Light and Spectra. COLOR λ, nm COLOR λ, nm violet 405 yellow 579 blue 436 orange 623 green 492 red 689

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1 Light and Spectra INTRODUCTION Light and color have intrigued humans since antiquity. In this experiment, you will consider several aspects of light including: a. The visible spectrum of colors (red to violet) b. Bright line spectra as emitted by an excited gas or solid c. The relationship between color, wavelength, frequency and energy. BACKGROUND AND THEORY Light is a form of energy called electromagnetic radiation. It has wavelength and frequency. Wavelength, λ (lambda), is the distance between adjacent wave crests. Visible light has wavelengths in the 400 nm to 700 nm range (1 nm = 10 9 m). Frequency, ν (nu), tells how many waves pass by a point in a second. Violet light, with a wavelength of 400 nm, has a frequency of 7.5 x per second, many more waves than are shown in the diagram. λ λ difference in energy between lower and higher electron energy levels. When the electrons return to lower energy levels, energy is emitted in the form of a photon. Since atoms have a number of energy levels available, there are a number of different energies that can be absorbed and released. For example, the visible portion of the spectrum for mercury contains the following colors: COLOR λ, nm COLOR λ, nm violet 405 yellow 579 blue 436 orange 623 green 492 red 689 The energies of the emitted wavelengths can be correlated with the energies of electron transitions from higher to lower energy levels. During an energy high energy low electronic transition, the greater the difference between the energy levels, the larger is the energy released, and the shorter is the wavelength of the photon emitted. In the energy level diagram of a generic atom shown below, the left-hand arrow represents an electron releasing energy as it drops from the 6th to the 2nd energy level. n = 6 n = 5 ν = 2 waves per time line ν = 3 waves per time line The energy of electromagnetic radiation varies directly with the frequency, and inversely with the wavelength. Thus, violet light is of higher energy than red light, which has wavelengths in the 650 nm range. White light, such as that from an ordinary incandescent light bulb, is a mixture of wavelengths in the visible range. When white light strikes a prism or diffraction grating, the light is dispersed into a continuous spectrum of visible colors. A non-continuous spectrum occurs when an electric current passes through a gaseous element in a gas discharge tube or when metal ions are put into a flame. This type of spectrum, called a bright line spectrum, is not continuous but instead contains only certain colors at particular wavelengths. A bright line spectrum is due to transitions of electrons between energy levels in atoms or ions. When an atom or ion absorbs energy, such as from a flame or electrical source, it absorbs only certain discrete amounts of energy. These amounts are the n = 4 n = 3 n = 2 The energy is released in the form of a photon. This photon will have more energy (and have a shorter wavelength) than the photon given off in the transition represented by the right-hand arrow. In terms of energy, (n 6 n 2 ) > (n 3 n 2 ) The larger energy would be in the violet-blue portion of the spectrum. The smaller energy would be in the red-orange portion of the spectrum. 1

