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Demostrate that molar heat capacity of a crystal at a temperature Tlt Theta, where Theta is the Debye temperature, is defined by Eq. (6.4f) |
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Answer» Solution :We use the formula (6.4d) `U= 9R Theta[(1)/(8)+((T)/(Theta))^(4) int_(0)^(Theta//T)(x^(3)DX)/(e^(x)-1)]` `= 9 R Theta[(1)/(8)+{int_(0)^(OO)(x^(3)dx)/(e^(x)-1)}((T)/(Theta))^(4)-((T)/(Theta))^(4) int_(0)^(oo) int_(Theta//T)^(oo)(x^(3)dx)/(e^(x)-1)]` In the LIMIT `Tlt lt Theta`, the third term in the bracket is exponentially small together with its derivatives. Then we can drop the last term `U= Const+(9R)/(Theta^(3))T^(4) int_(0)^(oo) (x^(3)dx)/(e^(x)-1)` Thus `C_(v)=((delU)/(delT))_(v)=((delU)/(delT))_(Theta)=36R(T/(Theta))^(3) int_(0)^(oo) (x^(3)dx)/(e^(x)-1)` Now from the table in the book `int_(0)^(oo)(x^(3)dx)/(e^(x)-1)=(pi^(4))/(15)`. Thus `C_(v)=(12pi^(4))/(5)((T)/(Theta))^(3)` Note: Call the `3^(rd)` term in the bracket above- `U_(3)`. Then `U_(3)=((T)/(Theta))^(4) int_(Theta//T)^(oo)(x^(3))/(2 sin h(x//2)).e^(-x//2)dx` The maximum value of `(x^(3))/(2 sin h(x)/(2))` is a finite+ve qauntity `C_(0)` for `0 le x lt oo`. Thus `U_(3) le 2C_(0)((T)/(Theta))^(4)e^(-Theta//2T)` we see that `U_(3)` is exponentially small as `T rarr 0`. So is `(dU_(3))/(dT)`. |
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