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

The diffusivity is given as
`8.064* 10^(-16) m^2/s`

Step-by-step 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*10^(-4)m^2/s\\Q_d=240.6 kJ/mol=2.406 * 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^{(-Q_d)/(RT)}\\D=(2.0* 10^(-4))e^{(-2.406* 10^5)/(8.314* 1103)}\\D=8.0640* 10^(-16) m^2/s`

So the diffusivity is given as
`8.064* 10^(-16) m^2/s`