Chapter 8, Problem 8.12QAP

Interpretation Introduction

(a)

Interpretation:

The two expressions for Doppler broadening and Doppler half-width needs to be shown equivalent to each other.

Concept introduction:

The equation for the half-width for Doppler broadening Δλ₀ of an atomic line can be used to study line broadening in a low − pressure laser-induced plasma.

Explanation of Solution

The change in wavelength at the center of the emission line can be represented as follows:

$\Delta {\lambda }_0(T)={\lambda }_0\sqrt{\frac{8KT\text{ ln 2}}{M{C}^2}}$

Here,

$\begin{array}{l}{\lambda }_0=\text{The wavelength at the center of the emission line}\\ K=\text{Boltzman's constant}\\ \text{T = Temperature}\\ \text{M= Atomic mass}\\ \text{C= Velocity}\end{array}$

Similarly, the Doppler half-width can be calculated as follows:

$\Delta {\nu }_D=2{\left[\frac{2KT\text{ ln 2}}{M}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{{\nu }_m}{C}$

Here,

$\begin{array}{l}\Delta {\nu }_D=\text{The Doppler half width}\\ {\nu }_m=\text{The frequency at the line maximum}\end{array}$

Also,

$\begin{array}{l}\Delta {\nu }_D=2{\left[\frac{2KT\text{ ln 2}}{M}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{{\nu }_m}{C}\\ \Delta {\nu }_D=\sqrt{\frac{8KT\text{ ln 2}}{M{C}^2}}{\nu }_m\\ \text{Hence, above two expressions are equivalent}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The half-width for Doppler broadening needs to be determined for 4s to 4p transition for nickel atom.

Concept introduction:

Doppler bordering is happened due to the Doppler effect caused by a distribution of velocities of atomic molecules.

Answer to Problem 8.12QAP

The half-width = 7934 nm and $3.78\times {10}^{13}{s}^{-1}$.

Explanation of Solution

Given information:

$\begin{array}{l}K=\text{Boltzman's constant=1}\text{.38}\times {\text{10}}^{\text{-23}}{m}^2{\text{ kg s}}^{\text{-2}}{K}^{-1}\\ \text{T = Temperature = 20000 K}\\ \text{M= Atomic mass = 58}\text{.69 amu}\\ \text{C= Velocity=}3\times {10}^8{\text{ m s}}^{\text{-1}}\end{array}$

Calculation:

The Doppler half-width can be calculated as follows:

$\begin{array}{c}\Delta \lambda ={\left[\frac{2KT}{M}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{C}\\ \Delta \lambda ={\left[\frac{2\times \text{1}\text{.38}\times {\text{10}}^{\text{-23}}\times 20000}{58.69\times 1.66\times {10}^{-27}}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{3\times {10}^8}\\ \Delta \lambda =7.934\times {10}^{-6}\text{ m}=793\text{4 nm}\\ \Delta \lambda =3\times {10}^8{\text{ m s}}^{\text{-1}}/7.934\times {10}^{-6}\text{mm}\\ \Delta \lambda =3.78\times {10}^{13}{\text{s}}^{-1}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The natural line width for the above transition needs to be determined, assuming that the lifetime of the excited state is $5\times {10}^{-8}s$.

Concept introduction:

Natural line width is associated with the decay time (Natural life-time) and it is a minimum line width that does not contain effects such as collisional and Doppler broadening.

Answer to Problem 8.12QAP

Natural line width = $1.05\times {10}^{-27}m$

Explanation of Solution

Natural line width can be calculated as follows:

$\begin{array}{l}\Gamma \text{=}\frac{\hslash }{\tau }=\frac{h}{4\pi \tau }\\ \text{Here,}\\ \tau =\text{ Life time}\end{array}$

Putting the values,

$\begin{array}{c}\text{Natural}\text{line}\text{width}=\frac{6.6\times {10}^{-34}}{4\pi \times 5\times {10}^{-8}}\\ \text{Natural}\text{line}\text{width}=1.05\times {10}^{-27}\text{ m}\end{array}$

Interpretation Introduction

(d)

Interpretation:

To show that the relativistic expression is consistent with the mentioned equation given for the low atomic speeds.

Concept introduction:

When compared with the speed of light, atomic speed is very low.

Explanation of Solution

$\begin{array}{l}\frac{\Delta \lambda }{\lambda }=\frac{1}{\sqrt{\frac{C-V}{C+V}}}-1\\ \end{array}$

When the atomic speed very low V is considerably small when compared to the c, that of the speed of light. Hence the above mentioned equation could be written as shown below. Hence, at low velocities, relativistic kinetic energy reduces to classical kinetic energy. No object with mass can achieve the speed of light because an infinite amount of energy input and an infinite amount of work is required to accelerate a mass to the speed of light.

$V=\sqrt{\frac{8kT}{\pi m}}$

Interpretation Introduction

(e)

Interpretation:

The speed of an iron atom the 4s to 4p transition at 385.9911 nm should be determined.

