91Ó°ÊÓ

We need to find the moment of inertia (I) of the rotating molecule using the formula \(I = \mu R^2\), where \(\mu\) is the reduced mass of the molecule and \(R\) is the bond length. Given that, Bond length, \(R = 160.92 \times 10^{-12}\) m Reduced mass, \(\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 127}{128}\), where mass of hydrogen (H) is 1 atomic unit and mass of iodine (I) is 127 atomic unit. Reduced mass, \(\mu = \frac{1 \times 127}{128} = \frac{127}{128} \) atomic unit. Since we know the atomic mass unit is 1u = 1.66054 x 10^{-27} kg, we can convert the reduced mass to kg: \(\mu = 1.66054 \times 10^{-27} \times \frac{127}{128} = 2.156218 \times 10^{-27}\) kg Now, we can compute the moment of inertia (I): \(I = \mu R^2 = 2.156218 \times 10^{-27} (160.92 \times 10^{-12})^2 = 5.575554 \times 10^{-46}\) kg m²

The rotational constant \(B\) relates to the moment of inertia (I) according to the formula \(B = \frac{h}{8\pi^2Ic}\), where \(h\) is the Planck's constant, and \(c\) is the speed of light in vacuum. Given that, Planck's constant, \(h = 6.62607015 \times 10^{-34}\) Js, and Speed of light in vacuum, \(c = 2.99792458 \times 10^{8}\) m/s. Now, we can calculate the rotational constant (B): \(B = \frac{6.62607015 \times 10^{-34}}{8 \pi^2 (5.575554 \times 10^{-46})(2.99792458 \times 10^{8})} = 1.336418 \times 10^{-11}\) J

The zero-point energy, \(E_0\), is the energy of the lowest rotational energy level when the quantum number \(J = 0\). We can calculate the zero-point energy, \(E_0\), using the formula, \(E_0 = \frac{1}{2}hB\), where \(h\) is the Planck's constant and \(B\) is the rotational constant we calculated in Step 2. Now, let's find the zero-point energy: \(E_0 = \frac{1}{2} (6.62607015 \times 10^{-34})(1.336418 \times 10^{-11}) = 4.430932 \times 10^{-45}\) J

The smallest quantum of energy that can be absorbed by the molecule is the energy difference between the first two rotational energy levels (transition from \(J=0\) to \(J=1\)). The energy of a rotational level is given by the formula, \(E_J = J(J + 1)hB\), where \(J\) is the quantum number. The energy difference between the first two levels is: \(\Delta E = E_1 - E_0 = hB(1) - \frac{1}{2}hB = \frac{1}{2}hB\) Now, let's find the smallest quantum of energy absorbed: \(\Delta E = \frac{1}{2} (6.62607015 \times 10^{-34})(1.336418 \times 10^{-11}) = 4.430932 \times 10^{-45}\) J The results are: a) Zero-point energy associated with the rotation is \(4.430932 \times 10^{-45}\) Joules. b) The smallest quantum of energy that can be absorbed by the molecule in a rotational excitation is \(4.430932 \times 10^{-45}\) Joules.