Chapter 9, Problem 9.12QAP

Interpretation Introduction

(a)

Interpretation:

Relative intensity of potassium at flame center and height 2 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

For this calculation following formulas will be used-

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

Energy of potassium is calculated by-

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

And relative intensity formula-

$\frac{I_X}{I_Y}=\frac{{\left(\frac{N_j}{N_o}\right)}_{h_2}}{{\left(\frac{N_j}{N_o}\right)}_{h_1}}$

Answer to Problem 9.12QAP

Ratio of excited state and ground state of potassium = $2.22\times {10}^{-4}$

Relative intensity of potassium at 2cm= 1.0

Explanation of Solution

Boltzmann equation will be used to calculate the ratio of excited to the ground state which is given as-

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

Where,

N_j = number of ions in excited state

N_o = number of ions in ground state

Ej = energy difference of excited state and ground state

g_j = statistical weight for excited state

g_o = statistical weight for ground state

Energy of potassium is calculated by −

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

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

Ej= energy difference

Wavelength of potassium is 766.5 nm when transition occur from 3p state to 3s state

$\begin{array}{l}h=\text{ 6}{\text{.62607×10}}^{\text{-34}}\text{J-s}\hfill \\ c={\text{ 3×10}}^{\text{8}}\text{m/s}\hfill \end{array}$

Now, energy of potassium, put the values in equation (2):

$E_j=\frac{\text{6}{\text{.62607×10}}^{\text{-34}}{\text{J-s×3×10}}^{\text{8}}\text{m/s}}{\text{766}{\text{.5×10}}^{\text{-9}}\text{m}}$

$E_j\text{for potassium atom}=2.59\times {10}^{-19}\text{J}$

From the diagram temperature at height 2 cm is 1700 ?

T= 1700^oC+273 = 1973 K

The value of g_o and g_j are 2 and 6 respectively for 3s and 3p state.

Therefore, the ratio is-

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{\text{-2}{\text{.59×10}}^{\text{-19}}\text{J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{J/K×1973K}}\right)$

$\frac{N_j}{N_o}\text{for potassium atom}=2.22\times {10}^{-4}$ At the height of 2 cm of flame.

It is assumed that the relative intensity of the potassium at 2 cm of height is 1.0.

So, relative intensity at 2 cm = 1.0

Interpretation Introduction

(b)

Interpretation:

Relative intensity of potassium at flame center and height 3 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

Calculation of ratio of excited state and ground state is done by following formula-

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

And relative intensity formula-

$\frac{I_X}{I_Y}=\frac{{\left(\frac{N_j}{N_o}\right)}_{h_2}}{{\left(\frac{N_j}{N_o}\right)}_{h_1}}$

Answer to Problem 9.12QAP

Ratio of excited state and ground state of potassium= $4.59\times {10}^{-4}$

Relative intensity of potassium at 3cm=2.07

Explanation of Solution

From the diagram temperature at 3cm height is 1863^oC

Or,

$T={1863}^{\circ }C+273=213\text{6 K}$

Therefore, ratio is:

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{\text{-2}{\text{.59×10}}^{\text{-19}}\text{J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{J/K×2136K}}\right)$

$\frac{N_j}{N_o}\text{for potassium atom}=4.59\times {10}^{-4}$ At the height of 3 cm of flame for 3p state to ground state.

Relative intensity of potassium with emission line = 766.5nm at 3cm height.

$\frac{I_X}{I_Y}=\frac{{\left(\frac{N_j}{N_o}\right)}_{3cm}}{{\left(\frac{N_j}{N_o}\right)}_{2cm}}$

Where relative intensity at 2 cm = 1.00

$\frac{N_j}{N_o}\text{at 2 cm}=2.22\times {10}^{-4}$

$\frac{N_j}{N_o}\text{at 3 cm}=4.59\times {10}^{-4}$

$\frac{I_X}{I_Y}=\frac{4.59\times {10}^{-4}}{2.22\times {10}^{-4}}$

Relative intensity at 3 cm =2.07

Interpretation Introduction

(c)

Interpretation:

Relative intensity of potassium at flame center and height 4 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

Ratio of excited state and ground state is calculated by the following formula-

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

And relative intensity formula-

$\frac{I_X}{I_Y}=\frac{{\left(\frac{N_j}{N_o}\right)}_{h_2}}{{\left(\frac{N_j}{N_o}\right)}_{h_1}}$

Answer to Problem 9.12QAP

Ratio of excited state and ground state of potassium is $3.83\times {10}^{-4}$.

Relative intensity of potassium at 4cm= 1.72.

Explanation of Solution

From the diagram temperature at 4cm height is 1820^oC

Or, $T=1820℃+273=209\text{3 K}$

Therefore, ratio is-

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{\text{-2}{\text{.59×10}}^{\text{-19}}\text{J}}{\text{1}{\text{.38×10}}^{\text{-23}}\text{J/K×2093K}}\right)$

$\frac{N_j}{N_o}\text{for potassium atom}=3.83\times {10}^{-4}$ At the height of 4 cm of flame for 3p state to ground state.

Relative intensity of potassium with emission line = 766.5nm at 3cm height.

$\frac{I_X}{I_Y}=\frac{{\left(\frac{N_j}{N_o}\right)}_{4cm}}{{\left(\frac{N_j}{N_o}\right)}_{2cm}}$

Where relative intensity at 2 cm = 1.00

$\frac{N_j}{N_o}\text{at 2 cm}=2.22\times {10}^{-4}$

$\frac{N_j}{N_o}\text{at 4 cm}=3.83\times {10}^{-4}$

$\frac{I_X}{I_Y}=\frac{3.83\times {10}^{-4}}{2.22\times {10}^{-4}}$

Relative intensity at 4 cm =1.72

Interpretation Introduction

(d)

Interpretation:

Relative intensity of potassium at flame center and height 5 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

Ratio of excited state and ground state is calculated by the following formula-

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

And relative intensity formula-

$\frac{I_X}{I_Y}\frac{{\left(\frac{N_j}{N_o}\right)}_{h_2}}{{\left(\frac{N_j}{N_o}\right)}_{h_1}}$

Answer to Problem 9.12QAP

Ratio of excited state and ground state of potassium= $2.50\times {10}^{-4}$.

Relative intensity of potassium at 5cm= 1.13.

Explanation of Solution

From the diagram temperature at 4cm height is 1725^oC

Or,

$T=1725℃+273=19\text{98 K}$

Therefore, ratio is-

$\frac{N_j}{N_o}=\frac{6}{2}\exp \left(\frac{-2.59\times {10}^{-19}J}{1.38\times {10}^{-23}J/K\times 1998K}\right)$

$\frac{N_j}{N_o}\text{for potassium atom}=2.50\times {10}^{-4}$

At height of 5 cm of flame for 3p state to ground state.

Relative intensity of potassium with emission line = 766.5nm at 3cm height.

$\frac{I_X}{I_Y}=\frac{{\left(\frac{N_j}{N_o}\right)}_{5cm}}{{\left(\frac{N_j}{N_o}\right)}_{2cm}}$

Where relative intensity at 2 cm = 1.00

$\frac{N_j}{N_o}\text{at 2 cm}=2.22\times {10}^{-4}$

$\frac{N_j}{N_o}\text{at 5 cm}=2.50\times {10}^{-4}$

$\frac{I_X}{I_Y}=\frac{2.50\times {10}^{-4}}{2.22\times {10}^{-4}}$

Relative intensity at 5 cm =1.13