Question

For $\mathrm{Na}$atoms and ${\mathrm{Mg}}^{+}$ions, compare the ratios of the number of particles in the 3p excited state to the number in the ground state in (a) a natural gas–air flame (2100 K). (b) a hydrogen-oxygen flame (2900 K). (c) an inductively coupled plasma source (6000 K).

Short answer

(a)

Ratio of particle present in an atom in 3p states and in ground state of Na at 2100K = $2.7\times {10}^{-5}$

Ratio of particle present in an atom in 3p states and in ground state of role="math" localid="1645609983184" ${\text{Mg}}^{+}$at 2100K=$6.9\times {10}^{-11}$

(b)

Ratio of particle present in an atom in 3p states and in ground state of Na at 2900K =$6.6\times {10}^{-4}$

Ratio of particle present in an atom in 3p states and in ground state of data-custom-editor="chemistry" ${\mathrm{Mg}}^{+}$at 2100K=$5.98\times {10}^{-8}$

(c)Ratio of particle present in an atom in 3p states and in ground state of Na at 6000K=$5.1\times {10}^{-2}$

Ratio of particle present in an atom in 3p states and in ground state of data-custom-editor="chemistry" ${\mathrm{Mg}}^{+}$at 6000K=$5.7\times {10}^{-4}$

Step-by-step solution

Part(a) Step1. Given information

Compare the ratios of the number of particles in the 3p excited state to the number in the ground state for sodium and ${\text{Mg}}^{+}$in natural gas air flame(2100K)

Part(a) Step2. Boltzmann equation 

For the calculation of the ratio, the Boltzmann equation is used. The equation is as follows:-

$\frac{N_j}{N_a}=\frac{g_j}{g_0}\text{exp}\left[\frac{-E_j}{kT}\right]$

Here, $N_j$and $N_0$are the number of atoms in an excited state and the ground state, respectively, $k$is Boltzmann’s constant, T is the absolute temperature,$\frac{g_j}{g_0}$is the ratio of statistical weight for excited state to the statistical weight for ground state and $E_j$is the energy difference between the excited state and the ground state.

Part(a) Step3. The Formula for the calculation of energy

Calculation of energy is done by the following formula:-

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

Here, h is the Planck's constant, $c$is the light velocity, $\lambda$is the wavelength.

Part(a) Step4.Calculation of energy for sodium atom  by using the above formula given in step 3.

$$\begin{gathered} E_j=\frac{6.6\times {10}^{-34}\mathrm{Js}\times 3.0\times {10}^8\mathrm{m}/\mathrm{s}}{589.3{\text{nm×10}}^{-9}{\displaystyle \frac{\mathrm{m}}{\mathrm{nm}}}} \\ =3.37\times {10}^{-19}\mathrm{J} \end{gathered}$$

Part(a).Step5.Calculate the ratio of number of particles in the 3p excited state to ground state for sodium atom by using the equation given in step2.

The statistical weights for the 3s and 3p quantum states are 2 and 6 of a sodium atom.

Substitute all the values in the equation given in step2 as follows:-

$$\begin{gathered} \frac{N_J}{N_0}=\frac{6}{2}\exp \left(\frac{-3.37\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 2100\mathrm{K}}\right) \\ =2.7\times {10}^{-5} \end{gathered}$$

Part(a) Step6. Calculation of energy for Mg+  by using the above formula given in step 3.

$$\begin{gathered} E_j=\frac{6.6\times {10}^{-34}\mathrm{Js}\times 3\times {10}^8\mathrm{m}/\mathrm{s}}{280.0\times {10}^{-9}\mathrm{m}} \\ =7.09\times {10}^{-19}\text{J} \end{gathered}$$

Part(a) Step7. Calculate the ratio of number of particles in the 3p excited state to ground state for Mg+ by using the equation given in step2. 

$$\begin{gathered} \frac{N_j}{N_0}=\frac{6}{2}\exp \left(\frac{-7.09\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 2100\mathrm{K}}\right) \\ =6.9\times {10}^{-11} \end{gathered}$$

Part(b) Step1. Given information

Compare the ratios of the number of particles in the 3p excited state to the number in the ground state for sodium and ${\text{Mg}}^{+}$in a hydrogen- oxygen flame (2900 K)

Part(b) Step2.Calculate the ratio of number of particles in the 3p excited state to ground state for Sodium by using the equation given in step2. 

$$\begin{gathered} \frac{N_j}{N_0}=\frac{6}{2}\text{exp}\left(\frac{-3.37\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 2900\mathrm{K}}\right) \\ {\text{=6.6×10}}^{-4} \end{gathered}$$

Part(b) Step3.Calculate the ratio of number of particles in the 3p excited state to ground state for Mg+ by using the equation given in step2. 

$$\begin{gathered} \frac{N_j}{N_0}=\frac{6}{2}\text{exp}\left(\frac{-7.09\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 2900\mathrm{K}}\right) \\ {\text{=5.9×10}}^{-8} \end{gathered}$$

Part(C) Step1.Given information

Compare the ratios of the number of particles in the 3p excited state to the number in the ground state for sodium and ${\text{Mg}}^{+}$in a hydrogen- oxygen flame (6000 K)

Part(C) Step12. Calculate the ratio of number of particles in the 3p excited state to ground state for sodium by using the equation given in step2. 

$$\begin{gathered} \frac{N_j}{N_0}=\frac{6}{2}\text{exp}\left(\frac{-3.37\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 6000\mathrm{K}}\right) \\ {\text{=5.1×10}}^{-2} \end{gathered}$$

Part(C) Step13. Calculate the ratio of number of particles in the 3p excited state to ground state for Mg+ by using the equation given in step2. 

$$\begin{gathered} \frac{N_j}{N_0}=\frac{6}{2}\text{exp}\left(\frac{-7.09\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 6000\mathrm{K}}\right) \\ {\text{=5.7×10}}^{-4} \end{gathered}$$