Question
For a linear chain of identical atoms of mass calculate the maximum value of the angular frequency of the longitudinal wave and the group velocity at
, given that the inter-atomic distance is 2.0 A and the spring constant is
.
Answer :
Word Count : 247
For a one-dimensional monoatomic linear chain, the dispersion relation for longitudinal waves is: [ \omega(k) = 2 \sqrt{\frac{K}{M}} , \bigg| \sin\frac{ka}{2} \bigg| ] where (K) is the spring constant, (M) is the atomic mass, (a) is the inter-atomic distance, and (k) is the wavevector. Given: [ M = 10^{-26} \text{ kg}, \quad a = 2.0 , \text{Å} = 2.0 \times 10^{-10} _______ ________ ______ _____ _______ ______ ___ _____.
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For a one-dimensional monoatomic linear chain, the dispersion relation for longitudinal waves is: [ \omega(k) = 2 \sqrt{\frac{K}{M}} , \bigg| \sin\frac{ka}{2} \bigg| ] where (K) is the spring constant, (M) is the atomic mass, (a) is the inter-atomic distance, and (k) is the wavevector. Given: [ M = 10^{-26} \text{ kg}, \quad a = 2.0 , \text{Å} = 2.0 \times 10^{-10} _______ ________ ______ _____ _______ ______ ___ _____.
_____ _______ ______ ___ ____ _________ ____ ________ _______ _____.
_____ ________ _____ __________ ______ ___ _____.
_________ _____ __________ _____ _______ ____ _____ __________ __________ ___.
____ _______ ______ _________ _______ ___ ___ ___ ___.
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