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Physics Question on Semiconductor electronics: materials, devices and simple circuits

A block of pure silicon at 300 K300\, K has a length of 10 cm10\, cm and an area of 1.0 cm21.0\, cm^2. A battery of emf 2 V2\, V is connected across it. The mobility of electrons is 0.14 m2 V−1 s−10.14 \,m^2 \,V^{-1}\, s^{-1} and their number density is 1.5×1016 m−31.5 \times 10^{16}\, m^{-3}. The electron current is

A

6.72×10−4 A6.72 \times 10 ^{-4}\, A

B

6.72×10−5 A6.72 \times 10^{-5}\, A

C

6.72×10−6 A6.72 \times 10^{-6}\, A

D

6.72×10−7 A6.72 \times 10^{-7}\, A

Answer

6.72×10−7 A6.72 \times 10^{-7}\, A

Explanation

Solution

As E=Vl=2V0.1mE = \frac{V}{l} = \frac{2V}{0.1 m} =20 Vm−1= 20 \,V m^{-1} A=1.0 cm2A = 1.0 \,cm^{2} =1.0×10−4m2= 1.0 \times 10^{-4}m^{2} ve=μeEv_{e} = \mu_{e} E =0.14×20= 0.14 \times 20 =2.8ms−1= 2.8 ms^{-1} The electron current is Ie=neAeveI_{e} = n_{e}Aev_{e} =(1.5×1016)×(1.0×10−4)×(1.6×10−19)×2.8= \left(1.5\times 10^{16}\right)\times \left(1.0\times 10^{-4}\right) \times \left(1.6\times 10^{-19}\right)\times 2.8 =6.72×10−7A = 6.72\times 10^{-7} A