Bragg Angles and Thermal Expansion

[Ichimaru]

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}

 

[nasu]

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.

 

[Kevinn]

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)
=1.88x10^-5