The energy gap of pure `Si` is `1.1 eV` The mobilities of electrons and holes are respectively `0.135 m^(2) V^(-1) s^(-1)` and `0.048 m^(2) V^(-1) s^(-1)` and can be taken as independent of temperature. The intrinsic carrier concentration is given by `n_(i) = n_(0) e^(-Eg//2kT)`.
Where `n_(0)` is a constant, `E_(g)` The gap width and `k` The Boltmann's constant whose vaue is `1.38 xx 10^(-23) JK^(-1)` The ratio of the electrical conductivities of `Si` at `600 K` and `300 K` is.

Explanation

The total electrical conductivity of a semiconductor is given by `sigma =e(n_(e)mu_(e)+n_(h)mu_(h))`
For an intrinsic semiconductor, `n_(e) = n_(h) = n_(i)`
We can thus write for the conductivity
`sigma =e(mu_(e) +mu_(h))n_(i)` or `sigma =e(mu_(e)+mu_(h))n_(0)e^(E_(g)//2kT)`
As the mobilities `mu_(e), mu_(h)` are independent of temperature, they can be regarded as constant. The ration of conductivities at `600 K` and `300 K` is
then, `(sigma_(600))/(sigma_(300)) =(e(mu_(e)+mu_(h))n_(0)e^(-E_(g)//2kx 600))/(e(mu_(e)+mu_(h))n_(0)e^(-E_(g)//2kx300)) = e^(-E_(g)//1200 k)`
As per given data `E_(g) = 1.1 eV`
`k = 1.38 xx10^(-23) JK^(-1)` or `((1.38 xx 10^(-23))/(1.6 xx 10^(-19))) eVK^(-1)`
`:. k = 8.625 xx 10^(-5) eVK^(-1)`
Solving we get the ratio of electrical conductivities is `4 xx 10^(4)`.