Chapter 29 Solutions - 3, 5, 7, 10, 12, 13, 15, 19, 23, 27, 32, 36, 39
  1. In relativity, the kinetic energy is defined by the equation KE = (m – mo)c2. As I mentioned in class, this expression looks strange, but it reduces to the classical expression for kinetic energy, KE = (1/2)mv2, you learned about last semester.
               Proof: Recall that m = mo/(1 – v2/c2)1/2. For small velocities, the square root term can be expanded in a rapidly  converging series. The series is
                              1/(1 – v2/c2)1/2 = 1 + (1/2)v2/c2 + Terms of higher order that can be neglected.
                        As a consequence, the KE becomes
                             KE = (mo + (1/2)mov2/ c2 – mo)c2 = (1/2)mo v2, as we had hoped

    To solve the problem simply, it is convenient to talk of the rest energy of the electron, expressed in electron volts instead of joules. Going from joules to electron volts means dividing the number of joules by 1.6 x 10-19 J/eV. Also, for convenience, it is customary to call 1 x 106 eV as 1 MeV, where the unit MeV is short for million electron volts. Thus, moc2 = (9.110 x 10-31 kg)(2.998 x 108 m/s)2/(1.6022 x 10-19 J/eV) = 5.11 x 105 eV = 0.511 MeV.The answer to the problem is now an easy matter, because much of the arithmetic is already done. Since the mass of the electron has been increased to 10mo, the total energy is 10moc2, and the kinetic energy added must have been 9moc2. But 9mo c2 = (9)(0.511 MeV) = 4.60 MeV or 4.60 x 106 eV. The electron must have been accelerated through a potential difference of 4.6 x 106 V.
  1. For ease of calculation, you need to get used to working with eV and MeV energy units. Assume that a proton is accelerated through a potential difference of 109 V. The proton therefore acquires a kinetic energy of 1000 MeV. Let us first find the rest energy of the proton in MeV. It is
               moc2 = (1.67264 x 10-27 kg) (2.998 x 108 m/s)2/(1.6022 x 10-19 J/eV) = 938.3 MeV.
    Thus, since the total energy is the sum of rest energy + KE, we have
               mc2 = moc2 + 1000 MeV = 1938 MeV. Then, mc2/moc2 = m/mo = 1938/938 = 2.066.
    The speed of the proton is computed using m = mo/ (1 – v2/c2)1/2. We can substitute the ratio just obtained for m/mo, then square both sides to get
               (v/c)2 = 1 – (1/2.0662) = 0.7657 → v/c = .875
  1. Lifting a 3.0 kg mass 5.0 m gives it an increase of gravitational potential energy. This increase is
               PE = mgh = (3.0 kg)(9.8 m/s)(5.0 m). = 147 J, which is the increase in mc2.
    Therefore, the increase in mass is 147/(3.0 x 108 m/s)2 = 1.63 x 10-15 kg. This amount is obviously very small compared to the rest mass of 3.0 kg. The ratio is only (1.63 x 10-15 kg)/3.0 kg = 5.4 x 10-15.

    Study Aid – How to compute the photon energy in eV, given wavelength in nm.
    Recall that the photon energy is given by E = hf = hc/(wavelength). Calculating the energy in joules is easy enough, but often it is useful to be able quickly to compute the energy in eV directly by knowing the wavelength in nm. The conversion from joules to eV is done by dividing by 1.6 x 10-19 J/eV. The transformation from meters to nanometers is accomplished by expressing wavelength in nm and multiplying by the factor 10 -9.

                         .

  1. According to the expression E = hf = hc/(wavelength), the energy per photon for light having a wavelength of 632.8 nm is 1240/632.8 = 1.96 eV. The conversion to joules is accomplished by multiplying by 1.6022 x 10-19 J/eV, giving 3.14 x 10-19 J.

    Since the beam power is 3.00 mW, or 3.0 x 10-3 J/s, the number of photons = (3.00 x 10-3 J/s)/(3.14 J/photon) = 9.4 x 1015 photons/s.
  1. The threshold occurs when the energy of the photon is just enough to “lift” an electron out of the energy well. That is the meaning of the term work function. Thus, the photon energy must be 4.5 eV or greater. From the above study guide, we see that the wavelength corresponding to 4.5 eV is
                    wavelength (in nm) = 1240/4.5 = 276 nm
    This wavelength is in the UV part of the spectrum.


