InterviewSolution
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InterviewSolution
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Assuming the propagation velocities of longitudinal and transverse vibrations to be the same and equal to `v`, find the Debye temperature (a) for a unidimensional crystal, i.e., a chain identical atoms, incorporating `n_(0)` atoms per unit length, (b) for a two-dimensional crystal, i.e., a plance square grid consisting of identical atoms, containing `n_(0)` atoms per unit area, (c ) for a simple cubic lattice consisting of identical atoms, containing `n_(0)` atoms per unit volume. |
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Answer» To detemine the Debye temperature we cut off the high frequency modes in such a way as to get the total number of modes correctly. (a) In a linear crystal with `n_(0)l` atoms, the number of modes of transverse vibrations in any given plane cannot exceed `n_(0)l`. Then `n_(0)l=(l)/(pi v) int_(0)^(omega_(0)) d omega=(l)/(piV)omega_(0)` The cut off frequency `omega_(0)` is related to the Debye temperature `Theta` by ` ħ omega_(0)=k Theta` Thus `Theta=(( ħ)/(k))pin_(0)v` (b) In a square lattice, the number of modes of transverse oscillations cannot exceed `n_(0)S` Thus `n_(0)S=(S)/(2 piv^(2)) int_(0)^(omega_(0)) omega d omega=(S)/(4pi v^(2))omega_(0)^(2)` or `Theta=(ħ)/(k)omega_(0)=((ħ)/(k))(sqrt(4pin_(0)))v` In a cubic crystal, the maximum number of transverse waves must be `2n_(0)V` (two for each atom). Thus `2n_(0)V=(V)/(pi^(2)v^(3))int_(0)^(omega_(0)) pmega^(2) d omega=(V omega_(0)^(3))/(3 pi^(2)v^(3))` Thus `Theta=((ħ)/(k))v(6pi^(2)n_(0))^(1//3)` |
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