Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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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
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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 1eV = 96.49kJ/mol .

Therefore, for Sodium,

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

wavelength = hcΔE = (6.6261×1034Js)(2.9979×108m/s)3.3642×1019J = 19.8643×1026m3.3642×1019 = 591nm

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 1eV = 96.49kJ/mol .

Therefore, for Copper,

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

wavelength = hcΔE = (6.6261×1034Js)(2.9979×108m/s)6.05556×10-19J = 19.8643×1026m6.05556×10-19 = 328nm

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 1eV = 96.49kJ/mol .

Therefore, for Bromine,

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

wavelength = hcΔE = (6.6261×1034Js)(2.9979×108m/s)12.8800×10-19J = 19.8643×1026m12.8800×10-19 = 154nm

To calculate N*/N0 at 2600K :

For Sodium,

N*N0 = g*g0eΔE/kT = 3e(3.3642×1019J)/(1.381×1023J/K)(2600K) = 2.6×104

For Copper,

N*N0 = g*g0eΔE/kT = 3e(6.05556×1019J)/(1.381×1023J/K)(2600K) = 1.4×107

For Bromine,

N*N0 = g*g0eΔE/kT = 23e(12.88×1019J)/(1.381×1023J/K)(2600K) = 1.8×1016

To calculate N*/N0 at 6000K :

For Sodium,

N*N0 = g*g0eΔE/kT = 3e(3.3642×1019J)/(1.381×1023J/K)(6000K) = 5.2×102

For Copper,

N*N0 = g*g0eΔE/kT = 3e(6.05556×1019J)/(1.381×1023J/K)(6000K) = 2.0×103

For Bromine,

N*N0 = g*g0eΔE/kT = 23e(12.88×1019J)/(1.381×1023J/K)(6000K) = 1.2×107

The completed table can be given as,

Element Na Cu Br
Excited state energy(eV)2.13.788.04
Wavelength (nm)591328154
Degeneracy Ratio ( g*/g0 )332/3
N*/N0 at 2600K in flame 2.6×104 1.4×107 1.8×1016
N*/N0 at 6000K in flame 5.2×102 2.0×103 1.2×107

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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