Atomic Spectra: In 1859, R. Bunsen and G. Kirchhoff developed the spectroscope to study the spectra of luminous sources. Arrangement of all types of electromagnetic radiation in order of decreasing wavelength is known as the electromagnetic spectrum. The different types of EMR differ from one another only in their wavelengths and frequencies. The visible region of the spectrum ranges from the longer wavelength red light to the shorter wavelength violet light. The range of colors on the visible spectrum is red, orange, yellow, green, blue, indigo, and violet. To help memorize the order, use the initials ROY G BIV. Also, infrared is close to red and ultraviolet is close to violet. White light (as in sunlight) consists of waves with all the wavelengths in the visible range. When a ray of white light is passed through a prism, the ray is spread out to form a continuous spectrum (a rainbow of colors with no blank spots). On an atomic level: When gases or vapors of a substance are heated in an electric arc or a Bunsen burner flame, the atoms absorb energy. The electrons will jump from their normal positions (the ground state) to outer levels, which are higher energy states. The atoms are then said to be in an excited state. When an electron falls back to a lower level, after being excited, it emits a definite amount of energy, which is given off as a quantum of light. Passing the light emitted by an element through a prism gives the atomic emission spectrum of the element. Each line in this spectrum corresponds to a characteristic frequency, wavelength, and energy. The emission spectrum of each element is unique, which makes it useful for identification. In 1884, Johann Balmer energized atoms of hydrogen gas and examined the light produced with a spectroscope. He found 4 prominent colored lines, but was unable to explain their origin. Balmer showed that the wavelengths of light for the visible region of the emission spectrum of hydrogen could be calculated using the Rydberg equation: with n = 1, 2, 3, ... and R = 2.178x10 -18 J (called the Rydberg constant).                      The energy level transitions for the hydrogen atom with a final level of n = 2 is known as the Balmer series and has the wavelengths given in the table below. Color Wavelength red 6.563 x 10-7 m blue-green 4.861 x 10-7 m blue 4.340 x 10-7 m violet 4.102 x 10-7 m It was not until 1913 that Neils Bohr satisfactorily explained the origin of the spectral lines. Proposals made by Max Planck in 1901 and then by Albert Einstein a few years later,  provided the clues Bohr needed. Wave-particle Duality of Light: Planck proposed the quantum theory of radiant energy to explain the different colors emitted from a solid when it is heated. He suggested that the atoms of a solid oscillate or vibrate with a given frequency at a given temperature, and this oscillation produces the radiation. He also suggested that instead of radiant energy being continuous, it could be absorbed or given off only in definite quantities called quanta (little bundles of energy). He related the energy delivered by a photon, E, to the frequency of the light, n , by the formula: E=nhn The proportionality constant "h" is called Planck's constant and is 6.63 x 10-34 Js and n = 1,2,3.... In other words, energy is quantized. High-energy radiation has a high frequency and, therefore, a short wavelength. Despite the idea that energy is quantized, Planck and other physicists continued to imagine that the emitted energy traveled in waves. However, the wave model could not explain the photoelectric effect, the flow of current when monochromatic light of sufficient energy shines on a metal plate. Einstein proposed that electromagnetic radiation itself can be viewed as a stream of particles called photons, each with energy hn. To calculate wavelength (l ): From Å to m: (Å)(1 x 10-10 m/Å)    **Note: 1 Å = 1 x 10-10 m Example: (5200 Å)(1 x 10-10 m/Å) = 5.20 x 10-7 m To calculate frequency (n ): n = c/l                   c=3.00 x 108 m/s Example: n = 3.00 x 108 m/s = 5.77 x 1014 s-1 (or Hz)                         5.20 x 10-7 m To calculate energy (E): E = h n                  h = 6.63 x 10-34Js (Planck's constant) Example: E = (6.63 x 10-34 Js)(5.77 x 1014 s-1) = 3.83 x 10-19J

Sample problem: A cook uses a microwave oven to heat a meal. the wavelength of the radiation is 12.0 cm. What is the energy of one photon of this microwave radiation? E = hn =  hc  = (6.63 x 10-34 Js)(3.00 x 108 m/s)   = 1.66 x 1024 J                  l                (12.0 cm)(1m/100cm) To be continued...                                  Top of Page                                                                                               ©December 2000 by Vicki Klawinski    Updated on January 2, 2001