Calculate the energies of the second four rotational levels (5 through 8) of \(^3H^{79}Br\) free to rotate in three dimensions, using its moment of inertia \(I=\mu R^2\).

A. \(E_5 = 3.87 \, \text{J}, \, E_6 = 8.63 \, \text{J}, \, E_7 = 15.35 \, \text{J}, \, E_8 = 23.92 \, \text{J}\)

B. \(E_5 = 7.13 \, \text{J}, \, E_6 = 12.59 \, \text{J}, \, E_7 = 19.77 \, \text{J}, \, E_8 = 28.68 \, \text{J}\)

C. \(E_5 = 1.94 \, \text{J}, \, E_6 = 5.52 \, \text{J}, \, E_7 = 10.33 \, \text{J}, \, E_8 = 16.48 \, \text{J}\)

D. \(E_5 = 4.65 \, \text{J}, \, E_6 = 9.23 \, \text{J}, \, E_7 = 15.98 \, \text{J}, \, E_8 = 24.86 \, \text{J}\)

To calculate the energies of the second four rotational levels of ^3H^79Br, use the formula for rotational energy E = l(l + 1)ħ²/2I for each level (5 through 8). You need additional information to determine ħ and μR², and thereby identify the correct energy values among the given options.

Explanation:

In order to calculate the energies of the second four rotational levels of ^3H^79Br, we must use the formula for rotational energy in a three-dimensional system, E = l(l + 1)ħ²/2I, where l is the quantum number representing the rotational level, ħ is the reduced Planck's constant, and I is the moment of inertia given as μR². In this case, we aren't given specific values for ħ or μR², so we can't calculate exact values. Instead, the question provides multiple energy options and we need additional information to select the correct one.

Most likely, this question is part of a broader problem set where the values for ħ and μR² are given or can be determined from prior information. Once these values are known, we can use the formula to calculate energy at each rotational level, substituting 5 through 8 for l in the equation.

For instance, for the 5th rotational level (l = 5), the energy E5 = 5(5 + 1)ħ²/2I. In the same manner, we can calculate for rotational levels 6, 7, and 8. To find the correct energy values from the provided options, these computations should match one of the given sets.

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