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I have a question that popped up in an old exam I just can't find a way to solve.

If you drop the temperature of this material from room temperature (I guess around 300K) to 0C (273K) the resistance drops with a factor of 3.5.

Is it: silicon, germanium or gallium arsenide ?

I've been googling, looking through the books for the last 2h and just can't find any way to solve this question.

Any idea?

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  • $\begingroup$ i could only find the formula: $\endgroup$
    – Sebastian Flückiger
    Commented Jan 31, 2012 at 16:14
  • $\begingroup$ p = p_o * e^(a*T) but this gives factors of sizes up to 10^295 which is rediculous :/ $\endgroup$
    – Sebastian Flückiger
    Commented Jan 31, 2012 at 16:15
  • $\begingroup$ Notice anything about the SIGN of the coefficient for semiconductors? $\endgroup$
    – 410 gone
    Commented Jan 31, 2012 at 16:36
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    $\begingroup$ It seems even Britney Spears does not know the answer to this question. I think you have to find a good reference with material properties for Si, Ge and GaAs, otherwise one can only guess what the coefficient a is for the different materials. $\endgroup$
    – Alexander
    Commented Jan 31, 2012 at 20:04
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    $\begingroup$ i finally found the answer =) for anyone who is ineterested: $\rho(T)=\rho_0\cdot e^{-\frac{E_g}{2k_BT}}$ so the factor would be $\frac{\rho(Room)}{\rho(273.15)}=e^{-\frac{E_g}{2k_B300K}+\frac{E_g}{2k_B273.15K}}$ to achieve 3,5 as factore we change the equation to find $E_g$ and get $E_g=-ln(3.5)\cdot 2k_B\cdot \frac{1}{\frac{1}{300K}-\frac{1}{273.15K}}\approx 0.65$ which is very close to 0.67, the bandgap of germanium. thanks for helping anyways =) $\endgroup$
    – Sebastian Flückiger
    Commented Jan 31, 2012 at 20:06

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i finally found the answer =) for anyone who is ineterested:

$\rho(T)=\rho_0\cdot e^{-\frac{E_g}{2k_BT}}$

so the factor would be

$\frac{\rho(Room)}{\rho(273.15)}=e^{-\frac{E_g}{2k_B300K}+\frac{E_g}{2k_B273.15K}}$

to achieve 3,5 as factore we change the equation to find $E_g$ and get

$E_g=-ln(3.5)\cdot 2k_B\cdot \frac{1}{\frac{1}{300K}-\frac{1}{273.15K}}\approx 0.65 eV$

which is very close to 0.67eV the bandgap of Germanium - hence germanium is the solution.

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