How does a spectrophotometer work?

UNDER KSS DISCUSSION WITH SHRI JKP SIR




SpectrophotometryGary L. Bertrand Electromagnetic radiation is characterized by its frequency (n) or its wavelength (l).These two are related by the velocity of light (c),
n=c/l. The electromagnetic spectrum ranges from high-energy cosmic rays (high frequency, short wavelength) to very low-energy microwaves (low frequency, long wavelength).
Visible light represents a very narrow section of this range with wavelengths between 400 nanometers (nm) for blue light to around 700 nm for red light. Shorter wavelengths fall into the ultraviolet region and longer wavelengths are in the infrared region.


White light is a mixture of all of the wavelengths in the visible range. When light strikes an object, it may be reflected, absorbed,transmitted, or diffracted.  A prism or a diffraction grating separates white light into its various colors.  If some of the light is absorbed, the reflected or transmitted light has the complementary color of the absorbed light.




A spectrophotometer uses an arrangement of prisms, mirrors, and slits to select light of a desired wavelength and to direct it toward a sample compartment and a detector.  The detector electronically measures the intensity of the light striking it.  A sample is placed in the light path, and the instrument compares the intensity of the light going through the sample (I) to the intensity observed without the sample (I_o).
The effect is measured either as Transmittance (T, the percentage of light that goes through the sample) or as the Absorbance (Abs, representing the amount of light absorbed by the sample):
T = 100(I/I_o)   ;  Abs = - log₁₀(T/100) = log₁₀(I_o/I) In the example above, a single sheet of the colored material transmits 70% of the light:
I/I_o= 0.70     ;    T = 70%   ;    Abs = - log₁₀(0.70)= 0.155  A second sheet transmits 70% of the light it receives, or 49% (0.70 x 70%) of the original light:
I/I_o= 0.49     ;    T = 49%   ;    Abs = - log₁₀(0.49)= 0.310 The third sheet transmits 70% of the light it receives, or 34.3 % (0.70 x 49%) of the original light:
I/I_o= 0.343     ;    T = 34.3%   ;    Abs = - log₁₀(0.343)= 0.465 The Absorbance is seen to be proportional to the number of sheets of the colored material.  This is Lambert's Law, the absorbance is directly proportional to the thickness or path length of the absorbing material.

A spectrophotometer is often used to study solutions.  A solution containing an absorbing material is compared to a reference solution of the same solvent and non-absorbing materials.  The transmittance of the reference solution is set to 100% (Abs = 0), then the relative transmittance of the solution is measured.
In the example above, the addition of a drop of red dye to one of the cells reduces the transmittance to 70%(Abs = 0.155). The addition of a drop to the second cell reduces the transmittance to 49%(Abs = 0.310) doubling the absorbance as is expected by Lambert's Law, since the pathlength of colored material is doubled.
However, the addition of the second drop to the first cell has exactly the same effect as adding it to the second cell.  In this case, the path length remains the same but the concentration of colored material is doubled, doubling the absorbance.  This is Beer'sLaw: at constant path length, the absorbance is directly proportional to the concentration of absorbing material.
The two laws are combined in the Beer-Lambert Law:
Abs = abC in which b is the path length, C is the concentration, and a is a constant which depends on the wavelength of the light, the absorbing material, and the medium (solvent and other components).  The constant a is called the extinction coefficient or the molar absorbtivity coefficient.
If there are several absorbing materials present, the effects are additive:
Abs = a₁b₁C₁ + a₂b₂C₂ + . . .


A graph of Absorbance vs Wavelength for a red dye shows a maximum at 525 nm:


For a blue dye, the maximum occurs near 625 nm:

If the both of these dyes are dissolved at the same concentrations to form a purple solution, the resulting graph shows both maxima
and at each wavelength, the absorbance of the purple solution is exactly equal to the sum of the absorbances of the red and blue solutions at that wavelength.  This can be seen clearly by looking at all three spectra at a wavelength of 525 nm.  The red dye shows an absorbance of 0.233, the blue dye has a small absorbance of 0.016, and the mixture has an absorbance of 0.249.
The absorbtivity coefficients can be calculated for the two dyes at wavelengths where the other will not interfere:

At 625 nm, the blue dye at 3.0 ppm has an absorbance of 0.318 in a cell of path length 1.00 cm. Therefore the absorbtivity coefficient (a) is:

a_blue,625 = 0.318/(3.0ppm x 1.00 cm) = 0.106 ppm^-1cm^-1. We will calculate the absorbtivity coefficient for the red dye at 510 nm to minimize the inflence of the blue dye.  (There are mathematical methods to optimize these calculations in overlapping regions, but that is beyond the scope of this discussion.)

At 510 nm, the red dye at 3.0 ppm has an absorbance of 0.183 in a cell of path length 1.00 cm. Therefore the absorbtivity coefficient (a) is:

a_red,510 = 0.183/(3.0ppm x 1.00 cm) = 0.061 ppm^-1cm^-1. These values may be used to calculate the concentrations of these red and blue dyes in other mixtures:

A different mixture of  these two dyes has Abs = 0.317 at 510 nm,
and Abs= 0.477 at 625 nm.

From the data at 510 nm, we calculate the concentration of red dye:

0.317 = (0.061 ppm^-1cm^-1)(1.00 cm)C_red
C_red = (0.317)/(0.061)= 5.2 ppm

From the data at 625 nm, we calculate the concentration of blue dye:

0.477 = (0.106 ppm^-1cm^-1)(1.00 cm)C_blue
C_blue = (0.477)/(0.106)= 4.5 ppm