InterviewSolution
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InterviewSolution
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In a chain of identical atoms the vibration frequency `omega` depends on wave number `k`as `omega= omega_(max) sin (ka//2)`, where `omega_(max)` is the maximum vibration frequency `omega`, a is the distance between neighbouring atoms. Making use of this dispersion relation, find the dependence of the number of longitudinal vibrations per unit. frequency interval on `omega` i.e., `dN//omega`, if the length of the chain is `l`. Having obtained `dN//omega`, find the total number `N` of possible longitudinal vibrations of the chain. |
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Answer» If the chain has `N` atoms, we can assume atom number `0` and `N+1` held ficed. Then the displacement of the `n^(th)` atom has the form `u_(n)=A("sin"(m pi)/(L),n a) sin omega t` Here `k=(m pi)/(L)`. Allowed frequencies then have the form `omega=omega_(max) "sin"(ka)/(2)` In our from only `+vek` values are allowed. The number of models in a wave number range `dk` is `dN=(Ldk)/(pi)=(L)/(pi)(dK)/(d omega)` But `d omega=(a)/(2) omega_(max)"cos"(ka)/(2)dk` Hence `(d omegaqa)/(dk)=(a)/(2)sqrt(omega_(max)^(2)-omega^(2))` So `dN=(2L)/(pia)(d omega)/(sqrt(omega_(max)^(2)-omega^(2)))` (b) The total number modes is `N= int_(0)^(omega_(max))(2L)/(pi a)(d omega)/(sqrt(omega^(2)_(max)-omega^(2)))=(2L)/(pi a).(pi)/(2)=(L)/(a)` i.e., the number of atoms in the chain. |
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