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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. |
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Answer» Solution :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 thenumber 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` |
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