PRINCIPLES OF INSTRUMENTAL ANALYSIS
PRINCIPLES OF INSTRUMENTAL ANALYSIS
7th Edition
ISBN: 9789353506193
Author: Skoog
Publisher: CENGAGE L
Question
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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-

NjNo=gjgoexp(EjkT) ......... (1)

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

Ej=hcλ ......... (2)

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

Ej= energy difference of excited state and ground state.

Expert Solution
Check Mark

Answer to Problem 9.14QAP

Ratio of particle present in an atom in 3p states and in ground state of Na at 2100K = 2.7×105

Ratio of particle present in an atom in 3p states and in ground state of Mg+ at 2100K= 6.9×1011

Explanation of Solution

Calculation will be done using the following formulas-

NjNo=gjgoexp(EjkT)

And

Ej=hcλ

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)

Ej=6.62607×10-34J-s×3×108m/s589.3×10-9m

EjforNa atom=3.37×1019J    ......(3)

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

EjforNa atom=3.37×1019J

gjgo=62, 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)

NjNo=62exp(-3.37×10-19 J1.38×10-23 J/K×2100 K)

NjNo=2.7×105 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

Ej=6.62607×1034Js×3×108m/s280.0×109m

EjforMg+atom=7.09×1019J    ......(4)

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

Ejfor Mg+atom=7.09×1019J

gjgo=62, statistical weight

T = 2100K

k= 1.38×10-23J/K

Put the above values in above equation-

NjNo=62exp(-7.09×10-19 J1.38×10-23 J/K×2100 K)

NjNo=6.9×1011 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-

NjNo=gjgoexp(EjkT) ......... (1)

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

Ej=hcλ ......... (2)

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

Ej= energy difference of excited state and ground state

Expert Solution
Check Mark

Answer to Problem 9.14QAP

Ratio of particle present in an atom in 3p states and in ground state of Na at 2900 K = 6.6×104

Ratio of particle present in an atom in 3p states and in ground state of Mg+ at 2900 K= 5.98×108

Explanation of Solution

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

EjforNa atom=3.37×1019J

gjgo=62 =3

T = 2900K

k= 1.38×10-23J/K

Put the above values in equ (1)

NjNo=62exp(-3.37×10-19 J1.38×10-23 J/K×2900 K)

NjNo=6.6×104

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

Ejfor Mg+ atom=7.09×1019J

gjgo=62 =3

T = 2900K

k= 1.38×10-23J/K

Put the above values in above equation-

NjNo=62exp(-7.09×10-19 J1.38×10-23 J/K×2900 K)

NjNo=5.9×108 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-

NjNo=gjgoexp(EjkT) ......... (1)

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

Ej=hcλ ......... (2)

Where,

h= Planck’s constant

c = light velocity

λ= wavelength

Ej= energy difference of excited state and ground state

Expert Solution
Check Mark

Answer to Problem 9.14QAP

Ratio of particle present in an atom in 3p states and in ground state of Na at 6000K = 5.1×102

Ratio of particle present in an atom in 3p states and in ground state of Mg+ at 6000K= 5.7×104

Explanation of Solution

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

EjforNa atom=3.37×1019J

gjgo=62

T = 7250K

k= 1.38×10-23J/K

Put the above values in equation (1)

NjNo=62exp(-3.37×10-19J1.38×10-23J/K×6000K)

NjNo=5.1×102

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

Ejfor Mg+ atom=7.09×1019J

gjgo=62, statistical weight

T = 6000K

k= 1.38×10-23J/K

Put the above values in above equation-

NjNo=62exp(-7.09×10-19J1.38×10-23J/K×6000K)

NjNofor Mg+ ion=5.7×104

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