Question

In a one dimensional diatomic crystal, the velocity of sound is 1500 m/s and the lattice
constant 4.0Å. The relationship between the mass of the atoms is M/m = 0.8. Calculate
the gap in the frequency at the Brillouin zone boundary.

Answer 1

Answer on Question #51564, Physics, Solid State Physics

In a one-dimensional diatomic crystal, the velocity of sound is $1500\mathrm{m / s}$ and the lattice constant $4.0\AA$ . The relationship between the mass of the atoms is $\mathrm{M / m} = 0.8$ . Calculate the gap in the frequency at the Brillouin zone boundary.

Solution

Fig.1

The gap in the frequency at the Brillouin zone boundary is given by Eq.(1) and (2).

$$\omega_ {1} = \sqrt {\frac {2 \beta}{M}}$$ $$\omega_ {2} = \sqrt {\frac {2 \beta}{m}}$$

The velocity of sound is $\nu = \sqrt{\beta / ma}$

Than

$$\omega_ {1} = \sqrt {2} \frac {v}{a} \sqrt {\frac {m}{M}} = \sqrt {2} \frac {1 5 0 0}{4 \cdot 1 0 ^ {- 1 0}} \sqrt {1 / 0 . 8} = 5 3. 0 3 \cdot 1 0 ^ {9} r a d / s$$ $$\omega_ {2} = \sqrt {2} \frac {v}{a} = \sqrt {2} \frac {1 5 0 0}{4 \cdot 1 0 ^ {- 1 0}} = 5 9. 3 1 \cdot 1 0 ^ {9} r a d / s$$

Answer:

The gap in the frequency is $\left[59.31\cdot 10^{9}rad / s;59.31\cdot 10^{9}rad / s\right]$

If the frequency is expressed in Hertz $(f = \omega /2\pi)$ [8.44GHz;9.44GHz]

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