1. What is the shortest wavelength x-ray produced by a 0.20-MV x-ray machine?
    Solution.
    The shortest wavelength x-ray corresponds to the highest frequency, which occurs when the energy of the photon is equal to the electron's kinetic enertgy.
    Ephoton = Kelectron
             hf = eV
           hc/λ = eV
               λ = hc/eV = (1240 eV·nm)/(e x 0.20 x 106 V) = 6.2 pm.
  2. In a photoelectric experiment using sodium, when incident light of wavelength 570 nm and intensity 1.0 W/m2 is used, the measured stopping potential is 0.28 V. (a) What would the stopping potential be for incident light of  wavelength 400.0 nm and intensity 1.0 W/m2? (b) What would the stopping potential be for incident light of wavelength 570 nm and intensity 2.0 W/m2? (c) What is the work function of sodium?
    Solution.
    (a) The work function for sodium is found from Einstein's photoelectric equation.
         Kmax = hf - Ф
             Ф = hf - Kmax
                 = hc/λ - Kmax
                 = (1240 eV·nm)/(570 nm) - 0.28 eV = 1.9 eV.
         The stopping potential for λ = 400.0 nm is
            eVs = hf - Ф
              Vs = (hc/e)(1/λ) - Ф/e
                   = (1240 eV·nm)/(e·400.0 nm) - 1.9 V = 1.2 V.
    (b) The stopping potential does not depend upon the intensity of the light, so it is still 0.28 V.
    (c)  Ф = 1.9 eV as found in part (a).
  3.  The Pashen series in the hydrogen emission spectrum is formed by electron transitions from ni > 3 to ni = 3. (a) What is the longest wavelength in the Pashen series? (b) What is the wavelength of the series limit ( the lower bound of the wavelength in the series)? (c) In what part or parts of the EM spectrum is the Pashen series found (IR, visible, UV, etc.)?
    Solution.
    (a) A photon with the longest wavelength   will have the least energy. The transition from the n = 4 state to the n = 3 state will release the least energy.
         E = E4 - E3 = E1/42  - E1/32 = (-13.6 eV/16) - (13.6 eV/9) = 0.6611eV
         Calculate the wavelength.
         E = hc/λ, 
         so λ = hc/E = (1240 eV·nm)/(0.611eV) = 1876 nm.
    (b) The energy released by a transition from the n = ¥ state to n = 3 state is  
         E = 0 - (-13.6 eV/32) = 1.511 eV.
         the wavelength of a photon with this energy is 
         λ = hc/E = (1240 eV·nm)/(1.511 eV) = 820.6 nm.
    (c) The range of wavelengths from 820.6 nm go 1876 nm is in the IR (infrared) part of the EM spectrum.