Frequency and Wavelength

Frequency and Wavelength - Self-Study Assignment You will...

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Unformatted text preview: Self-Study Assignment You will have a QUIZ on the attached pages on _____________________ . Your assignment is: READ the pages attached. WORK the examples in the lesson. Complete the pages as homework. To work the examples, • use a sheet of paper to cover below the * * * * * line, • try the problem on your paper, • then check your answer below the * * * * * line. Start early. This assignment will require 4-6 hours of work outside of class. Introduction to Chemistry Calculations ***** Module 19 – Light and Spectra ................................................................................... 381 Lesson 19A: Lesson 19B: Lesson 19B: Lesson 19C: Lesson 19D: Waves ................................................................................................................. 381 Wave Calculations and Consistent Units ....................................................... 386 Planck's Constant ............................................................................................. 391 The Hydrogen Atom Spectrum ....................................................................... 395 The Wave Equation Model .............................................................................. 402 Module 19 — Light and Spectra Module 19 — Light and Spectra ***** Lesson 19A: Waves Waves and Chemistry Electromagnetic energy includes gamma rays, x-rays, ultraviolet, visible, and infrared light, microwaves, and radio waves. Each of these types of energy occupies a different region of the electromagnetic spectrum. Chemical particles can both absorb and release electromagnetic energy. This absorption and release of energy can be a powerful tool in identifying chemical particles. Exposure to certain types of electromagnetic energy can also cause chemical particles to change and react. In some cases, the behavior of electromagnetic energy is best predicted by assuming that the energy is a particle, but in other cases, energy is best understood as a wave. Let us begin by investigating the properties of waves. Wave Terminology Crest 1 Wavelength (λ) 0 90 180 270 360 450 540 630 720 810 900 990 Trough The following are some of the components of a wave that are important in chemistry. 1. Wavelength is the distance between the crests of a wave, which is equal to the distance between the troughs of a wave. a. The symbol for wavelength is λ (the lower-case Greek letter lambda). b. Since a wave length is a distance, the units of wavelength are distance units, such as meters, centimeters, or nanometers. Page 381 Module 19 — Light and Spectra 2. Frequency is a number of events per unit of time. The unit for frequency is 1/time. For waves, frequency is the number of wave crests that pass a point per unit of time. a. In wave equations, the symbol for frequency is υ (the lower-case Greek letter nu). b. The SI unit for time, a fundamental quantity, is seconds. Because frequency is a derived quantity that is 1/time, the SI unit must be 1/seconds (s―1). The unit second―1 is also called a hertz (Hz). During calculations, it is best to write hertz as s―1 . Hertz and s―1 are equivalent and can cancel. c. When wave frequency is expressed as “cycles per second,” wave cycles are the entity being measured, and 1/seconds is the unit. When writing wave units, the term “wave cycle” or “cycle” is often included as a helpful label in conversion calculations, but is usually omitted as understood in equation calculations. 3. The speed of a wave is equal to its frequency times its wavelength. wave speed = λ υ = (lambda)(nu). Memorize the equation for wave speed in words, symbols, and names for the symbols. Wave Calculations Because wave relationships are often defined by multi-term equations, wave calculations are generally solved using equations rather than conversions. We will start with a simple problem that can be solve using both methods, but to practice with the equation that will be required for more complex calculations, solve the problem below using the equation method (for review, see Lesson 17D). Q. If ocean waves are traveling at 200. meters/minute and the crests pass a fixed point at a rate of 15.0 waves per minute, what is the wavelength, in meters? ***** Write the one equation learned so far for waves. Wave speed = λ υ List those three terms in a data table. After each term, write the data in the problem that corresponds to the term. Add a ? and the desired unit after the WANTED symbol. ***** Wave speed = 200 m/min. λ = ? meters υ = 15.0 wave cycles/min. = 15.0 min.