Calculating electron drift velocity

[jisbon]

Homework Statement

A Si wafer is doped with arsenic and has an electron concentration of
$3X10^{22}m^{-3}$ The electrical conductivity of the wafer is found to be 820
(ohm-m). Calculate the electron drift velocity and mobility when an
electric field of 600 W/m is applied to Si wafer.

Relevant Equations

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Just wanted to check in my workings to see if they are correct (seemed to be too short to me?)

Since electrical conductivity is 820 (ohm.m) which is = n*e*(mobility)
Mobility =0.17083?
And I can simply get drift velocity by multiplying mobility with an electric field (600V/m)?

Cheers

 

[etotheipi]

The long way might be $I = \frac{V}{R} = \frac{EL}{R} = EA \sigma$ so then $v_{d} =\frac{I}{neA} = \frac{E \sigma}{ne}$ which gives the same answer that you obtained of $0.17083$ multiplied by $600\text{V}\text{m}^{-1}$, for $v_{d} = \mu E$, though they're basically equivalent methods.

 

[jisbon]

The long way might be $I = \frac{V}{R} = \frac{EL}{R} = EA \sigma$ so then $v_{d} =\frac{I}{neA} = \frac{E \sigma}{ne}$ which gives the same answer that you obtained of $0.17083$ multiplied by $600\text{V}\text{m}^{-1}$, for $v_{d} = \mu E$, though they're basically equivalent methods.

Oh cool, thanks :)