  2. This question uses the same logic as in question 12. E (in eV) = 1240/546 = 2.27 eV.
  1. When a photon is absorbed,all of its energy is deposited in the absorbing object. There is no partial absorption. Now, the energy of 480 nm light is
               E = 1240/546 = 2.583 eV.
    Also, since the photoelectric threshold is at 546 nm, we know that the work function is
               W = 1240/546 = 2.271 eV.
    The energy supplied by the 480 nm light is therefore greater than the work-function energy. The excess is available as kinetic energy for the ejected electron. This kinetic energy is 2.583 eV – 2.271 eV = 0.312 eV.

    In order to compute the speed of the electron, we note that the kinetic energy 0.312 eV is very small compared to the relativistic rest-mass energy of 0.511 MeV. This means that the old classical equation for kinetic energy is good enough and easy to carry out after changing the energy units from eV to joules. If you choose to use the relativistic expression, you will get the same correct result, but it will take considerably more labor. Thus, using 1.6 x 10-19 J/eV, we find
               (0.312 eV)( 1.6 x 10-19 J/eV) = (½)(9.11 x 10-31 kg)v2
               v    2 = 0.1096 x 1012 → v = 3.3 x 105 m/s
  1. Notice that the photon energy is given as well as the energy levels involved in the transition. For the H-atom, the energy levels are given by E(in eV) = -(13.6)/n2. Therefore, E2 = -3.40 eV, and E6 = -0.378. The difference is 3.02 eV. This answer is entirely consistent with the photon energy
               E(photon) = 1240/410 = 3.02 eV.
  1. The Lyman series for helium involves transitions from higher states to the ground state at E = –54.4 eV. From the energy-level diagram for He, the transition from n = 2 to n = 1 represents an energy difference of –40.8 eV. This means that a photon having an energy of 40.8 eV is emitted. The wavelength is
               Wavelength = 1240/40.8 = 30.4 nm, which is in the ultraviolet.

  1. Since the difference in energy between levels is 0.293 eV, the wavelength of the emitted photon when the molecule falls to the next lowest vibrational level is 1240/0.293 = 4232 nm. This radiation is in the infrared.

  1. The power output of the reactor is 1.0 x 108 J/s. As a consequence, in one hour (3600 s) the electrical energy produced is (1.0 x 108 J/s)(3600) = 3.6 x 1011 Joules. But, because the efficiency is only 30%, the energy created from mass conversion is more, namely, E(total) = (3.6 x 1011 J)/(0.30) = 1.2 x 1012 J. From the relation E = mc2, the mass that must have been converted into energy is
               m = (1.2 x 1012 J)/(3.0 x 108 m/s)2 = 1.3 x 10-5 kg.
    That’s pretty impressive, isn’t it?

  1. The de Broglie wavelength of a particle of mass m, moving at a speed v is h/mv, where h is Planck’s constant. But the wavelength is also given in general by wavelength = v/frequency. Thus,
                    .
    Substituting the numbers gives
                    v2 = (6.63 x 10-34 Js)(5.0 x 1012 Hz)/(9.11 x 10-31 kg) = 36.4 x 108 m2/s2,
                   or     v = 6.0 x 104 m/s.

  1. This is one of those problems in which the key idea is to keep track of the units as a means of knowing what to do. Since we want the wavelength of the incident photons, and since the data involves power and energy, the obvious goal is first to find how much energy each photon carries, then convert this number to wavelength.
              Step 1) Convert 5.0 W/cm2 to eV/s/cm2 using 1 J = 1.6 x 10-19 eV.
                         5.0 W/cm2 = 5.0 J/s/cm2 = 5.0/(1.6 x 10-19) = 3.13 x 1019 eV/s/cm2.
               Step 2) Convert 1020 photons/min on 3 cm2 to photons /second/cm2.
                         rate = (1020)(1/60)(1/3) = 5.6 x 1017 photons/s/cm2.
               Step3) Combine the results of steps 1) and 2).
                        E    (photon) = (3.23 x 1019 eV/s/cm2)/ (5.6 x 1017 photons/s/cm2) = 56 eV/photon
               Step 4) Convert the photon energy of 56 eV to wavelengtb.
                        Wavelength = 1240/56 = 22 nm