Physics Solution Manual for 1100 and 2101

The total energy in ev of a hydrogen atom in the n 3

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Unformatted text preview: pression for f into Equation 29.2 gives E = hf = h c λ Applying this result to both sources, we have EB = h c λB EA = h and c λA Dividing the two expressions gives c h EB λB λA = = EA h c λB λA Using the given value for EA and the fact that λB = 3λA in this result shows that EB = EA λ λA = 2.1×10−18 J A = 7.0 ×10−19 J 3λ λB A ( ) 45. REASONING The de Broglie wavelength λ is related to Planck’s constant h and the magnitude p of the particle’s momentum. The magnitude of the momentum is related to the mass m and the speed v at which the bacterium is moving. Since the mass and the speed are given, we can calculate the wavelength directly. SOLUTION The de Broglie wavelength is λ= h p (29.8) The magnitude of the momentum is p = mv (Equation 7.2), which we can substitute into Equation 29.8 to show that the de Broglie wavelength of the bacterium is λ= h h 6.63 ×10−34 J ⋅ s = = = 1×10−18 m −15 p mv 2 × 10 kg ( 0.33 m/s ) ( ) 1540 PARTICLES AND WAVES 46. REASONING AND SOLUTION a. We know E = hc/λ for a photon. The energy of the photon is F × 10 J I = 8.0 × 10 1.60 E = 5.0 eV G H 1 eV J K −19 −19 J The wavelength is c h c 6.63 × 10 −34 J ⋅ s 3.00 × 10 8 m / s hc λ= = E 8.0 × 10 −19 J h= 2 .5 × 10 −7 m b. The speed of the 5.0-eV electron is v= c 2 8.0 × 10 −19 J 2E = m 9.11 × 10 −31 h= 1.3 × 10 kg 6 m/s The de Broglie wavelength is h λ= = mv 6.63 × 10 −34 J ⋅ s 9.11 c × 10 −31 c h kg 1.3 × 10 m / s 6 = 5.6 × 10 −10 m h 47. REASONING AND SOLUTION In the first case, the energy of the incident photon is given by Equation 29.3 as hf = KE max + W0 = 0.68 eV + 2.75 eV = 3.43 eV In the second case, a rearrangement of Equation 29.3 yields KE max = hf – W0 = 3.43 eV – 2.17 eV = 1.26 eV 48. REASONING The speed v of a particle is related to the magnitude p of its momentum by v = p/m (Equation 7.2). The magnitude of the momentum is related to the particle’s de Broglie wavelength λ by p = h/λ (Equation 29.8), where h is Planck’s constant. Thus, the speed of a particle can be expressed as v = h/(mλ). We will use this relation to find the speed of the proton. SOLUTION The speeds of the proton and electron are vproton = h mproton λproton and velectron = h melectron λelectron Chapter 29 Problems 1541 Dividing the first equation by the second equation, and noting that λelectron= λproton, we obtain vproton m m λ = electron electron = electron velectron mproton λproton mproton Using values for melectron and mproton taken from the inside of the front cover, we find that the speed of the proton is m vproton = velectron electron m proton 9.11×10−31 kg ( 6 3 = 2.45 ×10 m/s = 4.50 × 10 m/s ) −27 kg 1.67 ×10 49. REASONING The width of the central bright fringe in the diffraction patterns will be identical when the electrons have the same de Broglie wavelength as the wavelength of the photons in the red light. The de Broglie wavelength of one electron in the beam is given by Equation 29.8, λ electron = h / p , where p = mv . SOLUTION Following the reasoning described above, we find λ red light = λ electron λ red light = h melectron v electron Solving for the speed of the electron, we have v electron = h melectron λ red light = 6.63 × 10 –34 J ⋅ s = 1.10 × 10 3 m / s (9.11 × 10 –31 kg)(661 × 10 –9 m) 50. REASONING We will first calculate the potential energy of the system at each of the two separations, and then find the energy difference for the two configurations. Since the electric potential energy lost by the system is carried off by a photon that is emitted during the process, the energy difference must be equal to the energy of the photon. The wavelength of the photon can then by found using Equation 29.2 with Equation 16.1: E = hc / λ . 1542 PARTICLES AND WAVES SOLUTION The initial potential energy of the system is (see Equations 19.3 and 19.6) FI kq GJ r HK L × 10 (8.99 = ( 1.6 × 10 C) M N EPE 1 = eV1 = e 1 −19 9 O P Q N ⋅ m 2 / C 2 )( 8.30 × 10 –6 C) = 2 .84 × 10 –14 J 0.420 m The final potential energy is EPE 2 = eV2 = ( 1.6 × 10 −19 C) L × 10 (8.99 M N O P Q N ⋅ m 2 / C 2 )( 8.30 × 10 –6 C) = 7 .56 × 10 –15 J 1.58 m 9 The energy difference, and therefore the energy of the emitted photon, is ∆E = EPE 1 − EPE 2 = 2 .84 × 10 –14 J – 7 .56 × 10 –15 J = 2 .08 × 10 –14 J The wavelength of this photon is λ= hc (6.63 × 10 –34 J ⋅ s)(3.00 × 10 8 m / s) = = 9.56 × 10 –12 m –14 ∆E 2.08 × 10 J 51. REASONING Since the net external force acting on the system (the photon and the electron) is zero, the conservation of linear momentum applies. In addition, there are no nonconservative forces, so the conservation of total energy applies as well. Since the photon scatters at an angle of θ = 180.0° in Figure 29.10, the collision is "head-on." Thus, the motion takes place entirely along the horizontal direction, which we will take as the x axis, with the right as being the posit...
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