Physics 106 – Spring 2011 Recitation Exercises # 12, April. 20 & 27, 2010 1. How far away can a human eye distinguish two car headlights 2.0 m apart? Consider only diffraction effects and assume an eye diameter of 5.0 mm and a wavelength of 500 nm. Use the equation shown above in the figure, solve for L = as / ! = (5 " 10 #3 m)(2m) /(500 " 10 #9 m) = 2 " 10 4 m = 20,000m 2. Sodium undergoes the photoelectric effect; electron #1 is emitted when a photon of violet light falls on some sodium while electron #2 is emitted when an ultraviolet photon falls on sodium. What is different about the two electrons afterward? The electron from the UV will have more energy than the one from violet light because the incident UV light carries more energy (it has a smaller wavelength). 3. Gamma rays are high-energy (and therefore high frequency) photons. In certain nuclear reactions, gamma ray photons with energy of 0.511MeV (million electron volts) are produced. What is the frequency in Hz of these photons? By Planck’s relation, the photon energy is related to the frequency by E=hf so f=E/h. Putting in the numbers, f = E (0.511"106 eV) = = 1.24 "1020 Hz . !15 eV h (4.136 "10 Hz ) 7 4. Electrons striking the back of a TV screen travel at a speed of about 8x10 m/s. What is their de Broglie wavelength? != h 6.63 #10"34 Js = = 9.1#10"12 m "31 7 mv (9.11#10 kg )(8 #10 m / s ) 5. Determine the composition of the nuclei for the following isotopes. a) carbon-14 b) silver-108 c) plutonium-242 (a) Carbon-12: Z=6, A=12, so there are 6 protons (Z) and 8 (A-Z) neutrons. (b) Silver-108: Z=47, A=108, so there are 47 protons (Z) and 61 (A-Z) neutrons. (c) Plutonium-242: Z=94, A=242, so there are 94 protons (Z) and 148 (A-Z) neutrons. 6. A hydrogen atom has its electron in the n=2 state. a) How much energy would have to be absorbed by the atom for it to become ionized from this level? b) What is the frequency of the photon that could produce this result? c) What frequency of photon would be emitted if the electron in the n=2 state transitions to the ground state. d) If a large number of hydrogen atoms are in an n=4 state, what are the six possible energies of all the photons you would expect to be emitted by these atoms (note, the n=4 state could emit light in a multi step process, e.g. it could first transition to the n=2 state and then the n=1 state). a) From Figure 10.32, we get that the energy of the n=2 state of hydrogen is -3.4 eV. This is just the amount of energy needed to lift it to the unbound, or ionized, state. Thus E=3.4 eV. E 3.4eV = = 8.2 "1014 Hz . !15 h 4.136 "10 eV / Hz c) The energy of the photon is (-3.4eV)-(-13.6eV)=10.2eV. The frequency of this photon is therefore b) f = f = E 10.2eV = = 2.47 "1015 Hz !15 h 4.136 "10 eV / Hz d) The six possible transitions that could be produced starting with the n=4 state are: n = 4 ! n = 3 : E43 = ("0.9 eV) " ("1.5 eV) = 0.6 eV n = 4 ! n = 2 : E42 = ("0.9 eV) " ("3.4 eV) = 2.5 eV n = 4 ! n = 1: E41 = ("0.9 eV) " ("13.6 eV) = 12.7 eV n = 3 ! n = 2 : E32 = ("1.5 eV) " ("3.4 eV) = 1.9 eV n = 3 ! n = 1: E31 = ("1.5 eV) " ("13.6 eV) = 12.1 eV n = 2 ! n = 1: E21 = ("3.4 eV) " ("13.6 eV) = 10.2 eV 7. In each of the following nuclear reaction, determine the nucleus represented by X; use the standard notation to indicate the mass number, atomic number and chemical element: 91 1 (a) X +10 n #142 56 Ba + 36 Kr + 3 0 n He + 32 He # X + 2 11 H (b) 3 2 (c) 40 20 Ca + 42 He # X (d) 15 7 N +11 H # X + 42 He (e) X #14 7 N + e + 00 ! 0 " "1 In each case, the total mass number and the total atomic number must be the same on the left and right side of the reaction equations. Once you know the atomic number, you can look in the periodic table to find the correct symbol for the chemical element. 235 91 1 (a) 92 U +10 n #142 56 Ba + 36 Kr + 3 0 n He + 32 He #42 He + 2 11 H (b) 3 2 (c) 40 20 Ca + 42 He #44 22 Ti (d) 15 7 4 N +11 H #12 6 C + 2 He (e) 14 6 C #14 7 N + e + 00 ! 0 " "1 8. A piece of charcoal found in the fire pit of an archeological site is found to have about 1/8 of the carbon-14 content of a modern piece of wood. Estimate the age of that ancient fire pit? Assuming that when the wood for the charcoal was alive, it had the same carbon-14 content as modern wood, radioactive decay has reduced it by a factor of 8. The charcoal has therefore been sitting in the ground for three times the half-life of carbon-14. The half-life of carbon-14 is about 5700 years so the age of the sample is about 17,000 years. 9. The oxygen isotope 19 8 O has a half-life of 26s. If I start with 1024 atoms of this isotope, how long will it take, on average, before only 128 atoms are left? The fraction of atoms left is 128/1024=1/8=1/23. This means that the time is three times the half-life therefore the time is 3×26s=78s.