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A semiconductor is known to have an electron concentration of 5x10^(12) cm^(-3) and a hole concentration 8x10^(13) cm^(-3) . (i)Is the semiconductor n-type of p-type ? (ii)What is the resistivity of the sample. If the electron mobility is 23,000 cm^(2)v^(-1)s^(-1) ' and hole mobility is100cm^(2)v^(-1)s^(-1) ' Take charge on electron, e=.16x10^(-19)c. |
| Answer» Solution :(i)`n_(e) =5xx10^(12)cm^(-3)=5xx10^(18)m^(-3)``n_(e) =5xx10^(12)cm^(-3)=5xx10^(18)m^(-3)`` u_(e) =23000cm^(-2)v^(-1)s^(-1)=2.3m^(-2)v^(-1)s^(-1)` ` u_(e) =23000cm^(-2)v^(-1)s^(-1)=2.3m^(-2)v^(-1)s^(-1)`` u_(e) =100cm^(-2)v^(-1)s^(-1)=0.01m^(-2)v^(-1)s^(-1)` Since the SEMICONDUCTOR has greater HOLE density hence it is p-type.(ii)`n_(e) =5xx10^(12)cm^(-3)=5xx10^(18)m^(-3)` `n_(e) =5xx10^(12)cm^(-3)=5xx10^(18)m^(-3)`` u_(e) =23000cm^(-2)v^(-1)s^(-1)=2.3m^(-2)v^(-1)s^(-1)` ` u_(e) =23000cm^(-2)v^(-1)s^(-1)=2.3m^(-2)v^(-1)s^(-1)` ` u_(e) =100cm^(-2)v^(-1)s^(-1)=0.01m^(-2)v^(-1)s^(-1)` Now `((1)/(p))=e((n_(e)u_(e)+n_(H)u_(h)) =1.6xx10^(-19)[5xx10^(18)xx2.3+8xx10^(19)xx0.01]`=`1.6xx10^(-19)[1.15xx10^(19)xx2.3+0.08xx10^(19)]=1.968` or `RHO =((1)/(1.968))=0.508 Omega -m` | |