In a two-dimensional square lattice, frequency \omega of phonons in the long
wavelength limit changes linearly with the wave vector k. Then the density of
states of phonons is proportional to
Options:
(A) \omega
(B) \omega2
(C) √\omega
(D) 1
√\omega
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.43)**
**1. Solid State Physics Principles (Phonon Density of States in 2D)**:
- Two-dimensional lattice with dispersion relation $\omega(k) = v k$ in the long-wavelength limit.
- Density of states $D(\omega)$ in $d$ dimensions with dispersion $\omega \propto k^s$:
$D(\omega) \propto \omega^{\frac{d}{s} - 1}$
**2. Detailed Step-by-Step Derivation**:
- In $2$ dimensions ($d = 2$) with acoustic phonons ($s = 1$ since $\omega = v k$):
- Area of $k$-space shell: $d^2k = 2\pi k\, dk$.
- Number of states $N(k)\, dk = \frac{A}{(2\pi)^2} 2\pi k\, dk = \frac{A}{2\pi} k\, dk$.
- Substitute $k = \frac{\omega}{v}$ and $dk = \frac{d\omega}{v}$:
$D(\omega)\, d\omega = \frac{A}{2\pi v^2} \omega\, d\omega$
- Thus, the density of states $D(\omega)$ is directly proportional to $\omega$:
$D(\omega) \propto \omega$
**3. Official Verification**:
The verified official IIT Master Answer is **Option (A)** (`\omega`).
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