Answer to Question #86679 in Atomic and Nuclear Physics for Rabindra

Question

Show that the group velocity is zero at edge of the first Briclouin zone for a linear monoatomic chain.

Answer 1

So, the group velocity is:

vg = (dω/dk) = a(K/m)½cos(½ka)

vg = 0 at the BZ edge [k = ± (π/a)]

This tells us that a wave with λ corresponding to a zone edge wavenumber k = ± (π/a) will not propagate.