Concept introduction:

The rest wavelength of Nickel is 410 nm. The formula used is:

$\begin{array}{l}\frac{\Delta \lambda }{\lambda }=\frac{1}{\sqrt{\frac{(C-V)}{C+V}}}-1\\ \end{array}$

Answer to Problem 8.12QAP

$1.744\times {10}^{8}m{s}^{-1}=V$

Explanation of Solution

Given information:

$\begin{array}{l}\frac{\Delta \lambda }{\lambda }=\frac{1}{\sqrt{\frac{(C-V)}{C+V}}}-1\\ \end{array}$

Calculation:

$\frac{\Delta \lambda }{\lambda }=\frac{1}{\sqrt{\frac{C-V}{C+V}}}-1$

$\frac{\Delta \lambda }{\lambda }=\frac{1}{\sqrt{\frac{3\times {10}^{8}m{s}^{-1}-V}{3\times {10}^{8}m{s}^{-1}+V}}}-1$

$\frac{385.991nm}{410nm}=\frac{1}{\sqrt{\frac{3\times {10}^{8}m{s}^{-1}-V}{3\times {10}^{8}m{s}^{-1}+V}}}-1$

$0.941+1=\frac{1}{\sqrt{\frac{3\times {10}^{8}m{s}^{-1}-V}{3\times {10}^{8}m{s}^{-1}+V}}}$

$1.941=\frac{1}{\sqrt{\frac{3\times {10}^{8}m{s}^{-1}-V}{3\times {10}^{8}m{s}^{-1}+V}}}$

${\left(1.941\right)}^2=\frac{1}{\sqrt{\frac{3\times {10}^{8}m{s}^{-1}-V}{3\times {10}^{8}m{s}^{-1}+V}}}$

$\frac{3\times {10}^{8}m{s}^{-1}-V}{3\times {10}^{8}m{s}^{-1}+V}=\frac{1}{3.76}$

$\frac{3\times {10}^{8}m{s}^{-1}-V}{3\times {10}^{8}m{s}^{-1}+V}=\frac{1}{3.76}$

$\begin{array}{c}3.76\left(3\times {10}^{8}m{s}^{-1}-V\right)=3\times {10}^{8}m{s}^{-1}+V\\ 11.302\times {10}^8-3.76V=3\times {10}^{8}m{s}^{-1}+V\end{array}$

$\begin{array}{c}11.302\times {10}^{8}m{s}^{-1}-3\times {10}^{8}m{s}^{-1}=3.76V+V\\ 8.302\times {10}^{8}m{s}^{-1}=4.76V\end{array}$

$\frac{8.302\times {10}^{8}m{s}^{-1}}{4.76}=V$

$1.744\times {10}^{8}m{s}^{-1}=V$

$\begin{array}{c}\frac{\Delta \lambda }{\lambda }=\frac{1}{\sqrt{\frac{(C-V)}{C+V}}}-1\\ \frac{{\left(\Delta \lambda \right)}^2}{{\lambda }^2}=\frac{1}{\frac{(C-V)}{C+V}}-1\\ \end{array}$

Interpretation Introduction

(f)

Interpretation:

The fraction of a sample of iron atoms at 10,000 K that would have the velocity calculated in part e should be computed.

Concept introduction:

Natural line width is associated with the decay time. It is a minimum line width that does not contain effects such as collisional and Doppler broadening.

Answer to Problem 8.12QAP

$m=1.172\times {10}^{-35}kg$

Explanation of Solution

Given information:

$\begin{array}{l}V=1.744\times {10}^{8}m{s}^{-1}\\ k=1.38064852\times {10}^{-23}{m}^{2}kg{s}^{-2}{K}^{-1}\\ T=10,000K\\ \pi =\raisebox{1ex}{$22$}\!\left/ \!\raisebox{-1ex}{$7$}\right.\end{array}$

Calculation:

$\begin{array}{c}V=\sqrt{\frac{8kT}{\pi m}}\\ 1.744\times {10}^{8}m{s}^{-1}=\sqrt{\frac{8\times 1.38064852\times {10}^{-23}{m}^{2}kg{s}^{-2}{K}^{-1}\times 10,000K}{\raisebox{1ex}{$22$}\!\left/ \!\raisebox{-1ex}{$7$}\right.m}}\\ 1.744\times {10}^8=\sqrt{\frac{3.517\times {10}^{-19}}{m}kg}\\ 3.00\times {10}^{16}\text{=}\frac{3.517\times {10}^{-19}}{m}kg\\ \text{m =}\frac{3.517\times {10}^{-19}}{3.00\times {10}^{16}}kg\\ m=1.172\times {10}^{-35}kg\end{array}$

Interpretation Introduction

(g)

Interpretation:

A spreadsheet should be created to calculate the Doppler half-width $\Delta {\lambda }_D$ in nanometers for the nickel and iron lines cited in part b and e from 30000-10,000 K.

Concept introduction:

Doppler bordering is happened due to the Doppler effect caused by a distribution of velocities of atomic molecules.

Answer to Problem 8.12QAP

Refer the spreadsheet

Explanation of Solution

Given information:

$\begin{array}{c}\text{Atomic mass of nickel = 58}\text{.6934u}\\ \text{Atomic mass of iron = 55}\text{.845u}\\ \text{Temperature range = }30000-10,000K\\ C=\text{Speed of light}\\ \text{C = 3}\times {\text{10}}^{8}m{s}^{-1}\\ k=1.38064852\times {10}^{-23}{m}^{2}kg{s}^{-2}{K}^{-1}\end{array}$

Calculation:

$\begin{array}{l}\Delta \lambda ={\left[\frac{2kT}{M}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{C}\\ \end{array}$

Interpretation Introduction

(h)

Interpretation:

The four sources of pressure broadening should be listed by consulting the paper by Gornushkin et al. (note 10).

Explanation of Solution

The interaction of the surrounding particles with the radiating atom is the major source of pressure line broadening, which causes a phase shift and a frequency disturbance.

The most important cases of interaction are:

1. linear Starkeffect, p = 2;
2. resonance interaction between identical particles, p = 3;
3. quadratic Stark effect, p = 4,
4. van der Waals interaction, p = 6.

The superposition problems are avoided by two approximations:

1. ‘nearest neighbor approximation’, in this the considered interaction is interaction with the closest perturber.
2. The impact or collision concept, in which moving perturbers act sequentially in time.