Bragg Diffraction of X-Rays

          By Kyla Marie Frohlich

Objectives:
    The objectives of this laboratory were to detect and understand Bragg diffraction, to measure X-ray wavelength with NaCl crystal and to predict diffraction for other crystals.

Introduction:
  X-rays were discovered about 100 years ago by Wilhelm Konrad Rontgen.  X-rays are produced when electrons strike a metal anode at high energy.  The target anode must be at a forty-five degree angle to the electron beam. The X-rays are then emitted at a ninety degree angle to the electron beam.  We studied the wavelength spectrum or Bragg diffraction in this lab.  When X-rays penetrate crystals they are scattered by each layer at certain angles.  By using a Geiger-Mueller detector tube we are able to measure the X-rays at different angles.  These angles can them be used to calculate the X-ray wavelengths.

Procedure:

Data and Results:

    The detector tube was electronically moved in discrete intervals and connected to a counter circuit that would produce a graph on our computer.  We were looking solely at alpha and beta radiation.  This graph displayed the number of counts vs. twice the angle.  From this graph the wavelength of the X-rays were able to be calculated at each angle.  A picture of the halite graph and a close up of how the X-rays are produced are shown below.
 
 


 
 

    The small peaks around the first, second and third order peaks are background noise and were ignored for this lab.  These peaks were ignored because they were Bremstrahlung X-rays and not alpha or beta X-rays.  I predicted that the peaks for the LiF crystal would be larger because the crystal is more compact.  My prediction was correct.  Although the peak sizes changed for all three of these crystals, the wavelengths remained about the same!  The formula used to calculate the wavelengths was 2d*sin(theta/2).  A table of the wavelengths and the calculations are shown below.


NaC Crystal       First order peaks       Second order peaks     Third order peaks
Beta x-rays             28.3 deg                           58.7 deg
                                0.137 nm                          0.138nm                    Peak too small

Alpha X-rays         31.3 deg                             66.1 deg                        110.1 deg

                              0.152 nm                            0.153 nm                       0.153 nm


NaCl Crystal Sample Calculations             2dsin(theta/2) = n*wavelength

first order: 2*(2.81*10-8)*sin(28.3/2) = 0.137nm
second order: (2*(2.81*10-8)*sin(58.7/2))/2 = 0.138nm
third order: (2*(2.81*10-8)*sin(110.1/2))/3 = 0.153nm

    The same process was then repeated with a LiF crystal and a RbCl crystal.  The graphs for these crystals are not shown, but they look very similar to the NaCl graph of Counts vs. Detector Angle.


LiF crystal         First order peaks             Second order peaks
Beta X-rays             40.37 deg                             87.43 deg
                                0.139 nm                              0.139 nm

Alpha X-rays           44.68 deg                             99.87 deg
                                0.153 nm                              0.154 nm

LiF Sample Calculations                     2dsin(theta/2) = n*wavelength

first order: 2*(2.01*10-8)*sin(40.37/2) = 0.139nm
second order: (2*(2.01*10-8)*sin(87.43/2))/2 = 0.139nm

    When  the NaCl crystal, LiF crystal, and the RbCl crystal graphs are compared, it is easy to see that the number of counts, the detection angles, and the intensities of the peeks differ.  However, the wavelengths for the alpha and beta peeks for the first, second, and third order are almost identical!


RbCl crystal        First order peaks      Second order peaks     Third order peaks
Beta X-rays               24.13 deg                49.5 deg                            78.81 deg
                                  0.138 nm                 0.138 nm                           0.124 nm

Alpha X-rays               26.88 deg                55.5 deg                             89.25 deg
                                          0.153 nm                0.153 nm                           0.154 nm


RbCl Sample Calculations                     2dsin(theta/2) = n*wavelength

first order: 2*(3.29*10-8)*sin(24.19/2) = 0.138nm
second order: (2*(3.29*10-8)*sin(49.5/2))/2 = 0.138nm
third order: (2*(3.29*10-8)*sin(78.18/2))/3 = 0.153nm
 

     We were able to distinguish between the alpha and beta X-rays because they both give characteristic radiation.  The actual K alpha is 0.154 nm and the actual K beta is 0.138 nm for the first order.  These values were read off the machine and they tell us that the second peak is K alpha and the first peak is K beta.  The alpha peak was the larger of the two and the beta peak was the smaller.  Therefore, K alpha produced more photons than K beta.  Alpha radiation is produced when an electron falls from the L shell into the K shell.  Beta radiation is produced when an electron falls from the M shell into the K shell.  We can now take the known wavelengths and use them to calculate what our angles should have been.  The calculations are shown below.


                         Actual angle (deg)         Observed angle (deg)         % diff. (%)
K-Beta NaCl         14.2                                     14.15                             .35

K-Alpha NaCl       15.9                                     15.65                            1.57

K-Beta LiF            20.08                                   20.19                             .55

K-Alpha LiF         22.52                                    22.34                             .80

    The angles that we found experimentally in class were very close to what they should have been.  This is seen in the very small experimental error.
 
 

Other cool websites that deal with X-rays

Sky Stephens - Bragg Diffraction
Jesse Hagberg  X-Ray Diffraction