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[25 pts] (27.2) A flashbulb goes off at the origins as the origins of our usual reference frames A and B coincide. Frames A and B have a relative speed of 0.6 c alon[g their x-axes. A wavefront travels along the x-axis as observed in B. What time is recorded in B as this wavefront reaches xB = 1,000 m? Use these values of xB and tB to calculate the coordinates xA and tA that describe this event in A. Then use these values to calculate the speed of light in reference frame A ( vlight = xA/tA).  To find time tB we just use the definition of velocity v = c = xB/tB tB = xB/c   tB = 3.333 x 106 s Of course, we also have xB = 1000 m  Now we can apply the Lorentz Transformations to find xA and tA,      xA = 2 000 m      tA = 6.66 x 106 s Now we can find the velocity in the A-frame and determine if the Lorentz Transformations work. Is it, indeed, true that the velocity of light will be the same in both frames? Lets see, . . .   vA = 3.00 x 108 m/s And, indeed, this is the: speed of light! ========================================================== 2. [15 pts] (27.49) A spaceship travels away from Earth at a velocity of 2x108 m/s. A projectile is shot forward from the spaceship at 2.5 x 108 m/s relative to the spaceship. What is the velocity of the projectile relative to Earth?  This provides a more direct application of our velocity transformation equation,  with v = 2 x 108 m/s = (2/3) c vxMB = 2.5 x 108 m/s = (2.5/3) c       vxMA = 0.964 c ========================================================== 3. [10 pts] (28.15) What is the energy of the photons in the visible region of the electromagnetic spectrum? (Use l = 550 nm as an average value)? E = h f f l = c f = c / l E = h c / l E = (4.14 x 1015 eV s) (3.0 x 108 m/s) / 550 nm E = (4.14 x 10>15 eV s) (3.0 x 108 m/s) / 550 x 109 m E = 2.26 eV ========================================================== 4. [15 pts] (28.41) What is the wavelength of an electron moving at 1.5x108m/s (v = 0.5 c)? l = h/p = h / mv l = (6.626 x 1034 J s) / [(9.11 x 1031 kg)(1.5x108 m/s)] l = 4.85 x 10 10 m This wavelength is larger than a nuclear diameter and is a reasonable fraction of typical atomic spacing. ========================================================== 5. [15 pts] (29.1) Calculate the first wavelength of the Balmer series for hydrogen and tell what color it is. Balmer Series for Hydrogen  R = 1.097 x 107 m 1, Rydberg constant For n = 3 we have  = (1.097 x 107 m1) (0.1389) = 1.525 x 106 m1 ln=3 = 6.563 x 107 m ln=3 = 656.3 x 109 m ln=3 = 656.3 nm, and this is red light ========================================================== 6. [20 pts] (29.16) In state A the energy of an atom is 3.45 eV and in state B its energy is 5.67 eV. a) What is the energy of the photon emitted in the transition from A to B? b) What is the wavelength of this photon? c) What color is the light of this photon? Ephoton = DE = EA EB Ephoton = ( 3.45 eV) ( 5.67 eV) Ephoton = 2.22 eV E = hf = h(c/l) l = hc/E l = (6.626 x 1034 Js)(3.0 x 108 m/s) / 2.22 eV We need to convert E = 2.22 eV into units of J (or the equivalent) to finish carrying out this calculation  E = 3.55 x 1019 J l = (6.626 x 1034 Js)(3.0 x 108 m/s) / 3.55 x 1t019 J l = 5.60 x 107 m l = 560 x 109 m = 560 nm This is yellow light, near the middle of the range of visible light. ========================================================== Extra. (29.6) Calculate the energy, frequency, and wavelength of the photon that is emitted when a hydrogen atom undergoes a transition from its n = 2 state to its n = 1 state. 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