The total number of phonon modes in a solid of volume V is $\int_{\rm{0}}^{{\rm{ω_ D}}} {{\rm{g}}\left( {\rm{ω }} \right)\,} {\rm{dω }}$ = 3N, where N is the number of primitive cells, ω _Dis the Debye frequency and density of photon modes is g( ω ) = AV ω²(with A > 0 a constant). If the density of the solid doubles in a phase transition, the Debye temperature θ _D, will

Question

Prepp Team · Accepted Answer

Understanding Phonon Modes and Debye Temperature The question asks how the Debye temperature, $\theta_D$, of a solid changes when its density doubles during a phase transition. We are given a relationship involving the total number of phonon modes, the volume $V$, the Debye frequency $\omega_D$, and the density of modes $g(\omega)$. Relationship between Total Modes and Debye Frequency The total number of phonon modes in a solid of volume $V$ is given by the integral of the density of modes $g(\omega)$ up to the Debye frequency $\omega_D$: $\int_{\rm{0}}^{{\rm{&omega;_ D}}} {{\rm{g}}\left( {\rm{&omega; }} \right)\,} {\rm{d&omega; }} = 3N$ where $N$ is the number of primitive cells, and $g(\omega) = AV\omega^2$, with $A > 0$ being a constant. Substituting the expression for $g(\omega)$ into the integral, we get: $\int_{\rm{0}}^{{\rm{&omega;_ D}}} {{\rm{AV&omega; }}^2\,} {\rm{d&omega; }} = 3N$ $AV \int_{\rm{0}}^{{\rm{&omega;_ D}}} {{\rm{&omega; }}^2\,} {\rm{d&omega; }} = 3N$ $AV \left[ \frac{\omega^3}{3} \right]_0^{\omega_D} = 3N$ $AV \frac{\omega_D^3}{3} = 3N$ $AV \omega_D^3 = 9N$ This equation relates the constant $A$, the volume $V$, the Debye frequency $\omega_D$, and the total number of primitive cells $N$. Effect of the Phase Transition on Volume The problem states that the density of the solid doubles during a phase transition. Let the initial state be denoted by subscript 1 and the final state by subscript 2. Initial density = $\rho_1$, Initial volume = $V_1$ Final density = $\rho_2$, Final volume = $V_2$ We are given $\rho_2 = 2\rho_1$. Assuming the total mass $M$ of the solid remains constant during the phase transition, we have: $M = \rho_1 V_1 = \rho_2 V_2$ Substituting $\rho_2 = 2\rho_1$: $\rho_1 V_1 = (2\rho_1) V_2$ Since $\rho_1 \neq 0$, we can cancel it: $V_1 = 2V_2$ This means the final volume is half the initial volume: $V_2 = V_1/2$. Applying the Relation to Initial and Final States The total number of primitive cells $N$ in the solid remains constant during the phase transition (as the amount of material does not change). The constant $A$ is given in the definition of the density of modes $g(\omega)$, and based on how such problems are typically framed, we assume $A$ is a constant independent of the phase transition. For the initial state: $A V_1 \omega_{D1}^3 = 9N$ (Equation 1) For the final state (using the new volume $V_2$ and new Debye frequency $\omega_{D2}$): $A V_2 \omega_{D2}^3 = 9N$ (Equation 2) Equating Equation 1 and Equation 2 (since the right-hand side is the same): $A V_1 \omega_{D1}^3 = A V_2 \omega_{D2}^3$ Since $A \neq 0$, we can cancel it: $V_1 \omega_{D1}^3 = V_2 \omega_{D2}^3$ Now substitute $V_2 = V_1/2$: $V_1 \omega_{D1}^3 = (V_1/2) \omega_{D2}^3$ Since $V_1 \neq 0$, we can cancel it: $\omega_{D1}^3 = (1/2) \omega_{D2}^3$ Rearranging to find the ratio of final to initial Debye frequency: $\omega_{D2}^3 = 2 \omega_{D1}^3$ $\frac{\omega_{D2}^3}{\omega_{D1}^3} = 2$ $\left( \frac{\omega_{D2}}{\omega_{D1}} \right)^3 = 2$ $\frac{\omega_{D2}}{\omega_{D1}} = 2^{1/3}$ So, the Debye frequency increases by a factor of $2^{1/3}$. Change in Debye Temperature The Debye temperature $\theta_D$ is directly proportional to the Debye frequency $\omega_D$: $\theta_D = \frac{\hbar \omega_D}{k_B}$ where $\hbar$ is the reduced Planck constant and $k_B$ is the Boltzmann constant. These are constants. Thus, the ratio of final to initial Debye temperature is: $\frac{\theta_{D2}}{\theta_{D1}} = \frac{\hbar \omega_{D2} / k_B}{\hbar \omega_{D1} / k_B} = \frac{\omega_{D2}}{\omega_{D1}}$ Since $\frac{\omega_{D2}}{\omega_{D1}} = 2^{1/3}$, we have: $\frac{\theta_{D2}}{\theta_{D1}} = 2^{1/3}$ $\theta_{D2} = 2^{1/3} \theta_{D1}$ The Debye temperature will increase by a factor of $2^{1/3}$. Conclusion When the density of the solid doubles in a phase transition, its volume halves (assuming constant mass). With the given density of phonon modes and the total number of modes remaining constant, the Debye frequency, and thus the Debye temperature, increases by a factor of $2^{1/3}$. The correct option is that the Debye temperature will increase by a factor of $2^{1/3}$.