Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    \begin{equation}

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

    \end{equation}

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

    \begin{equation}

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

    \end{equation}

    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!

    \end{equation}
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Bragg Angles and Thermal Expansion
  1. Thermal expansion (Replies: 3)

  2. Thermal expansion (Replies: 2)

Loading...