―1 (speed units are distance over time) (the length of a wave is a distance) (frequency units are 1/time) When solving frequency calculations using equations, “wave cycles” is usually omitted as understood to be the object being measured. Solve the equation in symbols for the WANTED symbol, then substitute the DATA. Include the consistent units and check the unit cancellation. ***** Page 382 Module 19 — Light and Spectra SOLVE: Since Wave speed = λ υ λ in meters = speed = speed • 1 = 200. m • υ υ min. 1 = 13.3 meters 15.0 min.―1 Note in the unit cancellation in the denominator: min.• min.―1 = min.1• min.―1 = min.0 = 1 . Anything to the zero power equals one. Practice A 1. Write the SI units for a. Wavelength b. Frequency c. Energy d. Speed 2. Street lights containing sodium vapor lamps emit an intense yellow light at two close wavelengths. The more intense wave has a frequency of 5.09 x 1014 Hz. If light travels at the speed of 3.00 x 108 m • s―1 , what is the wavelength of this intense yellow wave in meters? (Use the equation method to solve.) Electromagnetic Waves The movement of electric charge creates electromagnetic waves. The waves propagate: they travel outward from the moved charge. The energy that was added to move the charge is carried outward by the waves. In a vacuum, all electromagnetic waves travel at the speed of light: 3.00 x 108 meters/second. The speed of light is the “speed limit of the universe:” the fastest speed possible for energy or matter. In wave calculations, the speed of light is given the symbol c. Electromagnetic waves slow when they travel through a medium that is denser than a vacuum, but when passing through air or other gases at normal atmospheric pressures, the speed of light does not slow sufficiently to affect most calculations in chemistry. For electromagnetic waves, this relationship will be true (and must be memorized): Speed of Light = c = λ υ = 3.00 x 108 m/s in vacuum or air Since c is a constant, υ and λ are inversely proportional. As wavelength goes up, frequency must go down. If υ goes up, λ must go down. Further, as long as we work in consistent units and in air or vacuum, since c is constant, a specific value for the frequency of an electromagnetic wave will always correlate to a specific value for its wavelength. The Regions of the Electromagnetic Spectrum The electromagnetic spectrum goes from very high to very low wavelengths and frequencies. Regions of the spectrum are assigned different names that help in predicting the types of interactions that the energy will display. However, all of these forms of energy Page 383 Module 19 — Light and Spectra are electromagnetic waves. The difference among the divisions of the spectrum is the length (or corresponding frequency) of the waves. The following table (no need to memorize) summarizes some of the general divisions of the electromagnetic spectrum. Frequency (s―1) Wavelength (m) Type of Electromagnetic Wave 1024 3 x 10─16 Gamma Rays 1021 3 x 10─13 1018 3 x 10─10 X-rays 1015 3 x 10─7 Ultraviolet, Visible, Infrared Light 1012 3 x 10─4 Microwaves 109 3 x 10─1 UHF Television Waves 106 300 Radio Waves Units For Frequency and Wavelength Measurements of wavelengths and frequencies often involve very large and very small numbers. Values are often expressed using SI prefixes such as gigahertz (GHz) or nanometers (nm). Prefixes needed most often are those for powers of three. Engineering Notation Scientific notation expresses a value as a significand between 1 and 10 times a power of 10. Engineering notation expresses values as a significand between 1 and 1,000 times a power of 10 that is divisible by 3. In wave calculations, answers are often preferred in engineering rather than scientific notation to ease conversion to the metric prefixes based on powers of three. Prefix Symbol Means tera T x 1012 giga- G x 109 mega- M x 106 kilo- k x 103 milli- m x 10―3 micro- μ x 10―6 nano- n x 10―9 pico- p x 10―12 Examples: Converting scientific to engineering notation, 5.35 x 10―4 m = 535 x 10―6 m in engineering notation ( = 535 micrometers = 535 μm) 9.23 x 1010 Hz = 92.3 x 109 Hz in engineering notation ( = 92.3 GHz ) To convert any exponential notation to engineering notation, adjust