Solutions to Week 14 Assignment

1. (a) The radius of ⁶⁴Cu is

                   r = (1.2 x 10^-15m)A^1/3 = (1.2 x 10^-15m)(64)^1/3 = 4.8 x 10^-15m =  4.8fm.

   (b) We find the value of A from

                   r = (1.2 x 10^-15m)A^1/3;

                  3.7 x 10^-15m = (1.2 x 10^-15m)A^1/3, which gives A = 29.

2. ⁶Li consists of three protons and three neutrons. We find the binding energy from the masses:

                 Binding energy =[3m(¹H) + 3m(¹n) - m(⁶Li)]c²

                                       = [3(1.007825u) + 3(1.008665u) - (6.015121u)]c²(931.5MeV/uc²

                                                   = 32.0MeV.

   Thus  the binding energy per nucleon is

                 32.0MeV/6 = 5.33MeV.

3. For alpha decay we have ²¹⁸₈₄Po® ²¹⁴₈₂Pb + ⁴₂He. The Q value is

                  Q = [m(²¹⁸Po) - m(²¹⁴Pb) - m(⁴He)]c²

                      = [(218.008965u) - (213.999798u) - (4.002602u)]c²(931.5MeV/uc²)

                      = 6.12MeV

    For beta decay we have ²¹⁸₈₄Po® ²¹⁸₈₅At + ⁰_-1e. The Q value is

                   Q = [m(²¹⁸Po) - m(²¹⁸At)]c²

                       = [(218.008965u - (218.00868u)]c²(931.5MeV/uc²)

                       = 0.27MeV.

4. The decay constant is

                 l = 0.639/T_1/2 = 0.639/(30.8s) = 0.0225s^-1.

    (a) The initial number of nuclei is

                N₀ = [(7.8 x 10^-6g)/(124g/mol)](6.02 x 10²³atoms/mol) = 3.8 x 10¹⁶nuclei.

   (b) When t = 2.0min, the exponent is

                l _t = (0.0225s^-1)(2.0min)(60s/min) 2.7,

         so we get

               N = N₀e^-lt = (3.8 x 10¹⁶)e^-2.7 = 2.5 x 10¹⁵nuclei.

   (c) The activity is

              lN = (0.0225s^-1)(2.5 x 10¹⁵) = 5.7 x 10¹³decays/s.

   (d) We find the time from

              lN = lN₀e^-lt;

              1 decay/s = (0.0225s^-1)(3.8 x 10¹⁶)e^-(0.0225/s), which gives t = 1.53 x 10³s = 2.5min.

5. The decay constant is

              l = 0.639/T_1/2 = 0.693/(53days) = 0.0131/day = 1.52 x 10^-7s^-1.

   We find the number of half-lives from

             (DN/Dt) /(DN/Dt)₀ = (1/2)ⁿ;

             (10decays/s)(350days/s) = (1/2)ⁿ, or nlog2 = log35, which gives n = 5.13.

   Thus the elapsed time is

             Dt = nT_1/2 = (5.13)(53days) = 272days = 9months.

   We find the number of nuclei from

             Activity = lN;

             350decays/s = (1.52 x 10^-7s^-1)N, which gives N =2.31 x 10⁹nuclei.

   The mass is

             m = [(2.31 x 10⁹nuclei)/(6.02 x 10²³atoms/mol)](7g/mol) = 2.7 x 10^-17kg.