Experiment 7 - EXPERIMENT 7 Optical Spectroscopy Objective:...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
7 – 1 EXPERIMENT 7 – Optical Spectroscopy Objective: Observe emission spectra from a variety of sources using a spectrometer. Design and implement a procedure capable of determining the composition of a solution that contains two or more ionic salts. Construct a partial energy-level diagram for hydrogen. Introduction: Optical spectroscopy involves the measurement and analysis of electromagnetic radiation (a.k.a. light). Many of the properties of light are conveniently described by means of a classical wave model. Within this model, light waves are characterized by such variables as frequency and wavelength. The frequency ( ν ), which describes the number of wave crests passing a given point per second for the light wave, is inversely proportional to the wavelength ( λ ), the distance between successive wave crests. The light we can see, visible light, corresponds to a very small portion of the electromagnetic spectrum, from about 400 nm (violet) to 800 nm (red). The light wave frequency and wavelength are related to one another by the equation c = λν (1) where c is the speed of the light wave. The speed of light in a vacuum is 3.00 × 10 8 m/s. For phenomena where the classical wave model of light proves insufficient, a particle model is invoked. In the particle model, light is composed of a stream of discrete particles called photons. The energy ( E ) of each photon is directly proportional to its frequency and inversely proportional to its wavelength, ! " hc h E = = (2) where the proportionality constant h is Planck’s constant (6.6261 × 10 –34 J s). Therefore, electromagnetic radiation can be described by either the wave or particle model; the model applied is the one that accurately describes the phenomenon being investigated. In spectroscopy, light is used as a means of probing matter. One means of probing matter with light uses the phenomenon of absorption. When an atom, molecule, or ion absorbs a photon, its energy increases. The energy change of the atom must be equivalent to the energy of the photon. Thus the absorbed wavelengths of light reveal the differences between energy levels: E 3 = –8.0 kJ/mol Δ E 3,1 = E 3 – E 1 = 31.5 kJ/mol E 3,2 λ 1 = hcN A / E 3,1 = 3800 nm E 2 = –17.0 kJ/mol E 2,1 = E 2 – E 1 = 22.5 kJ/mol E 3,1 2 = hcN A / E 2,1 = 5320 nm E 2,1 E 3,2 = E 3 – E 2 = 9.0 kJ/mol E 1 = –39.5 kJ/mol 3 = hcN A / E 3,2 = 13301 nm λ 1 λ 2 λ 3 Figure 7.1: For an atom with only three energy states as shown above, there are only three wavelengths of light that can be absorbed, each illustrated by an arrow. The values of the three absorbed wavelengths of light, as calculated from the
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Experiment 7: Optical Spectroscopy 7 – 2 energy-level differences, are shown as well. (Avogadro’s number, N A , is included in the equation for the wavelength because the energies are given per mole.) The converse of absorption is emission; when an atom emits a photon of light, its energy
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/10/2011 for the course CHEM 2090 taught by Professor Zax,d during the Spring '07 term at Cornell University (Engineering School).

Page1 / 11

Experiment 7 - EXPERIMENT 7 Optical Spectroscopy Objective:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online