A silicon block at 300,K has a length of 10,c | vedclass

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(D) Given: Length $l = 10\,cm = 0.1\,m$,Area $A = 1 	imes 10^{-4}\,m^2$,Voltage $V = 2\,V$,Mobility $\mu = 0.14\,m^2V^{-1}s^{-1}$,Electron density $n

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$6.72 	imes 10^{-7}\,A$

Vedclass · Answer

(D) Given: Length $l = 10\,cm = 0.1\,m$,Area $A = 1 	imes 10^{-4}\,m^2$,Voltage $V = 2\,V$,Mobility $\mu = 0.14\,m^2V^{-1}s^{-1}$,Electron density $n = 1.5 	imes 10^{16}\,m^{-3}$,Charge of electron $e = 1.6 	imes 10^{-19}\,C$.The current $I$ is given by $I = neAv_d$,where $v_d$ is the drift velocity.Since $v_d = \mu E$ and $E = V/l$,we have $v_d = \mu (V/l)$.Substituting this into the current equation: $I = neA \left( \frac{\mu V}{l} ight)$.$I = \frac{n e A \mu V}{l} = \frac{(1.5 	imes 10^{16}) 	imes (1.6 	imes 10^{-19}) 	imes (1 	imes 10^{-4}) 	imes 0.14 	imes 2}{0.1}$.$I = \frac{1.5 	imes 1.6 	imes 0.14 	imes 2 	imes 10^{16-19-4}}{10^{-1}}$.$I = \frac{0.672 	imes 10^{-7}}{10^{-1}} = 0.672 	imes 10^{-6} = 6.72 	imes 10^{-7}\,A$.

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