# Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion

13.

The carrier mobility in a semiconductor is 0.4 m2/Vs. Its diffusion constant at 300k will be (in m2/s).

 [A]. 0.43 [B]. 0.16 [C]. 0.04 [D]. 0.01

Explanation:

No answer description available for this question.

 Chestah said: (Oct 10, 2015) Isn't the answer letter D? D = ukT/q, where kT/q is 26 mV. 26 mV *0.4 m^2/Vs = 0.01 m^2/s.

 Ankit said: (Nov 22, 2015) You are right answer is D. Please explain if anybody have answer B?

 Abc said: (Mar 31, 2016) What is the right answer?

 Vijay said: (May 19, 2016) Right answer is option D.

 Manoj said: (Oct 14, 2017) Answer is D.

 Tup-Taguig said: (Mar 1, 2018) Yeah, it's D.

 Komal said: (May 16, 2018) 0.0104 is the answer.

 Lekhraj Rana said: (Jun 2, 2018) Yes,i.e .0104 answer. Dn/Ue(mobelity)= t/11600, Where as t =300 then 300/11600=26*10^-3, and Ue is .4, Then Dn(diffusion constant)=.4*26*10^-3=.0104.

 Mohan Mayank Chauhan said: (Apr 7, 2021) Option D is the correct answer.