solutions to Week 12 Assignment

1. The force from the electric field must balance the weight:

                 v = neV/d = mg;

                n(1.60 x 10^-19C)(340V)/(1.0 x 10^-2m) = (2.8 x 10^-15kg)(9.80m/s²), which gives n = 5.

2. (a) The potential energy on the first step is

               PE₁ = mgh = (58.0kg)(9.80m/s²)(0.200m) = 114J.

    (b) The potential energy on the second step is

               PE₂ = mg2h = 2PE₁ = 228J.

    (c) The potential energy on the third step is

               PE₃ = mg3h = 3PE₁ = 342J.

    (d) The potential energy on the nth step is

               PE_n = mgnh = nPE₁ = 114nJ.

    (e) The change in energy is

              DE = PE₂ - PE₆ = (2 - 6)(114J) = -456J.

3. (a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have

              KE = hf - W₀ = 0;

             W₀ = hc/l_max = (1.24 x 10³eV·nm)/(570nm) = 2.18eV.

   (b) The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy:

             KE_max = eV₀ = hf - W₀;

            (1e)V₀ = [(1.24 x 10³eV·nm)/(400nm)] - 2.18eV = 3.10eV - 2.18eV = 0.92eV,

       so the stopping voltage is 0.92V.

4. The photon with the longest wavelength has the minimum energy to create the masses:

            hf_min = hc/l_max =2m₀c²;

            (6.63 x 10^-34J·s)/l_max = 2(1.67 x 10^-27kg)(3.00 x 10⁸m/s)², which gives l_max = 6.62 x 10^-16m.

5. Because all the energies are much less than m₀c², we can use KE µ p²/2m₀, so

           l = h/p = h/[2m₀(KE)]^1/2 = hc/[2m₀c²(KE)]^1/2

             = (1.24 x 10³eV·nm)/[2(0.511 x 10⁶eV)(10eV)]^1/2 = 0.39nm.