100B SELF TEST 3

  1. A thin film of alcohol (n = 1.36) lies on a flat glass plate (n = 1.51). When monochromatic light whose wavelength can be changed, is incident normally, the reflected light is a minimum for l = 512nm and a maximum for l = 640nm. What is the thickness of the film?

    Solution.

      2. With respect to the incident wave, the wave that reflects from the top surface of the alcohol has a phase change of

                         f1 = p.        

      With respect to the incident wave, the wave that reflects from the glass at the bottom surface of the  alcohol has a phase change due to the additional path-length and a phase change of  p on reflection:

                         f2 = (2t/lfilm )2p + p.

              For constructive interference, the net phase change is

                        f = (2t/l1film )2p + p - p = m12p, m1 = 1, 2, 3, ..., or

                        t = 1/2l1film(m1) = 1/2(l1/nfilm)(m1), m1 = 1, 2, 3, ...

              For destructive interference, the net phase change is

                       f = (2t/l2film )2p + p - p = (2m2 + 1)p, m2 = 0,1, 2, ..., or

                       t = 1/4(l2/nfilm)(2m2 + 1), = 0, 1, 2, ...

             When we combine the two equations, we get

                      1/2(l1/nfilm)(m1) = 1/4(l2/nfilm)(2m1 + 1),

              or (2m2 + 1)/2m1 = l1/l2 = (640nm)/(512nm) = 1.25 = 5/4.

             Thus we see that m1 = m2 =2

             The thickness of the film is

                      t = 1/2l1film(m1) = 1/2[(640nm)/1.36)](2) = 471nm.
  2. A friend of yours travels by you in her fast sports vehicle at a speed of 0.580c. You measure it to be 5.80m long and 1.20m high. (a) What will be its length and height at rest? (b) How many seconds would you say elapsed on your friend's watch when 20.0s passed on yours? (c) How fast did you appear to be traveling to your friend? (d) How many seconds would she say elapsed on your watch when she saw 20.0s pass on her? (e) If your friend measures her vehicle's mass as 1.5 x 103kg what is its mass relative to you? (f) What is the vehicle's kinetic energy?

    Solution.

    (a) You measure  the contracted length. We find the rest length from

                     L = L0[1 - (v/c)2]1/2;

                     5.80m = L0[1 - (0.580)2]1/2, which gives  L0 = 7.12m.

    (b) We find the dilated time in the sports vehicle from

                    Dt = Dt0[1 - (v/c)2]1/2;

                    20.0s = Dt0[1 - (0.580)2]1/2, which gives Dt0 = 16.3s.

    (c) To your friend, you moved at the same relative speed: 0.580c.

    (d) She would measure the same time dilation: 16.3s.

    (e) We find the mass from

                   m = m0/[1 - (v/c)2]1/2 = (1.5 x 103kg)/[1 - (0.580)2]1/2 = 1.84 x 103kg.

    (f) We find the kinetic energy from

                  KE = mc2 - m0c2 = (m - m0)c2 = (1.84 x 103kg. - 1.5 x 103kg)\(3.00 x 108m/s)2 = 3.06 x 1019J.

  3. What is the maximum kinetic energy of electrons ejected from barium (W0 = 2.48eV) when illuminated by white light, l = 400 to 700nm?

    Solution.

    The photon of visible light with the maximum energy has the minimum wavelength:

                 hfmax = (1.24 x 103eV.nm)/l min = (1.24 x 103eV.nm)/(400nm) = 3.10eV.

    The maximum kinetic energy of the photoelectrons is

                KEmax = hf - W0 = 3.10eV - 2.48eV = 0.62eV.

  4. (a) Complete the following decays:

     (1)  22688Ra ®  22286Rn + (          )    ;

     (2)  146C ®  147N + (          ) + (           );

    (3) 18474W*  ®  18474W + (         ).

     (b)  74Be decays with a half-life of about 53d. It is produced in the upper atmosphere, and filters down onto the Earth's surface. If a leaf is detected to have 350 decays/s of 74Be, how long do we have to wait for the decay rate to drop to 10/s? Estimate the initial mass of 74Be on the leaf.

    Solution.

    (a) (1) (42He, or a); (2) (e-) +(n);  (3) (g).

    (b) The decay constant is 

               l = 0.693/T1/2= 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)/(350decays/s) = (1/2)n, or nlog 2 = log 35, which gives n = 5.13.

        Thus the elapsed time is

              Dt = nT1/2 =(5.13)(53days) = 272days » 9 months.

        We find the number of nuclei from

             Activity = lN;

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

        The mass is

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