Find the frequency factor of the following grain boundary diffusion, if the activation energy is given as 90KJ/mol and the diffusivity at the grain boundary at 1071K is given as 49*10-6m2/sec?
Find the frequency factor of the following grain boundary diffusion, if the activation energy is given as 90KJ/mol and the diffusivity at the grain boundary at 1071K is given as 49*10-6m2/sec? Correct Answer 1.2mm2/sec
It is found experimentally that diffusion along grain boundaries and free surfaces can be described by Db=Db’*exp (-Qb/RT), where Db the grain boundary diffusivity and Db’ the frequency factor. Qb is the experimentally determined values of the activation energy for diffusion. So in this case substituting the respective values we get, Db = 49*10-6*exp (-90000/R*1071) =1.2mm2/sec.
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Feb 20, 2025