the exponent and decimal position until the exponent is divisible by 3 and the significand is between 1 and 1,000. (To review moving the decimal, see Lesson 1A). Try this example. Q. Convert to engineering notation, then to metric-prefix notation: 5.27 x 10―11 m ***** Page 384 Module 19 — Light and Spectra A. 5.27 x 10―11 m = 52.7 x 10―12 m in engineering notation = 52.7 picometers or 52.7 pm 52.7 x 10―12 is the only way to write the given quantity that results in both an exponent divisible by 3 and a significand between 1 and 1,000. Engineering notation, like scientific notation, results in one unique expression for each numeric value, and this makes answers easy to compare and check. During electromagnetic wave calculations, you should work in general exponential notation, then, at the end, convert your answers to engineering notation and then prefix notation if needed. Practice B: Do every other question. Complete the rest during your next study session. 1. By inspection, convert these to units in engineering notation, without prefixes. a. 5.4 GHz b. 720 nm c. 96.3 MHz 2. Convert these first to engineering notation, then to measurements with metric prefixes in place of the exponential terms. a. 47 x 10―7 m b. 347 x 104 Hz c. 1.92 x 10―8 m d. 14,920 x 10―1 Hz e. 0.25 x 1011 Hz f. 7,320 m ANSWERS Practice A 1. a. Wavelength is a distance, and the SI unit for distance is the meter (m). b. Frequency is defined as 1/time, and the SI unit for time is the second, so the SI unit of υ is 1/s or s―1, which is called a Hertz. c. Energy The SI unit for energy is the joule (J) d. Speed is defined as distance over time, so the SI units are meters/second (m • s―1) 2. Wave speed = λ υ Speed = 3.00 x 108 m • s―1 λ in meters = ? υ = 5.09 x 1014 Hz s―1 ( During calculations, write Hz as s―1 ) ? = λ in meters = speed = 3.00 x 108 m • s―1 = 0.589 x 10―6 m = 5.89 x 10―7 m υ 5.09 x 1014 s―1 Page 385 Module 19 — Light and Spectra Practice B 1a. 5.4 GHz = 5.4 x 109 Hz or s―1 1b. 720 nm = 720 x 10―9 m 1c. 96.3 MHz = 96.3 x 106 s―1 2a. 47 x 10―7 m = 4.7 x 10―6 m = 4.7 μm (if exponent is made larger, make significand smaller) 2b. 347 x 104 Hz = 3.47 x 106 Hz = 3.47 MHz 2c. 1.92 x 10―8 m = 19.2 x 10―9 m = 19.2 nm 2d . 14,920 x 10―1 Hz = 1.492 x 103 Hz = 1.492 kHz 2e. 0.25 x 1011 Hz = 25 x 109 Hz = 25 GHz (significand must be between 1 and 1,000) 2f. 7,320 m = 7.32 x 103 m = 7.32 km ***** Lesson 19B: Wave Calculations and Consistent Units When using equations to solve science calculations, the units of measurements must be consistent: for each quantity in the equation, one unit must be chosen. The rule is To solve with equations, first convert to consistent units, solve in consistent units, then convert the consistent WANTED unit to other units if needed. When using a specified value and units for the gas constant R, our rule was: convert the DATA to the units used in the constant. When using specific heat capacities (c), our rule was: convert to the unit of the most complex term in the DATA. In wave calculations, we will observe both of these rules. Specific constants will be required, but those constants will generally contain the most complex units in the problem. To choose and convert to consistent units, our steps will be 1. Write the equation needed for the problem. 2. If the equation has constants, in the DATA table, first write each constant’s symbol, value and units, then below write the symbols for the variables. 3. In the DATA table, after each variable symbol, write “ = ? “and then its chosen consistent unit. Example: DATA: λ =? m = (a wavelength is a distance) To choose the consistent units to write after each variable symbol, apply these steps. a. If the equation has constants, after each variable write a unit that is appropriate for the symbol and is consistent with the units in the constants. Example: If c= λυ is the equation that is needed, list DATA: c = 3.00 x 108 m ∙ s―1 λ υ =? m = = ? s―1 = (list constants in the equation first) (wavelength is a distance; the distance unit in c is meters) (υ = 1/time, the time unit used in the constant c is seconds) Page 386 Module 19 — Light and Spectra b. If there are no constants are in the equation, label each variable with an appropriate unit matching the units used in the WANTED unit. 4. If the unit that is WANTED is not specified, a. pick a WANTED unit to match the units in the constants of the equation. b. If no constants are used, write after the WANTED unit the SI unit for the quantity. 