In an intrinsic semiconductor , the energy gap `E_(g)` is 1.2 eV . Its hole mobility is much smaller than electron mobility and independent of temperature . What is the ratio between conductivity of 600 K and that at 300 K ? Assume that the temperature dependence of intrinsic carrier concentration `n_i` is given by
n_(i)=n_(0) exp ((-E_(g))/(k_T)), where n_(0) is a constant and k_=8.62xx10^(-5)eV//K.

Explanation

`n_(i) = n_(o)` exp `(- (E_(g))/(2 k_(B) T))` , where `n_(o)` is a constant
For an intrinsic semiconductor `mu_(e) gt gt mu_(h) , n_(e) = n_(h) = n_(i)`
`sigma = e (n_(e) mu_(e) + n_(h) mu_(h)) = en_(i) (mu_(e) + mu_(h)) ~~ en_i mu_e`
But `n_(i) = n_(o)` exp `(- (E_(g))/(k_(B) T))`, where `E_(g) = (E_(g))/(2)`
`sigma = e mu_(e) n_(o)` exp `[(- E_(g))/(k_(B) T)]`
Here all the pre-exponential terms are assumed to be independent of temperature . So we can put a constant ,
`sigma_(o) = e mu_(o) n_(o)`
`therefore sigma = sigma_(o)` exp `[(- E_(g))/(k_(B) T)] , E_(g) = (E_(g))/(2) = (1.2)/(2) = 0.6` eV
`K_(B) = 8.62 xx 10^(-5) c V K^(-1)`
`therefore sigma_(660 K) = sigma_(o) "exp" (-0.6)/(8.62 xx 10^(-5) xx 600) , sigma_((300K)) = sigma _(o) " exp " (-0.6)/(8.62 xx 10^(-5) xx 300)`
`(sigma_((600K)))/(sigma_((300K))) = ((exp ((-0.6)/(8.62 xx 10^(5) xx 600)))/(exp ((-0.6)/(8.62 xx 10^(-5) xx 300)))) = exp [(0.6)/(8.62 xx 10^(-5)) ((1)/(300) - (1)/(600))]`
= exp `((0.6 xx 10^(5))/(8.62 xx 600)) = exp (11.6) = 1 xx 10^(5)`
This shows that the conductivity of a semiconductor increases rapidly with rise in temperature