To find the de Broglie wavelength of a neutron in thermal equilibrium with heavy water at temperature $T,$ we will use the de Broglie wavelength formula and the concept of thermal equilibrium.$\\$The de Broglie wavelength $\lambda$ is given by:$\\ \lambda = \frac{h}{p} \\$where $h$ is Planck's constant and $p$ is the momentum of the particle.$\\$For a particle in thermal equilibrium, the average kinetic energy is given by:$\\ \frac{1}{2} m v^2 = \frac{3}{2} k T \\$where $m$ is the mass of the neutron, $v$ is its velocity, and $k$ is Boltzmann's constant.$\\$From the kinetic energy equation, we can solve for $v: \\ m v^2 = 3 k T \\ v^2 = \frac{3 k T}{m} \\ v = \sqrt{\frac{3 k T}{m}} \\$The momentum $p$ is given by:$\\ p = m v \\$Substitute $v$ from the previous equation:$\\ p = m \sqrt{\frac{3 k T}{m}} \\ p = \sqrt{3 m k T} \\$Substitute $p$ into the de Broglie wavelength formula:$\\ \lambda = \frac{h}{\sqrt{3 m k T}} \\$Therefore, the de Broglie wavelength of the neutron is:$\\ \lambda = \frac{h}{\sqrt{3 m k T}} \\$This corresponds to Option 1.