5. In the DATA, write the data supplied for each symbol, then convert that DATA to the consistent units if needed. 6. First solve for the WANTED symbol in the consistent unit, then convert to a different WANTED unit if specified. The problem below will help you to understand and remember the rules above. Q. When neon gas at low pressure is subjected to high voltage electricity, it emits waves of light. One of the more intense waves in the visible spectrum has a wavelength of 640. nm, perceived by the eye as red light. What is the frequency of this light in terahertz (THz)? Solve using the steps above. ***** Answer This problem involves a frequency (υ) and a wavelength (λ) for light. We know that light travels at the speed of light (c), a constant. So far, we know only one equation that relates those three symbols. So, to start, your paper should look like this: c= λυ DATA: c = 3.00 x 108 m ∙ s―1 λ= υ= For the speed of light (c), in conversion calculations the unit m/s must be used as a ratio and written in the top/bottom format, but in equations, it will simplify unit cancellation if the units are written in the “on one line” format: 3.00 x 108 m ∙ s―1 . Now assign a consistent unit to each variable. Since this equation has a constant (c), after each variable symbol write a unit that both measures the variable and matches one of the units used in the constant. Do that step, then check below. ***** Page 387 Module 19 — Light and Spectra Your DATA should look like the Rule 2a Example above, minus the (comments). After the = sign for each variable, write the data for that variable that is supplied in the problem. Then, in the DATA table, convert the supplied units to the consistent units if needed. For the WANTED variable, after the assigned consistent unit, write the unit WANTED in the problem if it is not the consistent unit. Do those steps, and then check your answer below. ***** Your paper should look like this. c= λυ DATA: c = 3.00 x 108 m • s―1 λ = ? m = 640. nm = 640. x 10―9 m υ = ? s―1 but THz is WANTED For λ , meters is the chosen consistent unit, so you must convert nm to m. The easy way is to substitute what the prefix means. To SOLVE, First solve the equation for the WANTED symbol in symbols. Substitute the DATA into the solved equation using the consistent units, and solve including the units. If needed, convert to the unit WANTED in the problem. Modify your work if needed and finish. ***** SOLVE: ? = υ in s―1 then THz = c λ = 3.00 x 108 m • s―1 = 4.69 x 1014 s―1 640. x 10―9 m That solves in the consistent unit. To finish, convert to the WANTED unit. ***** 4.69 x 1014 s―1 Hz • 1 THz = 469 THz 1012 Hz Done! This method of choosing to solve in the units of the constants, is arbitrary. You can use any consistent units to solve. In physics, it is usually preferred (and simplifying) to solve all problems in SI units. In many chemistry calculations, constants will be stated in SI units, matching the physics practice. However, “solving in the units of the constants” will save a few steps if data is provided in mL, grams, kcals, electron volts, BTUs, or other non-SI units, as may be the case in some science calculations. Page 388 Module 19 — Light and Spectra Try Q2. How many wavelengths of the 640. nm red neon light above would fit into one centimeter? (one wave cycle = 1 wave = 1 wavelength) ***** If you are not sure how to proceed, list the data, try to assign symbols, and see if the symbols fit a known equation. ***** The wanted unit is waves/cm, which is the inverse of wavelength, not wavelength. Plus, none of the data has a frequency, so the data does not match the one equation we know so far. Note that the data includes an equality, and that all of the data can be listed as equalities or ratios. That’s a hint to try conversions to solve. ***** WANT: DATA: ? waves cm one wave = 1 wavelength (you want the waves per one cm, a ratio unit) (2 measures of the same object) one wavelength = 640. nanometers Though “waves” or “wave cycles” is usually left out of wave equation calculations as understood, including “waves” may help when using conversions. If needed, adjust your work and then finish the problem. ***** One way of several to SOLVE is ? waves = 1 wave • cm 1 wavelength 1 wavelength • 1 nm • 10―2 m = 15,600 waves 640. nm 1 cm cm 10―9 m How many waves of red light fit into a centimeter? Quite a few. Summary: Frequency and Wavelength 1. Wavelength: the distance between the crests of a wave. The symbol is λ (lambda). The units are distance units: the base unit meters, or nanometers, etc. 2. Frequency: the number of times a wave crest passes a point per unit of time. The symbol is υ (nu). The units are 1/time . SI frequency units = wave cycles per second = 1/seconds = s―1 = hertz (Hz). In calculations, write hertz as s―1 so that units will cancel properly. 