Question

Calculate the Debye temperature for iron in which the propagation velocities of longtitudinal and transverse vibrations are equal ot `5.85` and `3.23Km//s` respectivel.

Answer 1

We proceed as in the prevoius example. The total number of modes must be `3n_(0)v` (total trnasverse and one longitudinal per atom). On the other hand of transverse modes per unit frequency interval is by
`dN^(_|_)=(Vomega^(2))/(pi^(2)V_(_|_)^(3))d omega`
while the number of longitudinal moder per unit frequency interval is given by
`dN^(||)=(V omega^(2))/(2pi^(2)v_(||)^(3))d omega`
The total number per unit frequency interval is
`dN=(V omega^(2))/(2pi^(2))((2)/(V_(_|_)^(3))+(1)/(V_(||)^(3)))d omega`
If the high frequency cut off is at `omega_(0)=(kTheta)/( ħ)`, the total number of modes will be
`3n_(0)V=(V)/(6 pi^(2))((2)/(V_(_|_)^(3))+(1)/(v_(||)^(3)))((k Theta)/( ħ))^(3)`
Here `n_(0)` is the number of iron atoms per unit volume. Thus
`: Theta=(ħ)/(k)[18 pi^(2)n_(0)//((2)/(v_(_|_)^(3))+(1)/(v_(||)^(3)))]^(1//3)`
For iron
`n_(0)=N_(A)//(M)/(rho)=(rhoN_(A))/(M)`
`(rho=` density, `M=` atomic weight of iron `N_(A)=` Avogardo number).
`n_(0)= 8.389xx10^(22)per c c`
Substituting the data we get
`Theta= 469.1K`