Question

In order to produce a certain semiconductor, a process called doping is performed in which phosphorus is diffused into germanium. If = 2.010 and the activation energy is 240.6 , determine the diffusivity of in at 830. The gas constant is = 8.314. (Express your answer to three significant figures.)

Answer 1

Answer:

The diffusivity is given as $8.064\times 10^{-16} m^2/s$

Explanation:

From the given data as in the attached question found via the search (because the values were not clear in this one)

$D_o=2.0\times10^{-4}m^2/s\\Q_d=240.6 kJ/mol=2.406 \times 10^5 J/mol\\Gas Constant=R=8.314 J/mol K\\Temperature =830 C =830+273 =1103 K$

So  the Diffusion coefficient is given as

$D=D_oe^{\frac{-Q_d}{RT}}\\D=(2.0\times 10^{-4})e^{\frac{-2.406\times 10^5}{8.314\times 1103}}\\D=8.0640\times 10^{-16} m^2/s$

So the diffusivity is given as $8.064\times 10^{-16} m^2/s$