3. The speed of a wave is equal to its frequency times its wavelength. Wave speed = λ υ = (lambda)(nu). Page 389 Module 19 — Light and Spectra 4. Electromagnetic waves travel at the speed of light (symbol c). For all electromagnetic waves, c = λ υ = 3.00 x 108 m ∙ s―1 in vacuum or air. 5. To simplify solving wave calculations using equations, a. In the DATA table, list the constants first. Convert the DATA to consistent units: those the constants of the equation. If there are no constants, convert to the WANTED unit or to an SI unit. b. First solve in the consistent unit, then convert to the WANTED unit if needed. Practice (Additional practice with λ and υ will be provided in lessons that follow.) 1. If an AM radio station broadcasts a signal with a wavelength of 0.390 km, what is the frequency of the signal on a radio tuner, in kHz? 2. If there are 225 waves per centimeter, what is the wavelength of the waves in meters? ANSWERS 1. (The data is a λ and wanted is a υ. The equation that relates those two variables is:) c= λυ DATA: c = 3.00 x 108 m • s―1 λ = ? m = 0.390 km = 0.390 x103 υ = ? s―1 , then convert to kHz (list constants used in the equation first) m = 390 m 1 kHz = 103 Hz = 103 s―1 ( convert to units used in the constant) (listing metric conversions is optional) ? = υ in s―1 then kHz = c = 3.00 x 108 m • s―1 = 0.77 x 106 s―1 • λ 390 m 1 kHz = 770 kHz 103 Hz (Hertz and s―1 are equivalent and can cancel.) 2. (If you are not sure how to proceed, list the data, assign symbols, then see if the symbols fit a known equation.) WANTED: λ =? m DATA: 225 waves = 1 cm or ? meters/wave) (If no equation seems to fit the DATA, try conversions to solve. If needed, use that hint and finish. ***** Page 390 Module 19 — Light and Spectra If the WANTED unit is re-written as meters/wave, a ratio is wanted, and a ratio is in the data to start from. Arrange the given so that one unit is where is WANTED (Lesson 11B), but any order for these two conversions works.) ***** SOLVE: ? meters = 1 cm • wave 225 waves 10―2 m 1 cm = 4.44 x 10―5 m wave The wavelength is 4.44 x 10―5 m. ***** Lesson 19C: Planck’s Constant Energy and Frequency In 1900, the German physicist Max Planck, studying the black-body radiation emitted by objects at high temperature, discovered that energy is absorbed by or emitted from atoms in bundles of a small but constant size, or in multiples of that constant size. Planck’s discovery in equation form is written as ∆Eatom = ∆n ∙ h ∙ υ where n is an integer, and h is a number with units called Planck’s constant. Planck’s constant = h = 6.63 x 10―34 joule ∙ second Building on Planck’s work, in 1905 Albert Einstein proposed an explanation for the photoelectric effect: the observation that when UV light shines on a metal, the metal emits electrons. Einstein postulated that light can be considered to be made of small particles called photons. In Einstein’s formulation, electromagnetic energy has characteristics of both a wave and a particle, with photons acting as bundles of energy that travel at the speed of light. These bundles he called quanta. A single bundle is a quantum. The energy of each photon is correlated to its frequency as a wave. For photons, the general form of Planck’s equation that relates frequency and energy is Ephoton = h υ where h = 6.63 x 10―34 J ∙ s Because frequency must be positive, and Planck’s constant is small but positive, the equation E = hυ means that as the energy of a wave increases, its frequency must increase, and that higher frequency waves have higher energy. Page 391 Module 19 — Light and Spectra Since calculations using Planck’s constant involve electromagnetic waves, we can use our previous equation for the speed of those waves, solve that equation for υ: υ = c / λ , then write the photon energy equation as E = h υ c=λυ , or, substituting for υ , E= h∙ c λ These two general forms of Planck’s equation are equivalent. The first solves for energy in terms of frequency, the second in terms of wavelength. The first should be