Question

When a NaF molecule makes a transition from the l= 3 to the l= 2 rotational level with no change in vibrational quantum number or electronic state, a photon with wavelength 3.83 mm is emitted. A sodium atom has mass $3.82\times {10}^{-26}\mathrm{kg}$, and a fluorine atom has mass $3.15\times {10}^{-26}\mathrm{kg}$. Calculate the equilibrium separation between the nuclei in a NaF molecule. How does your answer compare with the value for NaCl given in Section 42.1? Is this result reasonable? Explain.

Short answer

The equilibrium separation between the nuclei in a NaF molecule is 0.193 nm.

Step-by-step solution

The rotational energy of particular level of diatomic molecule

The rotational energy of particular level of diatomic molecule is given by

$E_{\ell }=\ell (\ell +1)\frac{h^2}{4\pi 2l}$

Where, l is the rotational quantum number , lis the moment of inertia .

Reduced mass

It is the equivalent mass of system of two objects of masses $m_1$and $m_2$.

${\mathrm{m}}_{\mathrm{r}}=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

Planck’s quantum theory

1.Diffrent atoms and molecules can emit or absorb energy in discrete quantities known as quantum.

2. The energy of the radiation absorbed or emitted is directly proportional to the frequency of radiation.

Mathematically energy of radiation is given by

$\begin{array}{r}E=hv\\ E=\frac{hc}{\lambda }\end{array}$

Where, E is the energy of the radiation, h is the Planck’s constant , $\lambda$ is the wavelength of emitted light and c is the speed of light.

Moment of inertia

It is a measure of the resistance of a body to angular acceleration about a given axis that is equal to the products of element of mass in the body and square of the element’s distance from axis.

Mathematically it can be given by

$I=m{r}^2$

Where, is the mass of the particle and is the distance of particle from axis.

 

The calculation of energy of the emitted photon

Using

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

Put the value of constants in above equation

$$\begin{gathered} E=\frac{\left(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{3.83\times {10}^{-3}\mathrm{m}} \\ E=5.19\times {10}^{-23}\mathrm{J} \end{gathered}$$

Thus, the energy of emitted photon is$5.19\times {10}^{-23}J$.

The calculation of reduced mass

Using

${\mathrm{m}}_{\mathrm{r}}=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

Putting values of constants in above equation

$$\begin{gathered} {\mathrm{m}}_{\mathrm{r}}=\frac{\left(3.82\times {10}^{-26}\mathrm{kg}\right)\left(3.15\times {10}^{-26}\mathrm{kg}\right)}{\left(3.82\times {10}^{-26}\mathrm{kg}\right)+\left(3.15\times {10}^{-26}\mathrm{kg}\right)} \\ {\mathrm{m}}_{\mathrm{r}}=1.73\times {10}^{-26}\mathrm{kg} \end{gathered}$$

Thus, the reduced mass of the system is $1.73\times {10}^{-26}kg$.

Calculation of moment of inertia of NaF molecule

Using

$E_{\ell }=\ell (\ell +1)\frac{h^2}{4\pi 2l}$

The energy difference between quantum states l=2 and l=3

$$\begin{gathered} \mathrm{\Delta }E=[3\times 4-2\times 3]\frac{h^2}{4{\pi }^{2}2l} \\ \mathrm{\Delta }E=\frac{3h^2}{4{\pi }^{2}l} \end{gathered}$$

Here, the energy of the emitted photon equals the energy difference between the two states

$\Delta E=E$

Put the values of constants in above equation

$$\begin{gathered} \frac{2h^2}{4{\pi }^{2}I}=5.19\times {10}^{-23}\mathrm{J} \\ \mathrm{I}=\frac{2\left(6.63\times {10}^{-34}\right)}{5.19\times {10}^{-23}\times 4\times (3.14{)}^2} \\ \mathrm{I}=6.43\times {10}^{-46}\mathrm{kg}\cdot {\mathrm{m}}^2 \end{gathered}$$

Thus ,the moment of inertia of NaF molecule is$6.43\times {10}^{-46}kg.m^2$

The calculation of separation between nuclei in NaF molecule

Using

$\begin{array}{r}\mathrm{I}=m{r}^2\\ r=\sqrt{\frac{I}{m}}\end{array}$

Now put the values of constants in above equation

$$\begin{gathered} r=\sqrt{\frac{6.43\times {10}^{-46}\mathrm{kg}\cdot {\mathrm{m}}^2}{1.73\times {10}^{-26}\mathrm{kg}}} \\ r=1.93\times {10}^{-10}\mathrm{m} \end{gathered}$$

Hence,

The equilibrium separation between the nuclei in a NaF molecule is 0.193nm.

The equilibrium separation between the nuclei in a NaCl from section 42.1 is 0.24nm .

Our result is small than that for NaCl molecule which is reasonable since the fluorine atom is smaller than the chlorine atom.