Chapter 29, Problem 10P

To determine

The bond length for the $NaCl$ molecule.

Answer to Problem 10P

  $2.35\times {10}^{-10}m$

Explanation of Solution

Given data:

Three successive wavelength for rotational transitions are $23.1mm,11.6mm,\&7.71mm$ .

Formula used:

  $\Delta E=\frac{\text{hc}}{\lambda }$

  ${\text{E}}_{\text{rot}}=\text{L}\left(\text{L+1}\right)\left(\frac{{\hslash }^{\text{2}}}{2\text{I}}\right)$

  $I=m_{Na}{r}_1^2+m_{Cl}{r}_2^2$

Where,

  $\frac{{\hslash }^{\text{2}}}{2\text{I}}$ is characteristic rotational energy.

  $\hslash$ is reduced Planck’s constant.

  $c$ is speed of the light.

  ${\text{E}}_{\text{rot}}$ is rotational energy.

  $\text{m}$ is mass of atom.

  $\text{I}$ is moment of inertia.

  $\lambda$ is wavelength.

  $\text{L}$ is rotational angular momentum quantum number.

Calculation:

Finding the energies for the transitions form: $\Delta E=\frac{\text{hc}}{\lambda }$

  $\Delta {\text{E}}_1=\frac{1.24\times {10}^3\text{eV}\cdot \text{nm}}{23.1\times {10}^6\text{nm}}=5.36\times {10}^{-5}\text{eV}$

  $\Delta {\text{E}}_2=\frac{1.24\times {10}^3\text{eV}\cdot \text{nm}}{11.6\times {10}^6\text{nm}}=1.1\times {10}^{-4}\text{eV}$

  $\Delta {\text{E}}_3=\frac{1.24\times {10}^3\text{eV}\cdot \text{nm}}{7.71\times {10}^6\text{nm}}=16.1\times {10}^{-5}\text{eV}$

The rotational energy is ${\text{E}}_{\text{rot}}=\text{L}\left(\text{L+1}\right)\left(\frac{{\hslash }^{\text{2}}}{2\text{I}}\right)$ .

Since $\Delta {\text{E}}_3=3\Delta {\text{E}}_1,\Delta {\text{E}}_2=2\Delta {\text{E}}_1$ , so the three transitions must be from the $L=1,2\&3$ states.

The moment of inertia about the CM,

  

  $\Delta {\text{E}}_3=\frac{2\left(3\right){\left(1.055\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{2\text{I}}$

  $\text{I}=\frac{2\left(3\right){\left(1.055\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{2\times \left(16.1\times {10}^{-5}\times \text{1}\text{.60218}\times {\text{10}}^{\text{-19}}\text{J}\right)}=1.29\times {10}^{-45}\text{kg}\cdot {\text{m}}^{\text{2}}$

The positions of the atoms from the CM are

  $\therefore r_1=\frac{\left[m_{Na}\left(0\right)+m_{Cl}r\right]}{m_{Na}+m_{Cl}}=\frac{\left(35.5u\right)r}{\left(23.0u+35.5u\right)}=0.607r$

The bond length will be, $I=m_{Na}{r}_1^2+m_{Cl}{r}_2^2$

  $\therefore 1.29\times {10}^{-45}kg\cdot m^2=\left[23.0{\left(0.607r\right)}^2+35.5{\left(0.393r\right)}^2\right]\left(1.66\times {10}^{-27}kg/u\right)$

  $\therefore r=2.35\times {10}^{-10}m$

Conclusion: The bond length is $2.35\times {10}^{-10}m$ .