memorized, and the second either memorized (“for wavelength, the two constants are on top”) or derived as needed. Together, these equations mean that for electromagnetic waves, the three variables energy, frequency, and wavelength are directly correlated: If you know any one, you can calculate both of the other two. Further, it will always be true that as photon energies go up, the corresponding frequencies go up and wavelengths go down. Waves with higher energy have higher frequency and shorter wavelength. Calculations Using Planck’s Equation In these lessons, you will need to memorize the equations for waves and the value of the speed of light (which is used quite often in science), but the value for Planck’s Constant (h) will be supplied when needed in problems. When using Planck’s equation, to simplify problem-solving we will use the same rules as other equations. Identify the needed equation. In the DATA table, list the constants in the equation first. After each symbol for a variable, write a consistent unit. Choose the units of the constants if the equation has constants. If not, choose the WANTED unit if it is specified, or choose the SI unit for each quantity. In the DATA table, convert DATA to the consistent units. Solve for the WANTED symbol in the consistent unit, then convert to other WANTED units if needed. Try those steps on this problem. Q. Cosmic rays are high-energy radiation that enters the earth’s atmosphere from space. The energy of a single cosmic ray photon can be as high as 50. joules. What would be the frequency of this radiation in Hz? ( h = 6.63 x 10―34 J ∙ s ) ***** To decide which equation is needed to solve a problem, try this method: as you read the problem, write the symbol that fits the unit for each item of WANTED and DATA you encounter. Simply listing the symbols as you read will often quickly identify the equation that you need to solve. ***** Page 392 Module 19 — Light and Spectra 50 J = E, ? = υ . The fundamental equation that relates E and υ is ? ***** E = hυ Write a data table and solve. ***** DATA: h = 6.63 x 10―34 J ∙ s (list constants first, use their units) E in J = 50 J υ in s―1 = ? ? = υ ( in Hz ) = E = h then convert to Hz WANTED 50. J 6.63 x 10―34 J ∙ s = 7.5 x 1034 s―1 Hz Photons with this is extremely high energy and frequency are produced by nuclear processes in stars. Let’s try a problem with a more commonly encountered energy. Q2. A microwave oven warms food by producing radiation with a typical wavelength of about 12 cm. What is the energy of this wave? ( h = 6.63 x 10―34 J ∙ s ) ***** Answer As you read, you encounter a λ and a E. The equation that uses λ and E is ? ***** E= h ∙ c λ DATA: c = 3.00 x 108 m ∙ s―1 h = 6.63 x 10―34 J ∙ s (list the two constants first, use their units) E in J = ? λ in m = 12 cm = 12 x 10―2 m = 0.12 m ( h uses joules ) ( c uses m, c- = x 10―2) (In the DATA table, convert DATA to the consistent unit) ***** E = h ∙ c = ( 6.63 x 10―34 J ∙ s ) ( 3.00 x 108 m ∙ s―1 ) = 1.7 x 10―24 J λ 0.12 m Practice 1. Based on E = h υ and the rules for unit cancellation, what must the SI unit for Planck’s constant be? Why? Page 393 Module 19 — Light and Spectra 2. The human eye can generally see energy waves in the range of 400 to 700 nm. When hydrogen gas at low pressure is subjected to high voltage, it emits four waves of light in the visible region of the spectrum: one red, one blue-green, one blue-violet, and one violet. ( h = 6.63 x 10―34 J • s ) a. A photon of red light from the hydrogen spectrum has an energy of 3.03 x 10―19 J. What is the wavelength of this light in nanometers? b. The blue-green line consists of waves with a frequency of 615 THz. What is the energy of these waves? c. The blue-violet line has a wavelength of 434 nm. What is the frequency of these waves in Hz? ANSWERS 1. Since the SI unit for E is joules and for υ is s―1 , the units of E = h υ must be J = ( ? ) s―1. For unit cancellation to work, the units of h = ? must equal J • s (see Lesson 18D). 