How to calculate the activation energy of diffusion of carbon in iron? At room temperature, $298~\mathrm{K}$, the diffusivity of carbon in iron is 
  $9.06\cdot 10^{-26}\frac{m^2}{s}$.
  At $573~\mathrm{K}$ the diffusivity is $1.66\cdot 10^{-15}\frac{m^2}{s}$. What is the activation energy, $Q$?


I am wondering if my solution is correct:

$D=D_0\exp\left\{\frac{-Q}{RT}\right\}$ and $D_0$ is constant (is this correct), thus 

$\displaystyle \frac{D_{298}}{\exp\left\{-Q/(RT_{298})\right\}}=\frac{D_{573}}{\exp\left\{-Q/(RT_{573})\right\}} \implies Q\approx 122~\mathrm{\frac{kJ}{mol}}$

I would be very greatful if someone could check my math here as well, if the logic is correct, since this was a test problem.
Maybe I'm wrong but my calculation has given another result:
\begin{align}
D_{298} &amp;= D_0\cdot\exp\left\{\frac{-Q}{298~\mathrm{K}\cdot\mathcal{R}}\right\}\\
D_{573} &amp;= D_0\cdot\exp\left\{\frac{-Q}{573~\mathrm{K}\cdot\mathcal{R}}\right\}\\
\end{align}

Thus, going through the $\ln$ form, you get 
\begin{align}
\ln\{D_{298}\} &amp;= \ln\{D_0\} - \frac{Q}{298~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{573}\} &amp;= \ln\{D_0\} - \frac{Q}{573~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{298}\} - \ln\{D_{573}\} &amp;= -\frac{Q}{\mathcal{R}}\cdot\left(\frac{1}{298}-\frac{1}{573}\right)
\end{align}

And then I obtained $Q\approx 38~\mathrm{kJ/mol}$.