The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

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Answer 1

Question

Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Expert Solution & Answer

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Sodium,

E = 2.1 eV × (1.602 × 10-19 J/eV) = 3.3642 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)3.3642 × 10−19 J = 19.8643 × 10−26 m3.3642 × 10−19 = 591 nm

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Copper,

E = 3.78 eV × (1.602 × 10-19 J/eV) = 6.05556 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)6.05556 × 10-19 J = 19.8643 × 10−26 m6.05556 × 10-19 = 328 nm

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Bromine,

E = 8.04 eV × (1.602 × 10-19 J/eV) = 12.8800 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)12.8800 × 10-19 J = 19.8643 × 10−26 m12.8800 × 10-19 = 154 nm

To calculate N*/N0 at 2600 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 2.6 × 10−4

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.4 × 10−7

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.8 × 10−16

To calculate N*/N0 at 6000 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 5.2 × 10−2

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 2.0 × 10−3

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 1.2 × 10−7

The completed table can be given as,

Element	Na	Cu	Br
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( g*/g0 )	3	3	2/3
N*/N0 at 2600 K in flame	2.6 × 10−4	1.4 × 10−7	1.8 × 10−16
N*/N0 at 6000 K in flame	5.2 × 10−2	2.0 × 10−3	1.2 × 10−7

Bromine is not readily in absorption spectroscopy because the lowest excited state itself requires far-UV radiation for excitation. The excited state lies at such high energy which is not sufficiently populated to obtain the intensity for optical emission.

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Answer 2

Question

Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Expert Solution & Answer

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Sodium,

E = 2.1 eV × (1.602 × 10-19 J/eV) = 3.3642 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)3.3642 × 10−19 J = 19.8643 × 10−26 m3.3642 × 10−19 = 591 nm

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Copper,

E = 3.78 eV × (1.602 × 10-19 J/eV) = 6.05556 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)6.05556 × 10-19 J = 19.8643 × 10−26 m6.05556 × 10-19 = 328 nm

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Bromine,

E = 8.04 eV × (1.602 × 10-19 J/eV) = 12.8800 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)12.8800 × 10-19 J = 19.8643 × 10−26 m12.8800 × 10-19 = 154 nm

To calculate N*/N0 at 2600 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 2.6 × 10−4

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.4 × 10−7

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.8 × 10−16

To calculate N*/N0 at 6000 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 5.2 × 10−2

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 2.0 × 10−3

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 1.2 × 10−7

The completed table can be given as,

Element	Na	Cu	Br
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( g*/g0 )	3	3	2/3
N*/N0 at 2600 K in flame	2.6 × 10−4	1.4 × 10−7	1.8 × 10−16
N*/N0 at 6000 K in flame	5.2 × 10−2	2.0 × 10−3	1.2 × 10−7

Bromine is not readily in absorption spectroscopy because the lowest excited state itself requires far-UV radiation for excitation. The excited state lies at such high energy which is not sufficiently populated to obtain the intensity for optical emission.

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Answer 3

Question

Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Expert Solution & Answer

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Sodium,

E = 2.1 eV × (1.602 × 10-19 J/eV) = 3.3642 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)3.3642 × 10−19 J = 19.8643 × 10−26 m3.3642 × 10−19 = 591 nm

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Copper,

E = 3.78 eV × (1.602 × 10-19 J/eV) = 6.05556 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)6.05556 × 10-19 J = 19.8643 × 10−26 m6.05556 × 10-19 = 328 nm

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Bromine,

E = 8.04 eV × (1.602 × 10-19 J/eV) = 12.8800 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)12.8800 × 10-19 J = 19.8643 × 10−26 m12.8800 × 10-19 = 154 nm

To calculate N*/N0 at 2600 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 2.6 × 10−4

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.4 × 10−7

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.8 × 10−16

To calculate N*/N0 at 6000 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 5.2 × 10−2

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 2.0 × 10−3

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 1.2 × 10−7

The completed table can be given as,

Element	Na	Cu	Br
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( g*/g0 )	3	3	2/3
N*/N0 at 2600 K in flame	2.6 × 10−4	1.4 × 10−7	1.8 × 10−16
N*/N0 at 6000 K in flame	5.2 × 10−2	2.0 × 10−3	1.2 × 10−7

Bromine is not readily in absorption spectroscopy because the lowest excited state itself requires far-UV radiation for excitation. The excited state lies at such high energy which is not sufficiently populated to obtain the intensity for optical emission.

