Chapter 31 Solutions - 2, 5, 7, 13, 15, 17, 20, 21, 24, 25, 27, 31, 33

  1. The isotope contains a total of 99 neutrons + protons in the nucleus. Of these 99 so-called nucleons, 43 are protons and 56 are neutrons.

  1. The atomic weight of Co-59 is 58.93 u. Since 1 u = 1.6605 x 10-27 kg,
    1. The mass of a Co-59 atom = (58.93 u)(1.6605 x 10-27 kg/u) = 9.785 x 10-26 kg.
    2. The mass of a C-12 atom = (12.00 u)(1.6605 x 10-27 kg/u) = 1.993 x 10-26 kg.
    3. The mass of a Co-59 nucleus = mass of the atom – mass of the 27 electrons
                      = (9.785 x 10-26 kg) – (27)(9.11 x 10-31 kg/e) = 9.773 x 10-26 kg.
  1. The radius is approximately R = (1.2 x 10-15)A 1/3 m, where A is the mass number (for Co-60, A = 60).
               R = (1.2 x 10-15)601/3 m = 4.7 x 10-15 meters.
    The mass density is the mass/volume. This mass is approximately
              m     = (60 u)(1.66 x 10-27 kg/u) = 1.0 x 10-25 kg.
              V = (4/3)(pi)R3 = (4/3)(3.14)(1.2 x 10-15)360 m
    Substituting to find the mass density gives
              m/V = 2.3 x 1017 kg/m3.
    I think you should contemplate the enormity of this number. Since there are 106 cm3 in 1 m3, a 1 cm3 volume of this nuclear matter has a mass of 233 billion kg!
  1. The answer is simply the average of the two, using the weighting percentages given.
    Atomic mass = [(0.8022)(11.0093) + (0.1978)(10.0129)] u = 10.8122 u.
  1. The mass defect is found by calculating the difference between the total mass of the separate nucleons and the mass of the combined nucleons to form the atom. The lithium atom has 3 protons, 3 electrons and 4 neutrons. Be aware that in calculations, we sometimes use the mass of the “bare” nucleus, whereas other times we use the mass of the atom as a whole including the electrons. Nuclear mass implies the former; atomic mass implies the latter. In this problem, we can use either because of the following argument,
               The mass defect = the mass of 3 bare protons plus 4 neutrons – the mass of a Li nucleus.
    But, we can add 3 electrons to 3 bare protons to get 3 H atoms, and we can add 3 electrons to a lithium nucleus to get a lithium atom. Since the masses are subtracted, the electron masses cancel, and we can write
               The mass defect = the mass of 3 H atoms plus 4 neutrons – the mass of a Li atom.
    Each expression leads to the same answer for the mass defect. We will use the second form, since that’s how the data are given.
               Mass defect for Li = 4(1.008665 u) + 3(1.007825 u) – 7.016005 u = 0.0421 u.
               Mass defect per nucleon = (0.0421 u)/7 = 0.006019 u.
  1. In order to separate all the nucleons from the atom, we must supply an energy equal to the mass defect. Sn-120 (tin) has 50 protons and 70 neutrons. Thus,
               The mass defect of Sn-120 = the mass of 50 H atoms plus 70 neutrons – the mass of a Sn-120 atom.
               Defect = (50)(1.007825 u) + (70)(1.008665 u) –119.9022 u = 1.0956 u
    This is equivalent to a binding energy of (1.0956 u)(931 MeV/u) = 1020 MeV. The binding energy per nucleon is (1020 MeV)/120 = 8.50 MeV/nucleon. This is consistent with Figure 31.7 in your book.

  1. To find out if a particular nuclear reaction is possible, one first can check to see if charge is conserved and if the reaction is energetically possible. A reaction is energetically possible only if the daughter products have less mass than the parent. A neutron is known to decay to a bare proton plus an electron. Since the charge on a neutron is zero, the net charge after the reaction must also be zero, and that requirement is satisfied. Next we check out the energetics. The mass of a neutron is 1.008665 u. The mass of a bare proton plus an electron (which is the same as the mass of an Hatom) is 1.007825 u. The mass reduction is 0.000840 u. Therefore, the reaction can go. In the process an amount of energy is released equal to the mass energy of 0.000840 u. This energy appears as kinetic energy of the electron and proton as well as the energy of a neutrino, as mentioned in class.

    Energy released = (0.000840 u)(981 MeV/u) = 0.782 MeV.

  2. An alpha particle is a He nucleus, a beta-minus particle is a negative electron, and a gamma ray is a photon. A reaction must satisfy conservation of total charge, as well as conservation of nucleon number. Consider the first reaction. The parent nucleus has 231 nucleons of which 91 are protons. Therefore, since an alpha particle has 4 nucleons of which 2 are protons, the other daughter isotope must have 227 nucleons, including 89 protons. A look at the periodic chart leads to an identification of the element. Notice that the third involves the emission of a gamma ray. Since a gamma ray has no charge or mass, the nucleus is unchanged. It has just jumped down from an excited state to a lower energy state (such as the ground state). The three reaction are as follows

              

  1. The mass of a C-12 atom is 12.0000 u by definition. The mass of a Be-8 atom is 8.0053 u. The mass of a He-4 atom is 4.0026. Notice that I am using atomic masses as per the discussion in problem 15. Since the mass of C-12 is less than the sum of the masses of Be-8 and He-4, the reaction is forbidden by the energetics.

  2. In one hour, a total of 10 half-lives will have transpired. Thus, the original sample will have been halved 10 times. Since (1/2)10 = 1/1024, that is the fraction of the original sample remaining after one hour.

  1. If the activity reduces to 1/2 in 27 hours, then the half-life is by definition 27 hours. In 2 half-lives the activity is reduced to 1/4. In 3 half-lives, the activity is 1/8. In 4 half-lives, the activity is 1/16, and in 5 half-lives, the activity is 1/32. The answers are thus 81 hours (3 half-lives) and 135 hours (5 half-lives).

  1. To solve this problem, make use of Equations (31.3) and (31.4) on page 630 of your book. The approach is to find the decay constant (lambda) from the half-life, then to obtain the activity of the sample by multiplying the decay constant by the number of atoms in the sample.

    Step 1 – half-life in seconds = (46.3 days)(24 hrs/day)(3600 s/hr) = 4.0 x 106 s.
    Step 2 – decay constant = (0.693)/(half-life) = (0.693)/(4.0 x 106 s) = 1.73 x 10-7 /s.
    Step 3 – Find the number of Fe-53 atoms in the sample of 3.0 x 10-7 kg knowing the mass of each atom is (53)(1.66 x 10-27 kg/u) = 88 x 10-27 kg.
                    # atoms = (3.0 x 10-7 kg)/(88 x 10-27 kg/atom) = 34 x 1017 atoms.
    Step 4 – The activity (rate of decay in the sample) is
                    Activity = (1.73 x 10-7 /s)(34 x 1017 atoms) = 5.9 x 1011 atoms/s.

  1. In order to get an accurate value, you need to use the exponential law. We can get a reasonable approximation by noticing that 1/30 represents a little less than 5 half=lives. Thus, since 5 half-lives represent (5)(5730 yrs) = 28,650 years, the answer must be just a little less than 28,650 years. Your book gives the answer as about 29,000 years. Actually, a better answer is about 28,000 years