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Bragg Angles and Thermal Expansion

  1. Apr 8, 2014 #1
    Problem statement:

    The Bragg angles of a certain reflection from copper is 47.75◦ at 20◦C but is 46.60◦ at 1000◦C.
    What is the coefficient of linear expansion of copper? (Note: the Bragg angle θ is half of the
    measured diffraction (deflection) angle 2θ).

    Attempt at solution:

    Using \begin{equation} 2d sin( \theta )= n \lambda \end{equation} to find the ratio of d(T=1000) and d(T=20) and saying that this is equal to the lattice constant ratio for those temperatures I found that:


    \frac{a(T=1000)}{a(T=20)}=\frac{sin( \theta (T=20) )}{sin( \theta ( T= 1000))}


    Which when used in the equation for the linear expansion coefficient, kappa:


    \kappa = \frac{a(T=1000)}{a \Delta T}


    gives a value of 10^{-3} per kelvin, which is about 100 times too large when I compared it to the actual data. I know this is a basic question, but I can't see what I'm wondering what I'm doing wrong.

    Thanks in advance!

  2. jcsd
  3. Apr 9, 2014 #2
    The coefficient of thermal expansion is not given by that last formula.
    You should have a (Δa) in the formula, the difference between the lattice constants at the two temperatures.
  4. May 3, 2016 #3
    Maybe just a little late.. 2 years?
    I'm doing this question for PS204 study in DCU.

    Lo = n(lambda) / 2Sin47.75
    L = n(lambda) / 2Sin46.6
    change in L = L - Lo

    linear expansion coefficient = (1/Lo)(change in L / change in T)
    = (1 - Sin46.6/Sin47.75)(1/980)
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