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Answer 4

Question

Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Expert Solution & Answer

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Sodium,

E = 2.1 eV × (1.602 × 10-19 J/eV) = 3.3642 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)3.3642 × 10−19 J = 19.8643 × 10−26 m3.3642 × 10−19 = 591 nm

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Copper,

E = 3.78 eV × (1.602 × 10-19 J/eV) = 6.05556 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)6.05556 × 10-19 J = 19.8643 × 10−26 m6.05556 × 10-19 = 328 nm

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Bromine,

E = 8.04 eV × (1.602 × 10-19 J/eV) = 12.8800 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)12.8800 × 10-19 J = 19.8643 × 10−26 m12.8800 × 10-19 = 154 nm

To calculate N*/N0 at 2600 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 2.6 × 10−4

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.4 × 10−7

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.8 × 10−16

To calculate N*/N0 at 6000 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 5.2 × 10−2

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 2.0 × 10−3

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 1.2 × 10−7

The completed table can be given as,

Element	Na	Cu	Br
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( g*/g0 )	3	3	2/3
N*/N0 at 2600 K in flame	2.6 × 10−4	1.4 × 10−7	1.8 × 10−16
N*/N0 at 6000 K in flame	5.2 × 10−2	2.0 × 10−3	1.2 × 10−7

Bromine is not readily in absorption spectroscopy because the lowest excited state itself requires far-UV radiation for excitation. The excited state lies at such high energy which is not sufficiently populated to obtain the intensity for optical emission.

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Answer 5

Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Expert Solution & Answer

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Sodium,

E = 2.1 eV × (1.602 × 10-19 J/eV) = 3.3642 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)3.3642 × 10−19 J = 19.8643 × 10−26 m3.3642 × 10−19 = 591 nm

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Copper,

E = 3.78 eV × (1.602 × 10-19 J/eV) = 6.05556 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)6.05556 × 10-19 J = 19.8643 × 10−26 m6.05556 × 10-19 = 328 nm

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Bromine,

E = 8.04 eV × (1.602 × 10-19 J/eV) = 12.8800 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)12.8800 × 10-19 J = 19.8643 × 10−26 m12.8800 × 10-19 = 154 nm

To calculate N*/N0 at 2600 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 2.6 × 10−4

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.4 × 10−7

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.8 × 10−16

To calculate N*/N0 at 6000 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 5.2 × 10−2

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 2.0 × 10−3

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 1.2 × 10−7

The completed table can be given as,

Element	Na	Cu	Br
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( g*/g0 )	3	3	2/3
N*/N0 at 2600 K in flame	2.6 × 10−4	1.4 × 10−7	1.8 × 10−16
N*/N0 at 6000 K in flame	5.2 × 10−2	2.0 × 10−3	1.2 × 10−7

Bromine is not readily in absorption spectroscopy because the lowest excited state itself requires far-UV radiation for excitation. The excited state lies at such high energy which is not sufficiently populated to obtain the intensity for optical emission.

