Chapter 9, Problem 9.14QAP

Interpretation Introduction

(a)

Interpretation:

The ratios of particles present in an atoms or ions in 3p states and in ground state of Na atom and Mg⁺ needs to be compared when there is a natural gas air flame of temperature 2100 K.

Concept introduction:

Boltzmann equation is used for the calculation of the ratio. This equation tells that how much an atom or ion is populated as a function of temperature. This equation is given as-

$\frac{N_j}{N_o}=\frac{g_j}{g_o}\exp \left(\frac{-E_j}{kT}\right)$ ......... (1)

And the calculation of energy of atom and ion is done by the following formula-

$E_j=\frac{hc}{\lambda }$ ......... (2)

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

Ej= energy difference of excited state and ground state.

Answer to Problem 9.14QAP

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 Mg⁺ at 2100K= $6.9\times {10}^{-11}$

Explanation of Solution

Calculation will be done using the following formulas-

$\frac{N_j}{N_o}=\frac{g_j}{g_o}\exp \left(\frac{-E_j}{kT}\right)$

And

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

Energy difference between 3p excited state and ground state for Na atom-

Average Wavelength for the Na atom when transition occurs from 3p state to 3s state is 589.3 nm

h = 6.62607×10^-34J-s

c = 3×10⁸m/s

put above values in equation (2)

$E_j=\frac{\text{6}{\text{.62607×10}}^{\text{-34}}{\text{J-s×3×10}}^{\text{8}}\text{m/s}}{\text{589}{\text{.3×10}}^{\text{-9}}\text{m}}$

$E_j\text{forNa atom}=3.37\times {10}^{-19}\text{J}$    ......(3)

Ratio of 3p state to ground state of Na atom at natural gas flame, 2100K-

$E_j\text{forNa atom}=3.37\times {10}^{-19}\text{J}$

$\frac{g_j}{g_o}=\frac{6}{2}$, statistical weight of 3p state is 6 and for 3s state is 2.

T = 2100K

k= 1.38×10^-23 J/K

Put the above values in equation (1)

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{\text{-3}{\text{.37×10}}^{\text{-19}}\text{ J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{ J/K×2100 K}}\right)$

$\frac{N_j}{N_o}=2.7\times {10}^{-5}$ for Na atom

So, the energy difference between 3p and ground state for Mg⁺ ion-

Average Wavelength for the Mg⁺ atom when transition of ion occurs from 3p state to 3s state is 280.0 nm

$E_j=\frac{6.62607\times {10}^{-34}J-s\times 3\times {10}^{8}m/s}{280.0\times {10}^{-9}m}$

$E_{j}forM{g}^{+}atom=7.09\times {10}^{-19}J$    ......(4)

Ratio of 3p state to ground state of Mg⁺ ion at natural gas flame, 2100K-

$E_j{\text{for Mg}}^{\text{+}}\text{atom}=7.09\times {10}^{-19}\text{J}$

$\frac{g_j}{g_o}=\frac{6}{2}$, statistical weight

T = 2100K

k= 1.38×10^-23J/K

Put the above values in above equation-

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{\text{-7}{\text{.09×10}}^{\text{-19}}\text{ J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{ J/K×2100 K}}\right)$

$\frac{N_j}{N_o}=6.9\times {10}^{-11}$ for Mg⁺ ion.

Interpretation Introduction

(b)

Interpretation:

The ratios of particles present in an atoms or ions in 3p states and in ground state of Na atom and Mg⁺ needs to be compared when there is a hydrogen-oxygen flame of temperature 2900 K.

Concept introduction:

Boltzmann equation is used for the calculation of the ratio. This equation tells that how much an atom or ion is populated as a function of temperature. This equation is given as-

$\frac{N_j}{N_o}=\frac{g_j}{g_o}\exp \left(\frac{-E_j}{kT}\right)$ ......... (1)

And the calculation of energy of atom and ion is done by the following formula-

$E_j=\frac{hc}{\lambda }$ ......... (2)

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

Ej= energy difference of excited state and ground state

Answer to Problem 9.14QAP

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

Ratio of particle present in an atom in 3p states and in ground state of Mg⁺ at 2900 K= $5.98\times {10}^{-8}$

Explanation of Solution

Ratio of 3p state to ground state of Na atom at hydrogen-oxygen flame, 2900K-

$E_j\text{forNa atom}=3.37\times {10}^{-19}\text{J}$

$\frac{g_j}{g_o}=\frac{6}{2}$ =3

T = 2900K

k= 1.38×10^-23J/K

Put the above values in equ (1)

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{\text{-3}{\text{.37×10}}^{\text{-19}}\text{ J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{ J/K×2900 K}}\right)$

$\frac{N_j}{N_o}=6.6\times {10}^{-4}$

Ratio of 3p state to ground state of Mg⁺ ion at hydrogen- oxygen flame, 2900K-

$E_j{\text{for Mg}}^{\text{+}}\text{ atom}=7.09\times {10}^{-19}\text{J}$

$\frac{g_j}{g_o}=\frac{6}{2}$ =3

T = 2900K

k= 1.38×10^-23J/K

Put the above values in above equation-

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{\text{-7}{\text{.09×10}}^{\text{-19}}\text{ J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{ J/K×2900 K}}\right)$

$\frac{N_j}{N_o}=5.9\times {10}^{-8}$ for Mg⁺ ion

Interpretation Introduction

(c)

Interpretation:

The ratios of particles present in an atoms or ions in 3p states and in ground state of Na atom and Mg⁺ needs to be compared when there is an inductively coupled plasma source of 6000 K.

Concept introduction:

Boltzmann equation is used for the calculation of the ratio. This equation tells that how much an atom or ion is populated as a function of temperature. This equation is given as-

$\frac{N_j}{N_o}=\frac{g_j}{g_o}\exp \left(\frac{-E_j}{kT}\right)$ ......... (1)

And the calculation of energy of atom and ion is done by the following formula-

$E_j=\frac{hc}{\lambda }$ ......... (2)

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

Ej= energy difference of excited state and ground state

Answer to Problem 9.14QAP

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 Mg⁺ at 6000K= $5.7\times {10}^{-4}$

Explanation of Solution

Ratio of 3p state to ground state of Na atom at an inductively coupled plasma source, 6000K-

$E_j\text{forNa atom}=3.37\times {10}^{-19}\text{J}$

$\frac{g_j}{g_o}=\frac{6}{2}$

T = 7250K

k= 1.38×10^-23J/K

Put the above values in equation (1)

$\frac{N_j}{N_o}=\frac{6}{2}\mathrm{ex}\text{p}\left(\frac{\text{-3}{\text{.37×10}}^{\text{-19}}\text{J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{J/K×6000K}}\right)$

$\frac{N_j}{N_o}=5.1\times {10}^{-2}$

Ratio of 3p state to ground state of Mg⁺ ion at an inductively coupled plasma source, 6000K-

$E_j{\text{for Mg}}^{\text{+}}\text{ atom}=7.09\times {10}^{-19}\text{J}$

$\frac{g_j}{g_o}=\frac{6}{2}$, statistical weight

T = 6000K

k= 1.38×10^-23J/K

Put the above values in above equation-

$\frac{N_j}{N_o}=\frac{6}{2}\mathrm{ex}\mathrm{p}\left(\frac{-7.09\times 10^{-19}J}{1.38\times 10^{-23}J/K\times 6000K}\right)$

$\frac{N_j}{N_o}{\text{for Mg}}^{\text{+}}\text{ ion}=5.7\times {10}^{-4}$