2 Frequency and Energy The relationship between wavelength and frequency is: λν = C C, the speed of light, equals 3.00 x 10 8 m/s. The violet line with λ = 405 nm from the mercury spectrum would have a frequency calculated as follows: ν = C ν = 3.00 x108 m s 1 λ or 405x10 9 m = 7. 41x1014 s 1 Notice that λ is written as 405 x 10-9 m, rather than 405 nm. To find the energy of a photon, the Planck relationship is used: E = hν where h, the Plank constant, = 6.63 x J s. To find the energy of the violet line, ( )( 7. 41x10 14 s 1 ) = 4. 91x10 19 J E = 6. 63x10 34 J s This is the energy emitted when an electron in one particular energy level in a mercury atom drops down to a lower particular energy level x J might seem extraordinarily small, but in terms of a mole of atoms, there would be 296 kj mol -1. (Can you verify this?). This is a good time to make a general observation about quantities. You will hear people say, and perhaps you say, that seems like an awfully small amount. The response to this is, compared to what?. A mercury atom, one of the really heavy atoms, weighs about 3 x g. Compared to your weight, a mercury atom weighs nothing. Compared to a neon atom, the mercury atom weighs 10 times as much. A thousand percent heavier! The moral of the story is: Consider everything in perspective. EXPERIMENT Using the diffraction grating: Diffraction gratings split a light source into its component wavelengths. To use the grating, hold it so the MADE IN U.S.A. is on the bottom facing you. The right-side cardboard frame should be between your dominant eye and the light source, blocking the light source from your eye. The plastic should be to the left of your eye. You will see the spectrum by glancing into the plastic. Continuous Spectrum Find the tall, clear light bulb connected to a variable power transformer. Turn the transformer on with the dial set to high. Hold the diffraction grating as instructed above to observe the filament in the light bulb. A full spectrum is visible. Now t urn off the room lights, and observe the spectrum as the rheostat is turned down, then back up. Notice how the blues and greens diminish in intensity more than the reds diminish as the temperature of the filament decreases, then reappear as the temperature increases. The theory available 100 years ago predicted that as the temperature increased, the spectrum would completely shift over into the ultraviolet, with no reds or yellows or greens remaining. This inadequate theory had physicists mumbling about the ultraviolet catastrophe. Theoretical physicists looked to develop theories consistent with experimental data. Max Planck developed a new theory which has held up for close to a century. The equation: E = hν is part of the theory. Line spectra The conditions used in this part of the experiment, namely, high voltage through a low pressure sample of an element, produce a characteristic line spectrum for each element. Each element is labeled and set-up properly. Plug in the high voltage source if necessary. Hydrogen Spectrum: With the lights still out, observe the hydrogen spectrum. The closer you get, the clearer it appears. Observe the lines and draw them on the chart below. red yellow green blue violet Mercury Spectrum: Observe the lines and draw them on the chart below. Just as H and Hg differ, so each atom has a unique spectrum. red yellow green blue violet Neon Spectrum: Simply observe. Notice the number of lines in the red and yellow portion of the spectrum. Julie Enriquez Light source The unique spectrum given by each element provides a simple means of analysis for elements. After exciting a substance with a spark or in a flame, the position of the spectral lines produced gives the identity of the elements in the substance, and the intensity of the lines gives the quantity of the elements. Of course, instruments that do this have data bases built in that allow for the experimental data to be compared with previously compiled data. 2

3 CALCULATIONS Hg Spectrum. The one you observed in the lab is a low resolution spectrum. The spectral charts posted on the walls show higher resolution spectra of a number of substances. Look at a chart and sketch the spectral lines of mercury roughly to scale in the strip below: (numbers are λ in nm) Notice that the spectrum has three lines shown in the yellow region. Record the actual wavelengths from the chart, then calculate the frequency and energy of each of these three lines in the mercury spectrum: λ ν E Also, notice that the lines have different widths. This is typical of spectral lines, and spectroscopists rank lines according to their appearance and abundance. Hence the designations s, p, d, and f for electron orbitals (from sharp, principal, diffuse and fundamental. Hydrogen Spectrum. The hydrogen atom is the simplest atom and a detailed study of its spectrum leads to valuable insights. Shown in Figure 1 below are the energy levels in the hydrogen atom as calculated from the Schrödinger equation. The numbers on the left give the energy of each level in joules. The number on the right gives the principal quantum number of each level. The names along the top are the names of the scientists who first characterized the spectrum corresponding to these transitions. Shown in Figure 2 is a black and white representation of the experimentally observed spectrum of the hydrogen atom. IR is infrared, VIS is visible and UV is ultraviolet. The wavelength of each line in nanometers is listed below the line. Figure 2, which is experimental, can be calculated from Figure 1, which is theoretical Figure 1: Energy Levels and Some Possible Transitions for Hydrogen Atoms Lyman Ultraviolet Balmer Visible Paschen Infrared n = (13 or 14) n = 8 n = 7 n = 6 n = 5 n = 4 Energy/10-19 Joules n = 3 n = 2 n values Each arrow represents an energy release, E, as an electron goes from a higher energy level to a lower one. The energy is given off as photons. The photons are in the part of the spectrum indicated above the arrows n = 1 Figure 2: H Atom Spectrum Far IR IR Visible UV λ in nanometers

