# Transmittance: What It Is, Molecular Energy Diagram And Exercise

the transmittance Optical is the quotient between the emergent luminous intensity and the incident luminous intensity on a sample of translucent solution that has been illuminated with monochromatic light.

The physical process of the passage of light through a sample is called light transmission  and thetransmittance it is a measure of light transmission. Transmittance is an important value to determine the concentration of a sample that is generally dissolved in a solvent such as water or alcohol, among others.

An electro-photometer measures a current proportional to the light intensity that falls on its surface. To calculate transmittance, the intensity signal corresponding to the solvent alone is generally measured first and this result is recorded as Io.

Then the dissolved sample is placed in the solvent with the same lighting conditions and the signal measured by the electro-photometer is denoted as I, then the transmittance is calculated according to the following formula:

T = I / Ior

It should be noted that transmittance is a dimensionless quantity, because it is a measure of the luminous intensity of a sample in relation to the intensity of the solvent transmission.

## What is transmittance?

### Absorption of light in a medium

When light passes through a sample, some of the light energy is absorbed by the molecules. Transmittance is the macroscopic measure of a phenomenon that occurs at the molecular or atomic level.

Light is an electromagnetic wave, the energy it carries is in the electric and magnetic field of the wave. These oscillating fields interact with the molecules of a substance.

The energy carried by the wave depends on its frequency. Monochromatic light has a single frequency, while white light has a range or spectrum of frequencies.

All the frequencies of an electromagnetic wave travel in a vacuum at the same speed of 300,000 km / s. If we denote byc at the speed of light in vacuum, the frequency ratio F and wavelength λ it is:

c = λ⋅f

Since c is a constant, each frequency corresponds to its respective wavelength.

To measure the transmittance of a substance, the regions of the visible electromagnetic spectrum (380 nm to 780 nm), the ultraviolet region (180 to 380 nm) and the infrared region (780 nm to 5600 nm) are used.

The speed of propagation of light in a material medium depends on the frequency and is less than c. This explains the scattering in a prism with which the frequencies that make up white light can be separated.

### Molecular theory of light absorption

Atoms and molecules have quantized energy levels. At room temperature the molecules are at their lowest energy levels.

The photon is the quantum particle associated with the electromagnetic wave. The energy of the photon is also quantized, that is, a photon of frequencyF has energy given by:

E = h⋅f

where h is Planck’s constant whose value is 6.62 × 10 ^ -34 J⋅s.

Monochromatic light is a beam of photons of a given frequency and energy.

Molecules absorb photons when their energy matches the difference needed to bring the molecule to a higher energy level.

The energy transitions by absorption of photons in molecules can be of several types:

1- Electronic transitions, when the electrons of the molecular orbitals go to orbitals of higher energy. These transitions generally occur in the visible and ultraviolet range and are the most important.

2- Vibrational transitions, the molecular bonding energies are also quantized and when a photon from the infrared region is absorbed, the molecule goes to a higher state of vibrational energy.

3- Rotational transitions, when the absorption of a photon leads the molecule to a rotational state of higher energy.

## Molecular energy diagram

These transitions are best understood with a molecular energy diagram shown in Figure 2:

In the diagram the horizontal lines represent different molecular energy levels. Line E0 it is a fundamental or lower energy level. Levels E1 and E2 are excited levels of higher energy. The E0, E1, E2 levels correspond to the electronic states of the molecule.

The sublevels 1, 2, 3, 4 within each electronic level correspond to the different vibrational states corresponding to each electronic level. Each of these levels has finer subdivisions that are not shown to correspond to the rotational states associated with each vibrational level.

The diagram shows vertical arrows representing the energy of photons in the infrared, visible and ultraviolet ranges. As can be seen, infrared photons do not have enough energy to promote electronic transitions, while visible and ultraviolet radiation do.

When the incident photons of a monochromatic beam coincide in energy (or frequency) with the energy difference between molecular energy states, then absorption of photons occurs.

### Factors on which transmittance depends

According to what was said in the previous section, the transmittance will then depend on several factors, among which we can name:

1- The frequency with which the sample is illuminated.

2- The type of molecules to be analyzed.

3- The concentration of the solution.

4- The length of the path traveled by the light beam.

The experimental data indicate that the transmittance T decreases exponentially with concentration C and with the length L of the optical path:

T = 10-a⋅C⋅L

In the above expression to it is a constant that depends on the frequency and the type of substance.

## Exercise resolved

### Exercise 1

A standard sample of a certain substance has a concentration of 150 micromoles per liter (μM). When its transmittance is measured with light of 525 nm, a transmittance of 0.4 is obtained.

Another sample of the same substance, but of unknown concentration, has a transmittance of 0.5, when measured at the same frequency and with the same optical thickness.

Calculate the concentration of the second sample.

The transmittance T decays exponentially with the concentration C:

T = 10 -b⋅L

If the logarithm of the previous equality is taken, it remains:

log T = -b⋅C

Dividing member by member the previous equality applied to each sample and solving for the unknown concentration remains:

C2 = C1⋅ (log T2 / log T1)

C2 = 150μM⋅ (log 0.5 / log 0.4) = 150μM⋅ (-0.3010 / -0.3979) = 113.5μM

## References

1. Atkins, P. 1999. Physical Chemistry. Omega editions. 460-462.
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