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Calculate the relative populations of a spherical rotor at 298 K in the levels with J = 0 and J = 5, given that ᷉ B = 2.71 cm−1.
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- Calculate the relative populations of a linear rotor at 298 K in the levels with J = 0 and J = 5, given that ᷉ B = 2.71 cm−1.Find the oscillation amplitude of a diatomic molecule (CO) for n=1,2,3,4 & 10,when k=1926 N\m and reduced mass= 6.86 aDevelop an expression for the value of J corresponding to the most highly populated rotational energy level of a diatomic rotor at a temperature T remembering that the degeneracy of each level is 2J + 1. Evaluate the expression for ICl (for which ᷉ B = 0.1142 cm−1) at 25 °C. Repeat the problem for the most highly populated level of a spherical rotor, taking note of the fact that each level is (2J + 1)2-fold degenerate. Evaluate the expression for CH4 (for which ᷉ B = 5.24 cm−1) at 25 °C. Hint: To develop the expression, recall that the first derivative of a function is zero when the function reaches either a maximum or minimum value.
- Calculate the relative populations of a linear rotor in the levels with J = 0 and J = 5, given that B = 2.71 cm-1 and a temperature of 298 K.The rotational constant of 12C16O is 57.65 GHz. Calculate the value of J for the most populated level at (a) 300 K and (b) 1000 K.What is the most highly populated rotational level of Cl2 at (i) 25 °C, (ii) 100 °C? Take ᷉ B = 0.244 cm−1.
- Consider the diatomic molecule AB modeled as a rigid rotor (two masses separated by a fixed distance equal to the bond length of the molecule). The rotational constant of the diatomic AB is 25.5263 cm-1. (a) What is the difference in energy, expressed in wavenumbers, between the energy levels of AB with J = 10 and J = 6? (b) Consider now a diatomic A'B', for which the atomic masses are ma 0.85 mA and mB' 0.85 mB and for its bond length ra'B' = 0.913 rAB. What is the difference in energy, expressed in wavenumbers, between the energy levels of the A'B' molecule with J = 9 and J = 7?You have a vibrational degree of freedom that can be treated like a harmonic oscillator with a spacing between energy levels of 538,03 cm1, What is the probability that this degree of freedom is in state v = 1 at a temperature of 1,138.2 K. Give your answer with at least 3 significant figures.Calculate the energies of the first four rotational levels of 1H127I free to rotate in three dimensions; use for its moment of inertia I = μR2, with μ = mHmI/(mH + mI) and R = 160 pm. Use integer relative atomic masses for this estimate.
- The energy levels (in cm-1) of diatomic anharmonic oscillator are described by the following expression: Ey = (v + 1/2 )we - (v + 1/2 )< Using this equation, derive expressions for the energies of transition Aɛ for the fundamental vibration and the 2nd harmonic vibration for a diatomic molecule.Pure rotational Raman spectra of gaseous C6H6 and C6D6 yield the following rotational constants: ᷉ B(C6H6) = 0.189 60 cm−1, ᷉ B(C6D6) = 0.156 81 cm−1. The moments of inertia of the molecules about any axis perpendicular to the C6 axis were calculated from these data as I(C6H6) = 1.4759 × 10−45 kg m2, I(C6D6) = 1.7845 × 10−45 kgm2. Calculate the CC and CH bond lengths.Assuming a harmonic potential, the fundamental transition for a diatomic molecule is 1603 cm-1. What is the energy, in wavenumber, of the v=3 level of the molecule?