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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. |
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Answer» Solution :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. |
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