Question

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)

Answer 1

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)`.