4 Hydrogen Spectrum Calculations According to spectral theory, each line in the spectrum comes as an electron drops from a higher to a lower energy level. In Figure 1 on page 57, the various vertical arrows show a number of such possible transitions. The energy of a transition, for example, the one indicated by the left hand vertical arrow in Figure 1, can be calculated by subtracting the energy of the initial state from the energy of the final state (found on the left of the diagram): E = E final E initial ( ) ( x10 19 J) ( ) = 21. 8x10 19 J = 20. 9x10 19 J (The negative value indicates energy released.) This E is for an n 5 n 1 transition (see the right side of Figure 1 for n values). Each line in the hydrogen spectrum shown in Figure 2 has an energy which can be calculated using the Planck relationship. The far right hand line has λ = 97.3 nm, or 97.3 x 10-9 m. To calculate the energy, first find the ν of the line: To give consistency to the appearance of the energies calculated, all energies will be shown with as the exponent. Then the energy of the 97.3 nm line is 20.4 x J. This is very close to the absolute value of the E for the n 5 n 1 transition previously calculated, which was 20.9 x J. Upon close inspection of the left and right y-axis coordinates of Figure 1, it is seen that for the n 4 n 1 transition, represented by the second from the left vertical arrow: E = ( 21. 8x10 19 J) ( 1. 36x10 19 J) = x10 19 J This is exactly the same value as the Planck calculation for the λ = 97.3 nm given just above. So we can conclude that the spectral line at 97.3 nm is caused by an electron dropping from n = 4 to n = 1 in the hydrogen atom. Your understanding of the calculations and discussion on this page are crucial for the completion of the calculation table below. If you have difficulty with the calculations, reread this section. If the difficulties continue, see the instructor. Then ν = 3.00 x108 m s x10 9 m = 3. 08x1015 s 1 E = ( 6. 63x10 34 J s) ( 3. 08x10 15 s 1 ) = 2.04 x10 18 J The calculations just performed are summarized in the first line of the table below. Continuing with the wavelengths from Figure 2, complete the first three columns in the table. By analyzing the energy levels in Figure 1, find which transition is responsible for each spectral line, and indicate the transition in the last column. CALCULATION TABLE λ in nm ν in seconds 1 E in Joules n high n low x x

5 Name Grade Date The Rydberg equation is a famous relationship that first showed the order of the hydrogen spectrum. To calculate the frequency, the equation has the form: 1 ν = R H 2 n n 2 In this equation, n 1 is the same as n low in the table, and n 2 is the same as n high in the table. Use the values for ν, n low and n high from the first entry in the calculation table on the previous page, and solve this equation for R H (3 significant digits). Use the values of ν, n low and n high from the 5th entry and again calculate R H. R H (from 1) = s -1 R H (from 5) = s -1 Now use this value of R H (averaged if they differ) and the values of n 1 and n 2 in the 10th entry to calculate ν for the 10th data. ν (calculated) = s -1 Compare this calculated value to the value of the observed ν in the second column of the 10th entry. ν (experimental) = s-1 Does the calculated value agree with the observed value? If not, repeat your calculations. QUESTION When the Sun s spectrum is carefully observed, a great number of spectral lines are apparent. If the lines are viewed in groups, various elements are identified to cause specific groups of lines. For example, each of the lines seen in the mercury spectrum occurs in the Sun s spectrum. Helium is named after Helios, the Greek name for the Sun, because evidence for its existence was observed in 1868 by examining the sun s spectrum. Since helium was not found on earth until 1895, how do you suppose observers knew they had found a new element in 1868? 5

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