Chapter 21, Problem 21.17P

Interpretation Introduction

Interpretation:

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

Explanation of Solution

Calculation of Wavelength for Sodium;

To calculate wavelength the formula that can be used is,

$wavelength=\frac{hc}{\Delta E}$

In the problem statement it is given that $\text{1}\text{eV}\text{=}\text{96}\text{.49}\text{kJ/mol}$.

Therefore, for Sodium,

$\begin{array}{l}\text{E}\text{=}\text{2}\text{.1}\text{eV}\text{×}\left(\text{1}\text{.602}\text{×}{\text{10}}^{\text{-19}}\text{J/eV}\right)\\ \text{=}\text{3}\text{.3642}\times {\text{10}}^{\text{-19}}\text{J}\end{array}$

$\begin{array}{l}wavelength=\frac{hc}{\Delta E}\\ =\frac{\left(6.6261\times {10}^{-34}\text{Js}\right)\left(2.9979\times {10}^8\text{m/s}\right)}{3.3642\times {10}^{-19}\text{J}}\\ =\frac{19.8643\times {10}^{-26}\text{m}}{3.3642\times {10}^{-19}}\\ \text{=}\text{591}\text{nm}\end{array}$

Calculation of Wavelength for Copper;

To calculate wavelength the formula that can be used is,

$wavelength=\frac{hc}{\Delta E}$

In the problem statement it is given that $\text{1}\text{eV}\text{=}\text{96}\text{.49}\text{kJ/mol}$.

Therefore, for Copper,

$\begin{array}{l}\text{E}\text{=}3.78\text{eV}\text{×}\left(\text{1}\text{.602}\text{×}{\text{10}}^{\text{-19}}\text{J/eV}\right)\\ \text{=}\text{6}\text{.05556}\times {\text{10}}^{\text{-19}}\text{J}\end{array}$

$\begin{array}{l}wavelength=\frac{hc}{\Delta E}\\ =\frac{\left(6.6261\times {10}^{-34}\text{Js}\right)\left(2.9979\times {10}^8\text{m/s}\right)}{\text{6}\text{.05556}\times {\text{10}}^{\text{-19}}\text{J}}\\ =\frac{19.8643\times {10}^{-26}\text{m}}{\text{6}\text{.05556}\times {\text{10}}^{\text{-19}}}\\ \text{=}\text{328}\text{nm}\end{array}$

Calculation of Wavelength for Bromine;

To calculate wavelength the formula that can be used is,

$wavelength=\frac{hc}{\Delta E}$

In the problem statement it is given that $\text{1}\text{eV}\text{=}\text{96}\text{.49}\text{kJ/mol}$.

Therefore, for Bromine,

$\begin{array}{l}\text{E}\text{=}8.04\text{eV}\text{×}\left(\text{1}\text{.602}\text{×}{\text{10}}^{\text{-19}}\text{J/eV}\right)\\ \text{=}\text{12}\text{.8800}\times {\text{10}}^{\text{-19}}\text{J}\end{array}$

$\begin{array}{l}wavelength=\frac{hc}{\Delta E}\\ =\frac{\left(6.6261\times {10}^{-34}\text{Js}\right)\left(2.9979\times {10}^8\text{m/s}\right)}{\text{12}\text{.8800}\times {\text{10}}^{\text{-19}}\text{J}}\\ =\frac{19.8643\times {10}^{-26}\text{m}}{\text{12}\text{.8800}\times {\text{10}}^{\text{-19}}}\\ \text{=}\text{154}\text{nm}\end{array}$

To calculate ${\text{N*/N}}_{\text{0}}$ at $\text{2600}\text{K}$:

For Sodium,

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =3e^{-\left(3.3642\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(2600\text{K}\right)}\\ =2.6\times {10}^{-4}\end{array}$

For Copper,

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =3e^{-\left(6.05556\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(2600\text{K}\right)}\\ =1.4\times {10}^{-7}\end{array}$

For Bromine,

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =\frac{2}{3}{e}^{-\left(12.88\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(2600\text{K}\right)}\\ =1.8\times {10}^{-16}\end{array}$

To calculate ${\text{N*/N}}_{\text{0}}$ at $\text{6000}\text{K}$:

For Sodium,

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =3e^{-\left(3.3642\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(6000\text{K}\right)}\\ =5.2\times {10}^{-2}\end{array}$

For Copper,

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =3e^{-\left(6.05556\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(6000\text{K}\right)}\\ =2.0\times {10}^{-3}\end{array}$

For Bromine,

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =\frac{2}{3}{e}^{-\left(12.88\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(6000\text{K}\right)}\\ =1.2\times {10}^{-7}\end{array}$

The completed table can be given as,

Element	$\text{Na}$	$\text{Cu}$	$\text{Br}$
Excited state energy(eV)	2.1	3.78	8.04
Wavelength (nm)	591	328	154
Degeneracy Ratio ( $g*/g_0$)	3	3	2/3
${\text{N*/N}}_{\text{0}}$ at $\text{2600}\text{}\text{K}$ in flame	$2.6\times {10}^{-4}$	$1.4\times {10}^{-7}$	$1.8\times {10}^{-16}$
${\text{N*/N}}_{\text{0}}$ at $\text{6000}\text{K}$ in flame	$5.2\times {10}^{-2}$	$2.0\times {10}^{-3}$	$1.2\times {10}^{-7}$

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.