2a. (Part (a) involves E and λ . The equation that relates those variables is) E= h ∙ c λ DATA: h = 6.63 x 10―34 J ∙ s c = 3.00 x 108 m ∙ s―1 (list the two constants, convert DATA to those units) E = ? J = 3.03 x 10―19 J λ = ? m = then convert to nm WANTED ***** 1 nm = 10―9 m ( h uses joules ) ( c uses meters, solve consistent first ) (listing metric conversions is optional in DATA ) λ (in m) = h ∙ c = ( 6.63 x 10―34 J • s ) ( 3.00 x 108 m • s―1 ) = 6.56 x 10―7 m E 3.03 x 10―19 J λ (in nm) = 6.56 x 10―7 m = 656 x 10―9 m (engineering notation) = 656 nm or λ (in nm) = 6.56 x 10―7 m • 1 nm = 6.56 x 102 nm = 656 nm 10―9 m Page 394 Module 19 — Light and Spectra 2b. (Part (b) involves E and υ . The equation that relates those symbols is) E=hυ DATA: h = 6.63 x 10―34 J • s (list constants first. Use their units for WANTED and DATA) E in J = ? υ SOLVE: 2c. in s―1 = 615 THz = 615 x 1012 Hz = 615 x 1012 s―1 ( h uses seconds ) E (in J) = h υ = ( 6.63 x 10―34 J • s ) ( 615 x 1012 s―1 ) = 4.08 x 10―19 J (The problem has λ and υ. The equation that relates λ and υ for electromagnetic waves is) c= λυ DATA: c = 3.0 x 108 m • s―1 λ in m = 434 nm = 434 x 10―9 m υ in s―1 = ? , then convert to Hz (list constants used in the equation first) (nano means x 10―9 ) SOLVE: ? = υ in s―1 , then Hz = c = 3.00 x 108 m • s―1 = 0.00691 x 1017 s―1 = 6.91 x 1014 Hz λ 434 x 10―9 m ***** Lesson 19D: The Hydrogen Atom Spectrum Bohr and Atomic Spectra When atoms are heated to high temperatures, or vaporized and electrified, they glow: they give off light. Viewed through a prism or a diffraction grating, this light separates into thin lines of colors in the order of the colors of the rainbow: energy waves at sharply defined wavelengths. This is called the line spectrum of the atom. Each atom has a characteristic line spectrum. Astronomers can analyze the light from distant stars to identify the atoms that are present in the stars. In 1913, the Danish physicist Neils Bohr proposed that the light of an atom’s spectrum results from electrons falling from higher to lower fixed energy levels inside an atom, like marbles falling down a staircase (but one in which each step has a different rise). Page 395 Module 19 — Light and Spectra Based on a simple mathematical equation, En = ― 2.18 x 10―18 J/atom n2 Bohr’s model predicted exactly the observed wavelengths of the hydrogen atom spectrum. The energy levels inside the H-atom will be the basis for explaining the energy of the electrons in all other atoms as well. Understanding the behavior of the electrons in atoms will help in predicting the reactions of atoms and larger particles: the central focus of chemistry. Bohr’s Model For the Hydrogen Atom A neutral hydrogen atom consists of one proton at the center of the atom and one electron outside the nucleus. (A hydrogen nucleus may also include one or two neutrons, but these neutral particles do not affect the key electrical interaction between the positive proton and the negative electron.) Nearly all of the volume of the atom is due to the space occupied by the electron’s movement around the nucleus. The hydrogen electron can be described as existing at energy levels that are similar to a staircase. The H-atom electron must be found on one of the energy levels of the staircase: it cannot rest between steps. The steps of the H-atom staircase are uneven, but they follow a consistent pattern. The first step up from the bottom step is large, but the rise for each subsequent step is smaller. The staircase has an infinite number of steps, numbered from n = 1 to n = ∞ . Step n = 1 is the bottom step and n= ∞ is at the top of the staircase. In nature, systems tend to go to their lowest potential energy. The H-atom electron is therefore normally found at the bottom step n = 1, where it is said to be in its ground state. However, if energy is added to the H-atom, such as from heat, energy waves, or high voltage, the electron can be promoted up the staircase. If the H-atom electron is above the bottom step, it is unstable and said to be in an “excited state.” An H-atom electron promoted to an upper step is unstable because it is not at its lowest possible potential energy. The electron will therefore tend to fall back down the staircase. It falls like a marble, either hitting every step or skipping some steps, until it reaches the bottom step n = 1. The electron may “pause” on any step, but the electron cannot pause between steps. The electron falling down the staircase must lose energy. To do so, it emits energy waves (photons). The energy waves it emits are the lines of the H-atom spectrum. To calculate the energy of the lines in the spectrum, let’s add energy values to this H-atom model. The H Energy Levels Using this equation: En = ― 21.8 x 10―19 J (per each atom) n2 calculate and fill-in each missing En value for the steps of the H-atom listed below. To simplify upcoming calculations, we will write all of the En values as “ x 10―19 J .