Sebastian Flückiger
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 Apr 18 awarded Famous Question Feb 5 awarded Popular Question Nov 7 awarded Notable Question Feb 11 awarded Popular Question Jul 22 awarded Scholar Jul 22 accepted Degeneracy of Energy Levels for 2 identical particles in a One Dimensional box Jul 20 awarded Supporter Jul 20 comment Degeneracy of Energy Levels for 2 identical particles in a One Dimensional box thank you for this unbelievable thourough answer! it will take some time for me to process and understand it i will come back on it :) Jul 20 comment Degeneracy of Energy Levels for 2 identical particles in a One Dimensional box thx for editing, altough: it is definitly NOT homework. its an exercise from an old exam. (at a certain point there is no such thing as 'homework' anymore) ;) Jul 20 asked Degeneracy of Energy Levels for 2 identical particles in a One Dimensional box Feb 1 comment change of resistance in semiconductors due to temperature change wheras conductivity and resistance indirectly proportional so it does not really matter which one you calculate =) if the resistance drops with a factor 3.5 the conductivity goes up by the same factor, and vice versa. Feb 1 awarded Editor Feb 1 revised change of resistance in semiconductors due to temperature change added 96 characters in body Feb 1 awarded Teacher Feb 1 answered change of resistance in semiconductors due to temperature change Feb 1 comment change of resistance in semiconductors due to temperature change i couldnt because i had too little reputation. i will as soon as i get to a conputer (just cell here). rho is the specific resistance @akhmeteli Jan 31 awarded Student Jan 31 comment change of resistance in semiconductors due to temperature change haha thank you for this funny link :P as said - i was able to solve it myself =) Jan 31 comment change of resistance in semiconductors due to temperature change 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 =) Jan 31 comment change of resistance in semiconductors due to temperature change oh yes clearly silicon (sry im swiss, in german its called 'silicium' confusing..) and yes it should be -aT but still i dont get to 3.5 onany, maybe iused the wrong a..