Answer:
a
e(k) = \frac{2a}{c} * sin (\frac{k*a}{2} )
b
G_{v} = \frac{d e(k ) }{dk } = \frac{a^2}{c} * cos (\frac{k* a}{2} )
Explanation:
From the question we are told that
The velocity of transverse waves in a crystal of atomic separation is
[tex]b_y = c \frac{sin (\frac{k*a}{2} )}{\frac{k*a}{2} }[/tex]
Generally the dispersion relation is mathematically represented as
[tex]e(k) = b_y * k[/tex]
=> [tex]e(k) = c \frac{sin(\frac{k*a}{2} ) }{ \frac{k*a}{2} } * k[/tex]
=> [tex]e(k) = c * \frac{sin (\frac{k_a}{2} )}{ \frac{a}{2} }[/tex]
=> [tex]e(k) = \frac{2a}{c} * sin (\frac{k*a}{2} )[/tex]
Generally the group velocity is mathematically represented as
[tex]G_{v} = \frac{d e(k ) }{dk } = \frac{a^2}{c} * cos (\frac{k* a}{2} ) [/tex]
