Chapter 8, Problem 32P

To determine

To Find: The effective distance between the atoms.

Answer to Problem 32P

  $d=1.2\times {10}^{-10}\text{m}$

Explanation of Solution

Given information:

An oxygen molecule consists of two oxygen atoms whose total mass is $5.3\times {10}^{-26}\text{kg}$ and whose moment of inertia about an axis perpendicular to the line joining the two atoms, midway between them, is $1.9\times {10}^{-49}\text{}\text{kg}\cdot {\text{m}}^{\text{2}}$ .

Formula used:

  $I=M_1{r}_1^2+M_2{r}_2^2$

Here,

  $I$ is the moment of inertia,

  $M_1,M_2$ are the mass of atoms,

  $r_1,r_2$ is the distance of atoms from axis.

Calculation:

An oxygen molecule consists of two oxygen atoms,

With total mass $m=5.3\times {10}^{-26}\text{kg}$

Moment of Inertia of the molecule about an axis perpendicular to the line joining the two atoms, midway between them, $I=1.9\times {10}^{-46}\text{kg}\cdot {\text{m}}^{\text{2}}$ .

As the two atoms of oxygen are similar type, and the axis of rotation is midway between the two atoms,

Therefore, $r_1=r_2=r$

Mass of oxygen atom 1 is $M_1$ and atom 2 is $M_2$

  $\begin{array}{l}I=M_1{r}_1^2+M_2{r}_2^2\\ =M_1{r}^2+M_2{r}^2\\ =\left(M_1+M_2\right)r^2\\ 1.9\times {10}^{-46}=5.3\times 10r{}^2\\ r^2=\left(\frac{1.9\times {10}^{-46}}{5.3\times {10}^{-26}}\right){\text{m}}^{\text{2}}\\ =3.585\times {10}^{-21}{\text{m}}^{\text{2}}\\ r=5.9875\times {10}^{-11}\text{m}\end{array}$

Distance between the two atoms $d=2r$

  $\begin{array}{l}=2\left(5.9875\times {10}^{-11}\right)\text{m}\\ =1.197\times {10}^{-10}\text{m}\\ d=1.2\times {10}^{-10}\text{m}\end{array}$

Conclusion:

The value is $d=1.2\times {10}^{-10}\text{ m}$ .