A semiconductor is known to have an electron concentration of `5xx10^(12) cm^(-3)` and a hole concentration `8xx10^(13)cm^(-3) ` Is the semi-conductor n-type or p-type? What is the resistivity of the sample, if the electron mobility is `23,000 cm^(2) V^(-1)s^(-1)`? Take charge on electron, `e=1.6xx10^(-19) C`.

Explanation

(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`