Energy of photon = h!
h - Plank’s constant ! - frequency (sec-1)
"! = c
c - speed of light
As " increases, the frequency and energy of the photon decreases.
Common " units µm - 10-6 m nm - 10-9 m A - 10-10 m
Type wavelength Interaction
# < 10 nm nuclear emission X-ray < 10 nm atomic ionization UV 10-380 nm electronic transitions Vis 380-800 nm electronic transitions IR 800nm-100µm bond interaction Radio meters nuclear absorption
Light is absorbed by a substance only when the energy corresponds to some energy need or transition of a substance.
Changes in our substance can be
• Electronic
• Vibrational
• Rotational
The last two are only seen for molecules.
Absorption of light is a complicated process.
Each electronic state is subdivided into a number of vibrational sub-levels.
In turn, each vibrational sub-level is further divided into rotational sublevels.
Electronic levels (UV interactions)
Vibrational sublevels (IR interactions)
Rotational sublevels (IR interactions)
With atoms, our simplest case, we still have a relative complex absorption process.
Even for hydrogen, we end up with a complex line spectrum due to the major electronic transitions and the sublevels -s, p, d, f.
With molecules, we not only have electronic but vibrational and rotational sub-levels. Interactions with other molecules and a solvent will also have an effect. These result in ‘band’ spectra.
In reality, there are too many transitions to distinguish with conventional spectrophotometric equipment.
With UV, Vis, IR - absorption occurs over a range. By scanning and measuring absorption over a range of wavelengths, we can produce a spectrum.
UV Visible Near
IR IR
Molecular absorption spectra example
Because of differences in equipment, we typically obtain spectra for UV/Vis on the same equipment and IR on a second instrument.
Information obtained in each area varies.
UV/Vis - Electronic transitions IR - Bond interactions
Spectra also can vary significantly for different compounds.
dP dN - = K P
where:
P = radiant intensity
N = number of absorbing species in the light path
K = proportionality constant
dP P P
0
= - K N
0 dN
P Po
ln = - K N
b Po
P c
P
P is a measure of the light that passes through o the solution - transmitted
P
Po is also called the transmittance (T)
While it is absorbance that is used to produce a relationship with concentration, it is
transmittance that is directly measured.
While most instruments will produce both types of numbers, it is better to read transmittance if a meter is involved and then convert.
100 90 80 70 60 50 40 30 20 10 0
0.0 0.1 0.2 0.3 0.4 0.5 0.7 1.0
Absorbance
% Transmittance
Calculate the absorbance of a solution having a
%T of 89 at 400 nm.
%T = T x 100 so:
T = 89 / 100 = 0.89 A = -log(T) = -log(0.89)
= 0.051
Each instrument varies to some extent so it is best to determine the absorbtivity using standards.
Any concentration unit can be used (what ever is convenient, M, N, ppm,....)
If molar concentrations are used, we often use
$
to represent molar absorbtivity.
A
a = b c = 0.30 abs (2.00 cm)(4.5 ppm)
= 0.033 abs cm-1 ppm-1
A solution of Co(H2O)2+ has an absorbance of 0.20 at 530 nm in a 1.00 cm cell. $ is known to be 10 L n-1cm-1. What is its concentration.
A = abc
Where: A = 0.20, a = 10, b = 1cm c = A / (ab) = 0.020 M
The absorbance of an unknown MnO4- solution is 0.500 at 525 nm. When measured under identical conditions, a 1.0x10-4 M MnO4- is found to have an absorbance of 0.200.
Determine the concentration of the unknown.
=
=
Our unknown concentration can be found by:
cunknown = cknown
cunknown = (0.500/0.200) x 1.0x10-4 M = 2.5 x 10-4 M
This assumes that you are in the ‘linear range’
for the method.
Aunknown Aknown
-log = abc
log = abc -log(T) = abc -log(T) = A
A = abc
A = $bc
P Po Po
P
We always attempt to work at the wavelength of maximum absorbance ("max).
This is the point of maximum response so better sensitivity and lower detection limits.
We will also have reduced error in our measurement.
Small error
Large error
If we have a small variation of " during our measurement, there can be a large difference in response if we are not at the
"max.
identical variations in wavelength.
Your method should only be used in this range.
This response could be due to background, interference, or a lack of sensitivity
This could be due to self-absorption or insufficient light passing through the cell.
% T
concentration
Small %C - large %%T
Small %%T - large %C
Relative error vs. %T
Relative error
%T
80% 20%
When two or more species absorb light at the same wavelength, the resulting absorbance is the sum of all absorbances.
AT = a1b1c1 + a2b2c2
Since they are in the same sampling cell, then:
AT = (a1c1 + a2c2 ) b
We need to measure a metal-reagent complex (MR) which absorbs at 522 nm ($=1.18x104).
The solution also contains 1.00x10-4 M excess reagent (R) with an $ of 5.12x102 at 522 nm.
If the total absorbance is 0.727 at 522 nm in a 1.00 cm cell, what is the concentration of MR?
AT = $MR cMR + $R cR
0.727 = 1.18 x 104 cMR + (5.16 x 102)(10-4 M) cMR = 5.72 x 10-5 M
Two metal complexes (X & Y) demonstrate at least some absorption over the entire visible range.
A mixture was measured at two
" using a 1 cm cell and the following data was obtained.
A"1 = 0.533 A"2 = 0.590
Determine the concentration of each species.
$1 $2 X 3.55 x 103 5.64 x 102 Y 2.96 x 103 1.45 x 104
At "1
0.533 = (3.55 x 103) CX + (2.96 x 103) CY or
CX = 0.533-2.96 x 103 CY 3.55 x 103
At "2
0.590 = (5.64 x 102) CX + (1.45 x 104) CY Substituting for CX in this expression gives:
0.590 = + 1.45 x 104 CY
Solving for CY shows that CY = 3.60 x 10-5 M 5.64 x 102 (0.533 - 2.96 x 103 CY)
3.55 x 103
And
CX =
= 1.20 x 10-4 M
Two or three species can be determined with this approach. Beyond that, the errors tend to become too great.
0.533 - (2.96 x 103)(3.60 x 10-5) 3.55 x 103
Atoms, ions and molecules can be excited via a number of processes.
When they relax, they release the excess energy.
In some cases, the relaxation results in the emission of EM radiation.
The type of EM emission is often characteristic of the species.
