Solutions to Week 13 Assignment

1. We find the wavelength of the bullet from

                    l = h/p = h/mv

                       = (6.63 x 10^-34J·s)/(3.0 x 10^-3kg)(200m/s) = 1.1 x 10^-33m.

    The half-angle for the central circle of the diffraction patter is given by

                   sinq  = 1.22l/D. 

    For small angle, sinq  = tanq, so we have

                  r = Ltanq = Lsinq  = 1.22l/D;

                 0.50 x 10²m =  1.22L(1.1 x 10^-33m)/(3.0 x 10^-3m), which gives L = 1.1 x 10²⁸m.

    Diffraction effects are negligible for macroscopic objects.

2. We find the uncertainty in the energy of the free neutron from

                DE = h/Dt = (1.055 x 10^-34J·s)/(2.2 x 10^-6s) = 4.8 x 10^-29J = 3.0 x 10^-10eV.

    Thus the uncertainty in the mass is

               Dm = DE/c²  = 3.0 x 10^-10eV/c²

^3.. (a) We start with hydrogen and fill the level as indicated in the periodic table:

               1s²2s²2p⁶3s²3p⁶3d⁷4s².

          Note that the 4s² level is filled before the 3d level is started.

  (b) For Z = 36 we have

              1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶.

4. The third electron in lithium is in the 2s subshell, which is outside the more tightly bound filled 1s shell.

    This makes it appears as if there is a "nucleus" with a net charge +1e. Thus we use the energy of the hydrogen atom:

              E₂ = -(13.6eV)/n² = - (13.6eV)/2² = -3.4eV,

    so the binding energy is 3.4eV.

    Our assumption of complete shielding of the nucleus by the 2s electron is probably not correct.

5. With shielding provided by the remaining n =1 electron, we use the energies of the hydrogen atom with Z replaced by Z -1. The energy of the photon is

             hf = DE = -(13.6eV)(24 -1)²[(1/2²) - (1/1²)] = 5.40 x 10³eV.

    The wavelength of the photon is 

            l = (1.24 x 10³eV·nm)/DE = (1.24 x 10³eV·nm)/(5.40 x 10³eV) = 0.23nm.