Chapter 29, Problem 35GP

To determine

The moment of inertia.

Answer to Problem 35GP

  $\text{1}\text{.94}{\text{×10}}^{\text{-46}}\text{kg}\cdot {\text{m}}^{\text{2}}$

Explanation of Solution

Given info:

Length is $\text{0}\text{.121}\text{nm}$ .

One unit of mass is $\text{1}\text{.66}\text{×}{\text{10}}^{\text{-27}}\text{kg/u}$ .

Formula used:

  $\text{I= }{\displaystyle \sum {\text{mr}}^2}$

Where,

  $\text{I}$ is the moment of inertia, $\text{m}$ is the mass of atoms and $\text{r}$ is radius of atom.

Calculation:

  $\text{I= }{\displaystyle \sum {\text{mr}}^2}$

  $\text{I}\text{= 2}\left[\left(\text{16}\text{u}\right)\left(\text{1}\text{.66}\text{×}{\text{10}}^{\text{-27}}\text{kg/u}\right)\right]{\left(\text{6}{\text{.05×10}}^{\text{-11}}\text{m}\right)}^{\text{2}}$

  $\text{I}\text{=}\text{1}\text{.94}{\text{×10}}^{\text{-46}}\text{kg}\cdot {\text{m}}^{\text{2}}$

Conclusion:

The required value is $\text{1}\text{.94}{\text{×10}}^{\text{-46}}\text{kg}\cdot {\text{m}}^{\text{2}}$ .