Chemistry, asked by jangirkarishma1997, 6 months ago

Consider the hydrogen iodide diatomic molecule, HI, which has an equilibrium bond length of 1.2 Angstroms. The proton is much lighter than the iodine atom, so the moment of inertia can be approximated using only the mass of the proton, and has a value of 2.41*10^-47 km*m^2. The rotation of the HI molecule can be modeled as a particle on a ring, where the ground rotational state corresponds to the quantum number, n=0. Calculate the energy of the third excited rotational state (n=3). (unit of J, 2 sig figs, use 1.055*10^-34 Js for hbar)​

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Answered by sdas38005
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