Line broadening: The Doppler effect

When an atom moves towards a photon detector and emits radiation, the detector sees wave crest more often and detects radiation of higher frequency.

When an atom moves away from a photon detector and emits radiation, the detector sees wave crest less frequently and detects radiation of lower frequency.

Atomic motions occurs in every direction. An ensemble of atoms exhibit a Maxwell-Boltzmann velocity distribution.

A statistical distribution of frequencies, i.e., line broadening.∆λ/λ0 = v/c

Temperature effect: an example

Q: Calculate the ratio of Na atoms in the 3p excited states to the number in the ground state at 2500 k and 2510 k.

Answer:

go (3s) = 2; g* (3p) = 6

λ = 589.3 nm ΔE = 3.37 x 10-19 J.

@ 2500 K, N*/No = 1.72x10-4

@ 2510 K, N*/No = 1.79x10-4

kTEeg

g

N

N /

0

*

0

*

A temperature increase of 10 K results in a 4% increase in the number of excited Na atoms

4% increase in emission power

Implication: It is critical in atomic emission spectroscopy that the flame has stable temperatures.

Line broadening: The uncertainty effect

The uncertainty principle: 1 t

• The breadth of an atomic line would approach zero only if the lifetimes of the two states responsible for the transition approached infinity.

• The lifetimes of excited states are ~ 10-7 to 10-8 s.

Δν ~ 107 to 108

Δλ = (λ2 Δν)/c ~ 10-5 nm (10-4 Å) (Example 8-1)

5

Radiation source

• Monochromators generally can not isolate lines narrower than 10-3 to 10-2 nm. Continuum radiation sources are not suitable.

• A hallow-cathode lamp containing a vapor of the same element as that being analyzed.

Relative bandwidths of hallow-

cathode emission, atomic absorption

emission, and a monochromator.

The absorption line is broader than the

emission line due to (1) uncertainty

effect, (2) Doppler effect, and (3)

pressure effects due to collisions

between the same atoms and with

foreign atoms.

6

Inductively Coupled Plasma

• A spark from a Tesla coil ionizes the Ar gas and produces free electrons.

• The free electrons are accelerated by a radio-frequency field.

• The accelerated electrons collide with atoms and transfer their energy to the entire gas.

• A temperature of 6000-10,000 K in the plasma is maintained through electrons and ions absorbing energy from the induction coil.

• Electrons and ions interact with the fluctuating magnetic field and flow in the closed annular paths.

•A tangential flow of Ar cools the inside walls of the the center quartz tube and centers the plasma radially.

πσ σ

9

Common transitions producing X-ray

K series: electronic transitions

between higher energy levels and

the K shell.

L series: electronic transitions

between higher energy levels and

the L shell.

The energy between the L and K

levels is much larger than that

between the M and L levels. K

lines appear at shorter wavelength

than L lines.

The energy difference between Kα1

and Kα2 are small. typically only

one line is observed.

Similarly only one line is observed

for Kβ1 and Kβ2.

10

Fluorescence YieldCompeting processes:

1. Fluorescence process:

2. Auger process: When an electron is ejected from an inner shell and leaves a vacancy in an atom, a second ionization could occur. That is, energy that would otherwise have been released from the atom as an X-ray photon now goes toward ejecting an electron from an L or M shell.

X-ray fluorescence yield:

X-ray photons emitted from a given shell /the number of vacancies initially created in the shell.

Fluorescence yield increases steadily with Z, exceeds 50% for elements with Z>30.

wL never reaches 0.5 and fluorescence yields are still smaller for M and N shells.

11

What is the short-wavelength limit of the continuum produced by an X-Ray tube having a silver target and operated at 90KV?

