Question 705067

Given: Debye temperature $\Theta_D = 150 \text{ K}$. For a $d$-dimensional system, the specific heat capacity at low temperatures ($T \ll \Theta_D$) is given by $C_v \propto T^d$. Since it is a two-dimensional insulator, $d = 2$. Therefore, $C_v \propto T^2 \implies C_v = k T^2$ (where $k$ is a proportionality constant).

The heat required ($Q$) to raise the temperature from $T_i$ to $T_f$ is: $Q = \int_{T_i}^{T_f} C_v dT = \int_{T_i}^{T_f} k T^2 dT = \frac{k}{3} \left[ T^3 \right]_{T_i}^{T_f} = \frac{k}{3} (T_f^3 - T_i^3)$

Step 1: Heat required from $1 \text{ K}$ to $2 \text{ K}$ $Q_{1 \to 2} = \frac{k}{3} (2^3 - 1^3) = \frac{k}{3} (8 - 1) = \frac{7k}{3}$

Step 2: Heat required from $2 \text{ K}$ to $3 \text{ K}$ $Q_{2 \to 3} = \frac{k}{3} (3^3 - 2^3) = \frac{k}{3} (27 - 8) = \frac{19k}{3}$

Step 3: Calculating the Ratio $\text{Ratio} = \frac{Q_{1 \to 2}}{Q_{2 \to 3}} = \frac{7k/3}{19k/3} = \frac{7}{19}$

Therefore, the correct ratio is $7:19$, which matches Option 1.