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Draw the vibration energy of a crystal as a function of frequency (neglecting the zero-point vibrations). Consider two cases: T= Theta//21 and T=Theta//4, where Theta is the Debye temperature. |
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Answer» Solution :In the Debye MODEL `dB_(omega)=A omega^(2), 0LE omegale omega_(m)` Then `3N= int_(0)^(omega_(0))dN_(omega)=(A omega_(m)^(3))/(3)`.(Toltal no.of models is `3N`) Thus `A=(9N)/(omega_(m)^(3))` we GET `U=(9N)/(omega_(m)^(3))int_(0)^(omega_(m))(omega^(2).ħomega)/(e^(-ħ omega//KT-1))d omega` ignoring zero point energy `=9Nħ omega_(m) int_(0)^(1)(x^(3)dx)/(e^(ħ omega_(m)x//kT-1)),x=(omega)/(omega_(m))` `=9 R Theta int_(0)^(1)(x^(3)dx)/(e^(x Theta//T)-1) fo r 0 le x le 1` Thus `(1)/(9R Theta)(dU(x))/(dx)=(x^(3))/(e^(x Theta//T)-1) for 0lexle1` for `T= Theta//2`, this is `(x^(3))/(e^(2x)-1)`, for `T=(Theta)/(4)`, it is `(x^(3))/(e^(4x)-1)` |
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