Vehc

ho

o

uo, o: the maximum frequency and the low

wavelength limit

V: accelerating voltage

)(13.01090

12398

)(12398

)(1012398

10602177.1

/1099792.2)1062608.6(

3

10

19

834

o

o

o

A

AV

mV

CV

smJs

Ve

hc

12

• Line spectra in the UV and visible regions are produced when the radiating species

are individual atomic particles that are well separated in the gas phase

• Band spectra are encountered in spectral sources when gaseous radicals or small

molecules are present

• Continuum spectra is produced when solids are heated to incandescence.

Types of emission spectra

13Population inversion

14

15

nλ = (CB + BD)CB = dsin(i), BD = dsin(r)nλ = d(sin(i) + sin(r))

16

Resolving Power of Monochromators

R = λ/Δλ = nN

n = diffraction N = number of grating blazes

a) For a grating, how many lines per millimeter would be required for the first-order

diffraction line for λ = 400nm to be observed at reflection angle of 50

when the angle of incidence is 450.

b) Consider an infrared grating with 84 lines per millimeter and 15 mm of

illuminated area. Calculate the first-order resolution (λ/Δλ) of this grating.

18

19

Number of possible vibrational modes

• 3N-5 for linear molecules

• 3N-6 nonlinear molecules

N: number of atoms in a molecules

3N: the totol # of coordinates needed to specify the location of all N atoms

3 coordinates needed to specify the location of the center of the mass of the molecule

2 angles needed to specify the orientation of a linear molecule

3 angles needed to specify the orientation of a nonlinear molecule

Examples:

O2, N2, Cl2: 3N-5=3x2-5=1, only one stretching vibration mode

CO2: 3N-5=3x3-5=4

H2O: 3N-6=3x3-6=3

CO2

H2O

20

Example: estimate group frequency of IR absorption

Q: Calculate the approximate wavenumber and wavelength of the fundamental absorption

due to the stretching vibration of a carbonyl group C=O.

Solution:Group frequency:

frequency at which an

organic functional

group absorbs IR

radiation.

21

Vτ = λ/2

V: mirror moving velocity

τ: time required the mirror to λ/2

f: 1/τ = 2V/λ = 2V(ν/c)

ν: the frequency of the radiation

c: speed of light

P(δ) = B(ν) cos2πft

B(ν) : the radiation power of beam

P(δ) = B(ν) cos2π(2V(ν/c))t

V=δ/2t

P(δ)=B(ν) cos2πδ(ν/c)

P(δ)=∫B(ν)cos2πδ(ν/c)dν

B(ν)=∫P(δ)cos2π(ν/c)δdδ

∆ν = ν2 – ν1 = 1/δ

f = 2V(ν/c)= 2x1.5(cm/sec)ν/3x10-10cm/sec

f = 10-10ν

22

Resolution

∆ν = ν1 –ν2

P(δ)=∫B(ν)cos2πδ(ν/c)d

ν

B(ν)=∫P(δ)cos2π(ν/c)δd

δ

δν2 – δν1 = 1 or ν2 –ν1 = 1/δ = ∆ν

What length of mirror drive will provide a resolution of 0.1 cm-1

0.1 cm-1 = 1/δ δ= 10cm

Mirror motion = δ/2 = 5cm

Fig e:

P(δ) = B1(ν1)cos2πδν1 + B2(ν2)cos2πδν2

The resolution in wavenumbers

will improve on proportion to the

reciprocal of the distance that the

mirror travels

23

Resolution

∆ν = ν1 –ν2

P(δ)=∫B(ν)cos2πδ(ν/c)d

ν

B(ν)=∫P(δ)cos2π(ν/c)δd

δ

δν2 – δν1 = 1 or ν2 –ν1 = 1/δ = ∆ν

What length of mirror drive will provide a resolution of 0.1 cm-1

0.1 cm-1 = 1/δ δ= 10cm

Mirror motion = δ/2 = 5cm

Fig e:

P(δ) = B1(ν1)cos2πδν1 + B2(ν2)cos2πδν2

The resolution in wavenumbers

will improve on proportion to the

reciprocal of the distance that the

mirror travels