QB_PH_SEC7_010: Crystal Structure & Lattice Dynamics

For a 1D diatomic lattice consisting of two different atoms of masses $M_1$ and $M_2$ with nearest-neighbor spring constant $C$, what is the angular frequency $\omega$ of the optical phonon branch at the Brillouin zone center ($k = 0$)?

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Equations of Motion**: For displacements $u_n$ (atom $M_1$) and $v_n$ (atom $M_2$): $M_1 \ddot{u}_n = C(v_n + v_{n-1} - 2u_n)$ $M_2 \ddot{v}_n = C(u_{n+1} + u_n - 2v_n)$ 2. **Secular Determinant**: Substituting plane wave solutions $u_n = u e^{i(kna - \omega t)}$ and $v_n = v e^{i(kna - \omega t)}$: $\begin{vmatrix} 2C - M_1 \omega^2 & -2C \cos(ka/2) \\ -2C \cos(ka/2) & 2C - M_2 \omega^2 \end{vmatrix} = 0$ 3. **Dispersion Roots**: $\omega^2 = C\left(\frac{1}{M_1} + \frac{1}{M_2}\right) \pm C\sqrt{\left(\frac{1}{M_1} + \frac{1}{M_2}\right)^2 - \frac{4\sin^2(ka/2)}{M_1 M_2}}$ 4. **Optical Branch at $k = 0$**: Taking the plus sign with $\sin(0) = 0$: $\omega^2 = C\left(\frac{1}{M_1} + \frac{1}{M_2}\right) + C\left(\frac{1}{M_1} + \frac{1}{M_2}\right) = 2C\left(\frac{1}{M_1} + \frac{1}{M_2}\right)$ $\omega_{\text{optical}}(k=0) = \sqrt{2C\left(\frac{1}{M_1} + \frac{1}{M_2}\right)}$ 5. **Conclusion**: Option (A) is correct.