Question

For the flame shown in Figure 9-3, calculate the relative intensity of the 766.5-nm emission line for potassium in the flame center and at the following heights above the orifice (assume no ionization and use

the 2.0 cm for comparison).

(a) 2.0 cm (b) 3.0 cm (c) 4.0 cm (d) 5.0 cm

Short answer

(a) Ratio of excited state and ground state of potassium = 2.2 * 10 ^-4

Relative intensity of potassium at 2 cm= 1.0

(b) Ratio of excited state and ground state of potassium = 4.59 * 10^-4

Relative intensity of potassium at 3 cm=2.07

(c) Ratio of excited state and ground state of potassium =3.83 * 10^-4.

Relative intensity of potassium at 4 cm= 1.72.

(d) Ratio of excited state and ground state of potassium= 2.50* 10^-4.

Relative intensity of potassium at 5 cm= 1.13.

Step-by-step solution

Part (a) Step 1: Given Information

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.

Part (a) Step 2: Explanation

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

E_j = 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 –

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

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

E_j= energy difference

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

h= 6.62607 * 1034 J-s

c= 3 * 108 m/s

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

$E_j=\frac{6.62607\times {10}^{-34}J-s\times 3\times {10}^{8}m/s}{766.5\times {10}^{-9}m}$

E_j for potassium atom = 2.59 * 10^-19J.

From the diagram temperature at height 2 cm is 1700 ℃

T= 1700℃+273 = 1973 K

The value of go and gj are 2 and 6 respectively for 3s and 3p state.

Therefore, the ratio is-

$\frac{N_j}{N_o}=\frac{6}{2}exp\left(\frac{-2.59\times {10}^{-19}J}{1.38\times {10}^{-23}J/K\times 1973K}\right)$

$\frac{N_j}{N_o}$for potassium atom= 2.22 * 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

Part (b) Step 1: Given Information

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.

Part (b) Step 2: Explanation

From the diagram temperature at 3cm height is 1863℃

Or,

T = 1863 ^oC +273 = 2136 K.

Therefore, ratio is:

$\frac{N_j}{N_o}=\frac{6}{2}exp\left(\frac{-2.59\times {10}^{-19}J}{1.38\times {10}^{-23}J/K\times 2136K}\right)$

$\frac{N_j}{N_o}$for potassium atom = 4.59 * 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.

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

Where relative intensity at 2 cm = 1.00

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

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

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

Relative intensity at 3 cm =2.07

Part (c) Step 1: Given Information

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.

Part (c) Step 2: Explanation

From the diagram temperature at 4cm height is 1820℃

Or, T= 1820℃ +273 = 2093 K

Therefore, ratio is-

$\frac{N_j}{N_o}=\frac{6}{2}exp\left(\frac{-2.59\times {10}^{-19}J}{1.38\times {10}^{-23}J/K\times 2093K}\right)$

$\frac{N_j}{N_o}$ for potassium atom = 3.83 * 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.

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

Where relative intensity at 2 cm = 1.00

$$\begin{gathered} \frac{N_j}{N_o}\mathrm{at}2\mathrm{cm}=2.22\times {10}^{-4} \\ \frac{N_j}{N_o}\mathrm{at}4\mathrm{cm}=3.83\times {10}^{-4} \end{gathered}$$

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

Relative intensity at 4 cm =1.72

Part (d) Step 1: Given Information

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.

Part (d) Step 2: Explanation

From the diagram temperature at 4cm height is 1725℃

Or,

T= 1725℃ +273 = 1998 K

Therefore, ratio is-

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

$\frac{N_j}{N_o}$ for potassium atom = 2.50 * 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.

Where relative intensity at 2 cm = 1.00

$$\begin{gathered} \frac{N_j}{N_o}\mathrm{at}2\mathrm{cm}=2.22\times {10}^{-4} \\ \frac{N_j}{N_o}\mathrm{at}5\mathrm{cm}=2.50\times {10}^{-4} \end{gathered}$$

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

Relative intensity at 5 cm =1.13