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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
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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