In an intrinsic semiconductor the energy gap `E_(g)` is 1.2eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K ? Assume that the temperature dependence of intrinsic carrier concentration `n_(i)` is given by `n_(i)=n_(0)"exp"(-(E_(g))/(2k_(B)T))` where `n_(0)` is a constant.

Explanation

Suppose `n_(i)=n_(0)"exp"(-(E_(g))/(2k_(BT)))`
Here, `n_(0)` is constant.
Conductivity of semiconductor
`sigma=e(n_(e )mu_(e )+m_(h)mu_(h))`
`therefore sigma=n_(i)e(mu_(e )+mu_(h))`
(`because` For intrinsic semiconductor `n_(i)=n_(e )=n_(h)`)
and the mobility of the electrons is much higher than that of the mobility of hole, means `mu_(e ) gt gt mu_(h)`
`therefore sigma=n_(i)e mu_(e )`
but `n_(i)=n_(0)"exp"[(-Eg)/(2k_(B)T)]` is given.
`therefore sigma=n_(0)"exp"[(-Eg)/(2K_(B)T)]e mu_(e )`
but `e mu_(e )n_(0)` in independent to temperature hence putting constant `sigma_(0)` for it.
`sigma=sigma_(0)"exp" [(-Eg)/(2k_(B)T)]`
`therefore` At `T_(1)=600K, sigma_(1)=sigma_(0)"exp"[(-Eg)/(2k_(B)T_(1))]`
and at `T_(2)=300K, sigma_(2)=sigma_(0)"exp"[(-Eg)/(2k_(B)T_(2))]`
`therefore (sigma_(1))/(sigma_(2))=("exp"[(-Eg)/(2k_(B)T_(1))])/("exp"[(-Eg)/(2k_(B)T_(2))])="exp"[(Eg)/(2k_(B)){(1)/(T_(2))-(1)/(T_(1))}]`
`="exp"[(1.2)/(2xx8.62xx10^(-5)){(1)/(300)-(1)/(600)}]`
`="exp"[(0.6xx10^(5))/(8.62)xx(1)/(600)]`
`therefore (sigma_(1))/(sigma_(2))="exp"(11.6)`
`=e^(11.6)`
`=(2.718)^(11.6)`
`"log"(sigma_(1))/(sigma_(2))=11.6log(2.718)`
`=11.6xx0.4343`
`=5.1247`
Antilog of 0.1247
`therefore (sigma_(1))/(sigma_(2))=1.1332 xx 10^(5)`
`~~1.1xx10^(5)`
Therefore, the ratio between the conductivities is `1.09xx10^(8)`. The ratio shows that conductivity of semiconductor increases rapidly with temperature.