1H35Cl has a bond length of 127.5 pm. The isotopic masses are: 1H: 1.0078 amu, 35Cl: 34.9689 amu 1. Calculate the reduced mass ? and the moment of inertia I. 2. For the rotational level l=2, calculate the angular momentum |l| and the energy E.
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First, we need to calculate the reduced mass (µ) using the formula: µ = (m1 * m2) / (m1 + m2) where m1 and m2 are the isotopic masses of 1H and 35Cl, respectively. µ = (1.0078 * 34.9689) / (1.0078 + 34.9689) µ ≈ 1.0071 amu Show more…
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1H35Cl has a bond length of 127.5 pm. The isotopic masses are: 1H: 1.0078 amu, 35Cl: 34.9689 amu. 1. Calculate the reduced mass μ and the moment of inertia I. 2. For the rotational level l=2, calculate the angular momentum |l| and the energy E.
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Consider an $\mathrm{O}_{2}$ molecule rotating about its center of mass (as in Fig. 12.26). (a) Derive an expression for its moment of inertia $I$ in terms of $m$, the mass of an $\mathrm{O}$ atom, and $R_{0}$, the bond length. (b) Use Eq. (12.27) to find the lowest three rotational energy levels in $\mathrm{eV}$. (The bond length of $\mathrm{O}_{2}$ is $0.121 \mathrm{~nm}$.)
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