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In an intrinsic semiconductor the energy...

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 at `600K` and `300K`? Assume that temperature dependence intrinstic concentration `n_(i)` is given by
`n_(i)=n_(0) exp ((-E_(g))/(2k_T))`, where `n_(0)` is a constant and `k_=8.62xx10^(-5)eV//K`.

Text Solution

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Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as:
`n_(i)=n_(0)exp[-(E_(g))/(2K_(B)T)]`
Where,
`k_(B)` = Boltzmann constant `= 8.62 xx 10^(−5) eV//K`
T = Temperature
`n_(0)` = Constant
Initial temperature, `T_(1) = 300 K`
The intrinsic carrier-concentration at this temperature can be written as:
`n_(i_(1))=n_(0)exp[-(E_(g))/(2k_(B)xx300)]_(...(1))`
Final temperature, `T_(2) = 600 K`
The intrinsic carrier-concentration at this temperature can be written as:
`n_(i2)=n_(0)exp[-(E_(g))/(2k_(B)xx600)]_(...(2))`
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.
`(n_(i2))/(n_(i1))=(n_(0)exp[-(E_(g))/(2k_(B)xx600)])/(n_(0)exp[-(E_(g))/(2k_(B)xx300)])`
`=exp""E_(g)/(2k_(B))[(1)/(300)-(1)/(600)]=exp[(1.2)/(2xx8.62xx10^(-5))xx(2-1)/(600)]`
`=exp[11.6]=1.09xx10^(5)`
Therefore, the ratio between the conductivities is `1.09 xx 10^(5)`.
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