Solve the Mystery: Pure Silicon Resistivity at Zero

[kahwawashay1]

I noticed that the temperature coefficient of resistivity of pure silicon is a rather high negative number, so just out of curiosity I wanted to see at what temp the resistivity would drop to zero.

The formula is ρ-ρ0 = ρ0α(T-T0)
where ρ is the final resistivity, ρ0 is the reference resistivity, α is the temperature coefficient of resistivity, T is the final temp and T0 is the reference temp

My book gives the α at a reference temp of 293 K. At this temp, ρ0 is 2.5*10^3 and α is -70*10^-3. Therefore, if we set the final resistivity (ρ) to 0:

-ρ0 = ρ0α(T-T0)
-1/α = T-T0
T = -1/α + T0
= -1/(-70*10^-3) + 293
= 307 K

I have been told that this can't possibly be right, but no one will tell me exactly why.
I have been told that what I'm doing wrong is "assuming that resistivity does not change with temp", even though I obviously am taking that into consideration, since I am using an equation that says approximately how resistivity changes with temp.
I have also been told that the linear approximation equation I am using holds only for a limited range with respect to the reference temp, and I also know that it won't hold for huge temps, but 307 K is not too far from room temp. (and actually, according to my book, this equation holds "over a rather large temperature range")

Could someone please point out exactly what I am doing wrong?
There must be something wrong because I don't think silicon is a conductor at some 35-ish degrees Celsius..

 

[technician]

The equation you have quoted is a version of the variation of resistance/resistivity equation for a material ( metal) with a + ve temp coefficient ( resistance increases with temp)

R = Ro(1+αt) ( t is temp in C)
This equation is only an approximation for reasonably low temps and shows that resistance increases uniformly with temp.
A fuller form of the equation is R = Ro(1+αt+βt^2+...)
one thing for sure... It is not the equation to use for semiconductors ( silicon)