Questions: Use the diagram and information box below to answer the question that follows. bond length = 9.60 x 10^-11 m m0 = 2.66 x 10^-26 kg mH = 1.67 x 10^-27 kg +z out of page Which of the following values represents the moment of inertia of the molecule shown about the z axis through the O atom? (A) 1.17 x 10^-47 kg · m^2 (B) 1.91 x 10^-47 kg · m^2 (C) 5.04 x 10^-47 kg · m^2 (D) 3.08 x 10^-47 kg · m^2 (E) none of the choices available

Use the diagram and information box below to answer the question that follows.

bond length = 9.60 x 10^-11 m
m0 = 2.66 x 10^-26 kg
mH = 1.67 x 10^-27 kg
+z out of page

Which of the following values represents the moment of inertia of the molecule shown about the z axis through the O atom?
(A) 1.17 x 10^-47 kg · m^2
(B) 1.91 x 10^-47 kg · m^2
(C) 5.04 x 10^-47 kg · m^2
(D) 3.08 x 10^-47 kg · m^2
(E) none of the choices available

Transcript text: 1. Use the diagram and information box below to answer the question that follows. \[ \begin{array}{l} \text { bond length }=9.60 \times 10^{-11} \mathrm{~m} \\ m_{0}=2.66 \times 10^{-26} \mathrm{~kg} \\ m_{\mathrm{H}}=1.67 \times 10^{-27} \mathrm{~kg} \\ +\hat{z} \text { out of page } \end{array} \] Which of the following values represents the moment of inertia of the molecule shown about the $z$ axis through the $O$ atom? (A) $1.17 \times 10^{-47} \mathrm{~kg} \cdot \mathrm{~m}^{2}$ (B) $1.91 \times 10^{-47} \mathrm{~kg} \cdot \mathrm{~m}^{2}$ (C) $5.04 \times 10^{-47} \mathrm{~kg} \cdot \mathrm{~m}^{2}$ (D) $3.08 \times 10^{-47} \mathrm{~kg} \cdot \mathrm{~m}^{2}$ (E) none of the choices available

Solution

Solution Steps

Step 1: Determine the moment of inertia contribution from each hydrogen atom

The moment of inertia for a point mass is given by _I = mr_², where _m_ is the mass and _r_ is the distance from the axis of rotation. In this case, the mass of each hydrogen atom is _m__H = 1.67 × 10^-27 kg. The distance _r_ can be found using trigonometry.

_r = d_sin(52°) where _d_ is the bond length = 9.60 × 10^-11 m.

_r_ = (9.60 × 10^-11 m)sin(52°) = 7.56 × 10^-11 m

_I__H = _m__H_r_²= (1.67 × 10^-27 kg)(7.56 × 10^-11 m)²= 9.52 × 10^-47 kg⋅m²

Step 2: Determine the total moment of inertia

Since there are two hydrogen atoms, the total moment of inertia is simply twice the moment of inertia of a single hydrogen atom. The oxygen atom is located on the axis of rotation, so it does not contribute to the moment of inertia.

_I__total = 2_I__H = 2 × 9.52 × 10^-47 kg⋅m² = 1.90 × 10^-46 kg⋅m² which is approximately equal to $1.91 × 10^{-47} kg⋅m^2$