Answer 6

Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Expert Solution & Answer

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Sodium,

E = 2.1 eV × (1.602 × 10-19 J/eV) = 3.3642 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)3.3642 × 10−19 J = 19.8643 × 10−26 m3.3642 × 10−19 = 591 nm

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Copper,

E = 3.78 eV × (1.602 × 10-19 J/eV) = 6.05556 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)6.05556 × 10-19 J = 19.8643 × 10−26 m6.05556 × 10-19 = 328 nm

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Bromine,

E = 8.04 eV × (1.602 × 10-19 J/eV) = 12.8800 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)12.8800 × 10-19 J = 19.8643 × 10−26 m12.8800 × 10-19 = 154 nm

To calculate N*/N0 at 2600 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 2.6 × 10−4

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.4 × 10−7

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.8 × 10−16

To calculate N*/N0 at 6000 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 5.2 × 10−2

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 2.0 × 10−3

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 1.2 × 10−7

The completed table can be given as,

Element	Na	Cu	Br
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( g*/g0 )	3	3	2/3
N*/N0 at 2600 K in flame	2.6 × 10−4	1.4 × 10−7	1.8 × 10−16
N*/N0 at 6000 K in flame	5.2 × 10−2	2.0 × 10−3	1.2 × 10−7

Bromine is not readily in absorption spectroscopy because the lowest excited state itself requires far-UV radiation for excitation. The excited state lies at such high energy which is not sufficiently populated to obtain the intensity for optical emission.

Answer 7

Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Answer 8

Expert Solution & Answer

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Sodium,

E = 2.1 eV × (1.602 × 10-19 J/eV) = 3.3642 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)3.3642 × 10−19 J = 19.8643 × 10−26 m3.3642 × 10−19 = 591 nm

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Copper,

E = 3.78 eV × (1.602 × 10-19 J/eV) = 6.05556 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)6.05556 × 10-19 J = 19.8643 × 10−26 m6.05556 × 10-19 = 328 nm

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

wavelength = hcΔE

In the problem statement it is given that 1 eV = 96.49 kJ/mol .

Therefore, for Bromine,

E = 8.04 eV × (1.602 × 10-19 J/eV) = 12.8800 × 10-19 J

wavelength = hcΔE = (6.6261 × 10−34 Js)(2.9979 × 108 m/s)12.8800 × 10-19 J = 19.8643 × 10−26 m12.8800 × 10-19 = 154 nm

To calculate N*/N0 at 2600 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 2.6 × 10−4

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.4 × 10−7

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(2600 K) = 1.8 × 10−16

To calculate N*/N0 at 6000 K :

For Sodium,

N*N0 = g*g0e−ΔE/kT = 3e−(3.3642 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 5.2 × 10−2

For Copper,

N*N0 = g*g0e−ΔE/kT = 3e−(6.05556 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 2.0 × 10−3

For Bromine,

N*N0 = g*g0e−ΔE/kT = 23e−(12.88 × 10−19 J)/(1.381 × 10−23 J/K)(6000 K) = 1.2 × 10−7

The completed table can be given as,

Element	Na	Cu	Br
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( g*/g0 )	3	3	2/3
N*/N0 at 2600 K in flame	2.6 × 10−4	1.4 × 10−7	1.8 × 10−16
N*/N0 at 6000 K in flame	5.2 × 10−2	2.0 × 10−3	1.2 × 10−7

Bromine is not readily in absorption spectroscopy because the lowest excited state itself requires far-UV radiation for excitation. The excited state lies at such high energy which is not sufficiently populated to obtain the intensity for optical emission.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 21 - Prob. 21.AE Ch. 21 - Prob. 21.BE Ch. 21 - Prob. 21.CE Ch. 21 - Prob. 21.DE Ch. 21 - Prob. 21.EE Ch. 21 - Prob. 21.1P Ch. 21 - Prob. 21.2P Ch. 21 - Prob. 21.3P Ch. 21 - Prob. 21.4P Ch. 21 - Prob. 21.5P

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Concept explainers

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Explanation of Solution

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Chapter 21 Solutions

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Concept explainers

Interpretation: The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.

Explanation of Solution

Want to see more full solutions like this?

Chapter 21 Solutions

Interpretation:

The given table has to be filled using Boltzmann distribution and why Br is not readily observed in atomic emission or atomic absorption has to be explained.