Question

In higher-temperature sources, sodium atoms emit a doublet with an average wavelength of 1139 nm. The transition responsible is from the 4s to 3p state. Calculate the ratio of the number of excited atoms in the 4s level to the number in the ground 3s level in (a) an acetylene-oxygen flame (3000°C). (b) the hottest part of an inductively coupled plasma source (,9000°C).

Short answer

(a) Ratio of excited state 4s to ground state 3s for acetylene- oxygen flame of 3000 ℃ is $1.2\times {10}^{-5}$.

(b)Ratio of excited state 4s to ground state 3s for an inductively coupled plasma source of 9000℃ is$1.8\times {10}^{-2}$.

Step-by-step solution

Part(a) Step1. Given information

an acetylene-oxygen flame (3000°C)

average wavelength = 1139 nm

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, and are the number of atoms in an excited state and the ground state, respectively, is Boltzmann’s constant, T is the absolute temperature,is the ratio of statistical weight for excited state to the statistical weight for ground state and 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:-

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

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.

$Energy3sto4s\left(E_j\right)=energy3sto3p\left(E_{j1}\right)+energy3pto4s\left(E_{j2}\right)$

Therefore, from3s to 3pquantum state at this wavelength of sodium is 589.3 nm

role="math" localid="1645686205883" $$\begin{gathered} E_{j1}=\frac{6.6\times {10}^{-34}\mathrm{Js}\times 3.0\times {10}^8\mathrm{m}/\mathrm{s}}{589.3{\text{nm×10}}^{-9}{\displaystyle \frac{\mathrm{m}}{\mathrm{nm}}}} \\ =3.37\times {10}^{-19}\mathrm{J} \end{gathered}$$

Now from 3p to 4s quantum state at this wavelength of sodium is 1139 nm.

$$\begin{gathered} E_{j1}=\frac{6.6\times {10}^{-34}\mathrm{Js}\times 3.0\times {10}^8\mathrm{m}/\mathrm{s}}{1139{\text{nm×10}}^{-9}{\displaystyle \frac{\mathrm{m}}{\mathrm{nm}}}} \\ =1.75\times {10}^{-19}\mathrm{J} \end{gathered}$$

$$\begin{gathered} E_j=3.37\times {10}^{-19}\text{J+1.75}\times {10}^{-19}\text{J} \\ =5.12\times {10}^{-19}J \end{gathered}$$

Part(a).Step5.Calculate the ratio of excited atom to the unexcited atom at the temperature 3000°C.

The statistical weights for the 3s and 4s quantum states are 2 and 2 of a sodium atom.

$$\begin{gathered} T\left(\text{K}\right)={3000}^{°}\text{C+273} \\ \text{=3273K} \end{gathered}$$

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

$$\begin{gathered} \frac{N_J}{N_0}=\frac{2}{2}\exp \left(\frac{-5.12\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 3273\mathrm{K}}\right) \\ =1.2\times {10}^{-5} \end{gathered}$$

Part(b) Step1. Given information

Calculate the ratio of the number of excited atoms in the 4s level to the number in the ground 3s level in the hottest part of an inductively coupled plasma source (,9000°C).

Average wavelength of 1139 nm

Part(b) Step2.Calculate the ratio of excited atoms in the 4s level to the number in the ground 3s level in the hottest part of an inductively coupled plasma source (,9000°C).

$$\begin{gathered} T\left(\text{K}\right)={9000}^{°}\text{C+273} \\ \text{=9273K} \end{gathered}$$

By using the equation given in step2:-

$$\begin{gathered} \frac{N_j}{N_0}=\frac{2}{2}\text{exp}\left(\frac{-5.12\times {10}^{-19}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times 9273\mathrm{K}}\right) \\ {\text{=1.8×10}}^{-2} \end{gathered}$$