Question

For Na atoms and Mg⁺ ions, compare the ratios of the number of atoms or ions in the 3p excited state to the number in the ground state in

(a) a natural gas–air flame (1800 K).

(b) a hydrogen-oxygen flame (2950 K).

(c) an inductively coupled plasma source (7250 K).

Short answer

$$\begin{gathered} a)3.85\times {10}^{-6}\to \text{ForNaand}0.16\times {10}^{-12}\to {\text{ForMg}}^{+} \\ b)7.6\times {10}^{-4}\to \text{ForNaand}8.0\times {10}^{-8}\to {\text{ForMg}}^{+} \\ c)0.10\to \text{ForNaand}2.5\times {10}^{-3}\to {\text{ForMg}}^{+} \end{gathered}$$

Step-by-step solution

Given Information

(a) a natural gas–air flame (1800 K).

(b) a hydrogen-oxygen flame (2950 K).

(c) an inductively coupled plasma source (7250 K).

Explanation:

The energies for 3p states can be obtained from the emission wavelengths. For sodium we will use an average wavelength of 5893 $\stackrel{0}{A}$and for Mg⁺ , 2800 $\stackrel{0}{A}$is used.

1. The energy for excited state in case of sodium (Na):

localid="1645801209193" $$\begin{gathered} E_{y1}=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\text{Js}\times 3.00\times {10}^8\text{m/s}}{5893\times {10}^{-10}\text{m}} \\ {E}_{y1}=3.37\times {10}^{-19}\text{J} \end{gathered}$$

2. For Mg+ the energy of the excited state:

localid="1645801258685" $$\begin{gathered} E_{y2}=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\text{Js}\times 3.00\times {10}^8\text{m/s}}{2800\times {10}^{-10}\text{m}} \\ {E}_{y2}=7.09\times {10}^{-19}\text{J} \end{gathered}$$

From Boltzmann equation

The magnitude of the effect from the Boltzmann equation, which takes the form:

$\frac{N_j}{N_0}=\frac{g_j}{g_0}exp\left(\frac{-E_j}{KT}\right)$

where, Nj and N0 are the number of atoms in an excited state and the ground state, respectively, k is Boltzmann’s constant/K), T is the absolute temperature, and Ej is the energy difference between the excited state and the ground state.

Part (a) Step 4: Calculation

For 1800 K,

$\frac{N_j}{N_0}=3exp\left(\frac{-3.37\times {10}^{-19}\text{J}}{1.38\times {10}^{-23}\text{J/K}\times 1800\text{K}}\right)=3.85\times {10}^{-6}\to \text{ForNa}$

$\frac{N_j}{N_0}=3exp\left(\frac{-7.09\times {10}^{-19}J}{1.38\times {10}^{-23}\text{J/K}\times 1800\text{K}}\right)=1.16\times {10}^{-12}\to {\text{ForMg}}^{+}$

Part (b) Step 5: Calculation 

Similarly, a hydrogen-oxygen flame (2950 K).

$$\begin{gathered} \frac{N_j}{N_0}=3exp\left(\frac{-3.37\times {10}^{-19}\text{J}}{1.38\times {10}^{-23}\text{J/K}\times 2950\text{K}}\right)=7.6\times {10}^{-4}\to \text{ForNa} \\ \frac{N_j}{N_0}=3exp\left(\frac{-7.09\times {10}^{-19}\text{J}}{1.38\times {10}^{-23}\text{J/K}\times 2950\text{K}}\right)=8.2\times {10}^{-8}\to {\text{ForMg}}^{+} \end{gathered}$$

Part (c) Step 6: Calculation

Similarly, an inductively coupled plasma source (7250 K).

$$\begin{gathered} \frac{N_j}{N_0}=3exp\left(\frac{-3.37\times {10}^{-19}\text{J}}{1.38\times {10}^{-23}\text{J/K}\times 7250\text{K}}\right)=0.10\to \text{ForNa} \\ \frac{N_j}{N_0}=3exp\left(\frac{-7.09\times {10}^{-19}\text{J}}{1.38\times {10}^{-23}\text{J/K}\times 7250\text{K}}\right)=8.2\times {10}^{-8}\to {\text{ForMg}}^{+} \end{gathered}$$