min 43 beats 60 sec 1 min = 1.40 sec beat t = t 0 1 v c 1.40sec = 0.86sec 1 v c 1 v c = 0.86sec 1.40sec 1 v c = 0.61 v = 0.8c