The first two non-zero rotational energies of a N₂ molecule are 0.493 meV and 1.48 meV.The two nitrogen atoms are separated by A)0.110 nm B)0.192 nm C)0.214 nm D)0.096 nm E)0.089 nm

Question

Quizplus · Accepted Answer

The rotational energy levels of a diatomic molecule are given by $E_J = \frac{h^2}{8\pi^2I}J(J+1)$, where $E_J$ is the energy of the rotational level, $h$ is Planck's constant, $I$ is the moment of inertia of the molecule, and $J$ is the rotational quantum number (starting from 0 for the ground state). The moment of inertia $I$ is given by $I = \mu r^2$, where $\mu$ is the reduced mass of the molecule and $r$ is the distance between the two atoms.Given the first two non-zero rotational energies (0.493 meV and 1.48 meV), we can set up ratios to eliminate the need for Planck's constant and the reduced mass, focusing on the relationship between energy and the rotational quantum number.The ratio of the second energy to the first is approximately 3 ($1.48 / 0.493 \approx 3$), which corresponds to the ratio of the energy levels for $J=1$ and $J=2$ since $E_J \propto J(J+1)$, and $2(2+1)/1(1+1) = 3$.Using the given energies and solving for $r$ in the formula for $E_J$, we find that the distance that corresponds to the observed energy levels is approximately 0.110 nm, which matches choice A. This calculation involves using the specific values of $h$ and the mass of nitrogen to find the moment of inertia and subsequently the bond length $r$.

The First Two Non-Zero Rotational Energies of a N2 Molecule

The First Two Non-Zero Rotational Energies of a N₂ Molecule