Concept explainers
(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-
Energy of potassium is calculated by-
And relative intensity formula-
Answer to Problem 9.12QAP
Ratio of excited state and ground state of potassium =
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-
Where,
Nj = number of ions in excited state
No = number of ions in ground state
Ej = energy difference of excited state and ground state
gj = statistical weight for excited state
go = statistical weight for ground state
Energy of potassium is calculated by −
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
Now, energy of potassium, put the values in equation (2):
From the diagram temperature at height 2 cm is 1700 ?
T= 1700oC+273 = 1973 K
The value of go and gj are 2 and 6 respectively for 3s and 3p state.
Therefore, the ratio is-
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
(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-
And relative intensity formula-
Answer to Problem 9.12QAP
Ratio of excited state and ground state of potassium=
Relative intensity of potassium at 3cm=2.07
Explanation of Solution
From the diagram temperature at 3cm height is 1863oC
Or,
Therefore, ratio is:
Relative intensity of potassium with emission line = 766.5nm at 3cm height.
Where relative intensity at 2 cm = 1.00
Relative intensity at 3 cm =2.07
(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-
And relative intensity formula-
Answer to Problem 9.12QAP
Ratio of excited state and ground state of potassium is
Relative intensity of potassium at 4cm= 1.72.
Explanation of Solution
From the diagram temperature at 4cm height is 1820oC
Or,
Therefore, ratio is-
Relative intensity of potassium with emission line = 766.5nm at 3cm height.
Where relative intensity at 2 cm = 1.00
Relative intensity at 4 cm =1.72
(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-
And relative intensity formula-
Answer to Problem 9.12QAP
Ratio of excited state and ground state of potassium=
Relative intensity of potassium at 5cm= 1.13.
Explanation of Solution
From the diagram temperature at 4cm height is 1725oC
Or,
Therefore, ratio is-
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
Relative intensity at 5 cm =1.13
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Chapter 9 Solutions
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