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.