Chapter 8, Problem 76GP

(a)

To determine

To Calculate: The moment of inertia for the ${\text{H}}_{\text{2}}\text{O}$ molecule about an axis passing through the center of the oxygen atom which is perpendicular to the plane of the molecule.

Answer to Problem 76GP

  $\mathrm{I}=3.07\times {10}^{-45}\text{kg}\cdot {\text{m}}^{\text{2}}$

Explanation of Solution

Given data:

Length of bond, $\mathrm{R}=0.96\text{nm}=\text{0}\text{.96}\times {\text{10}}^{\text{-9}}\text{m}$

The angle of $\text{H}-\text{O}-\text{H}$ bond $=104°$

Formula used:

Write the expression for the moment of inertia.

  $\mathrm{I}=2\times \mathrm{m}{\mathrm{R}}^2$

Here,

  $\mathrm{I}$ is the moment of inertia,

  $\mathrm{m}$ is the mass,

  $\mathrm{R}$ is the radius.

Calculation:

The moment of inertia can be determined by the following equation,

  $\mathrm{I}=2\times \mathrm{m}{\mathrm{R}}^2$

Mass of the atom is $1.67\times {10}^{-27}\text{kg}$ .

  $\mathrm{I}=2\times 1.67\times {10}^{-27}\times {\left(0.96\times {10}^{-9}\right)}^2$

  $\text{kg}\cdot {\text{m}}^{\text{2}}$

  $\mathrm{I}=3.07\times {10}^{-45}\text{kg}\cdot {\text{m}}^{\text{2}}$

Conclusion:  The moment of inertia of the ${\text{H}}_{\text{2}}\text{O}$ molecule about an axis passing through the center of the oxygen atom which is perpendicular to the plane of the molecule is $\mathrm{I}=3.07\times {10}^{-45}\text{kg}\cdot {\text{m}}^{\text{2}}$ .

(b)

To determine

To Calculate: The moment of inertia for the ${\text{H}}_{\text{2}}\text{O}$ molecule about an axis passing through the center of the oxygen atom in the plane of the molecule bisecting the $\text{H}-\text{O}-\text{H}$ molecule.

Answer to Problem 76GP

  $\mathrm{I}=2.42\times {10}^{-45}\text{kg}\cdot {\text{m}}^{\text{2}}$ .

Explanation of Solution

Given data:

Length of bond $=0.96\text{nm}=\text{0}\text{.96}\times {\text{10}}^{\text{-9}}\text{m}$

The angle of H-O-H bond $\mathrm{\theta }=104°$

Radius $\mathrm{R}=\sin \left(\frac{104°}{2}\right)\times 0.96\times {10}^{-9}\text{m}$

Formula used:

Write the expression for the moment of inertia.

  $\mathrm{I}=2\times \mathrm{m}{\mathrm{R}}^2$

Here,

  $\mathrm{I}$ is the moment of inertia,

  $\mathrm{m}$ is the mass,

  $\mathrm{R}$ is the radius.

Calculation:

The moment of inertia can be determined by the following equation,

  $\mathrm{I}=3.07\times {10}^{-47}\text{kg}\cdot {\text{m}}^{\text{2}}$

  $\mathrm{I}=2\times \mathrm{m}{\mathrm{R}}^2$

  $\text{kg}\cdot {\text{m}}^{\text{2}}$

Mass of the Hydrogen atom is $1.67\times {10}^{-27}\text{kg}$ .

  $\mathrm{I}=2\times 1.67\times {10}^{-27}\times \sin \left(\frac{{104}^0}{2}\right)\times {\left(0.96\times {10}^{-9}\right)}^2$

  $\text{kg}\cdot {\text{m}}^{\text{2}}$

  $\mathrm{I}=2.42\times {10}^{-45}\text{kg}\cdot {\text{m}}^{\text{2}}$

Conclusion:  The moment of inertia of the ${\text{H}}_{\text{2}}\text{O}$ molecule about an axis passing through the center of the oxygen atom in the plane of the molecule bisecting the $\text{H}-\text{O}-\text{H}$ molecule $\mathrm{I}=2.42\times {10}^{-45}\text{kg}\cdot {\text{m}}^{\text{2}}$ .