“ If you need help, check the sample calculation below. Page 396 Module 19 — Light and Spectra n = ∞ ________________ E∞ = 0 Joules … n = 6 ________________ E6 = ― 0.606 x 10―19 J n = 5 ________________ E5 = ↓ Calculate the remaining En values n = 4 ________________ E4 = n = 3 ________________ E3 = n = 2 ________________ E2 = n = 1 ________________ E1 = ***** For the energy level n = 6, : E6 = ― 21.8 x 10―19 J = ― 0.606 x 10―19 J 62 Finish calculating the remaining energies if needed. ***** Your answers should match those at the right. (The numbers are accurate, but the spacing between the steps is not drawn to scale.) The movement of electric charge creates electromagnetic waves. When the H-atom electron moves falls from one level to another, it must release an energy wave with an energy equal to the difference in energy between those two levels. The H-atom Energy Levels n = ∞ ________________ E∞ = 0 J … n = 6 ________________ E6 = ― 0.606 x 10―19 J n = 5 ________________ E5 = ― 0.872 x 10―19 J n = 4 ________________ E4 = ― 1.36 x 10―19 J n = 3 ________________ E3 = ― 2.42 x 10―19 J n = 2 ________________ E2 = ― 5.45 x 10―19 J n = 1 ________________ E1 = ― 21.8 x 10―19 J In your notebook, calculate the difference in energy between energy levels E3 and E2. ***** Page 397 Module 19 — Light and Spectra E3 minus E2 = ― 2.42 x 10―19 J ― (― 5.45 x 10―19 J) = + 3.03 x 10―19 J In Lesson 19C, Problem 2a, you calculated the wavelength of a wave with this energy. What was the value that wavelength? According to Problem 1a, what color would your eye perceive this wave to be? ***** 656 nm. When the H-atom electron falls from level n= 3 to n = 2, it produces an energy wave at a wavelength of 656 nm that your eye sees as red. When the electron falls between other energy levels, it emits light with other colors (plus light in other regions of the electromagnetic spectrum that the eye cannot see). Practice: The H-Atom Spectrum Do Problems 1 and 3 below. Do more if you need more practice. In Lesson 19C, Practice 2, and in the lesson above, we calculated these values for the H-Atom Visible Spectrum Color Wavelength (nm) Frequency (Hz) red 656 blue-green 486 6.15 x 1014 blue-violet 434 6.91 x 1014 violet Energy (J/atom) Transition 3.03 x 10―19 Step 3 2 410 4.08 x 10―19 Use this table as data for the problems below. Add your answers to the table as you complete the following calculations. ( h = 6.63 x 10―34 J • s ) 1. As the H-atom electron falls from level n = 5 to n = 2, based on the values for the energy levels (En) that you calculated in this lesson, a. what would be the energy of the wave emitted? b. What would be the wavelength of the wave in nm? c. What would be the color of the wave? d. What information do these answers allow you to add to the table above? Page 398 Module 19 — Light and Spectra 2 Calculate the energy of the violet line of the H-atom visible spectrum a. using Planck’s equation. b. Use the energy-level diagram for the H-atom to determine which transition produces a photon with the energy of the violet line. 3. If sufficient energy is added to the H-atom in its ground state (with the electron at n = 1), the electron can be ionized: it can be taken an essentially infinite distance away from the proton that is attracting the electron. The energy needed to remove the electron is the energy needed to promote the electron from level n = 1 to n = ∞. This value is termed the ionization energy of the atom. a. Based on the energy level diagram for H, how much energy would be needed to be added to take away (ionize) an H-atom electron? b. If an ionized electron falls from level n = ∞ down to level n = 1 in one transition, what will be the energy of the wave emitted by the electron? c. How does this energy value compare to the energy values of the lines in the visible spectrum listed in the chart above? d. What will be the wavelength of this energy wave in nm? e. If the human eye can generally see energy waves in the range of 400 to 700 nm, will you be able to see this wave when looking at an H-atom spectrum? 4. One of the lines in the ultraviolet region of the spectrum of hydrogen has a frequency of 2.46 x 1015 Hz. Which transition does this represent? 5. Using alternate units, the ionization energy of hydrogen is 13.6 electron volts (eV) per atom. Convert this ionization energy to kilojoules per mole (kJ/mol). (Use 1 eV = 1.60 x 10―19 J) 6. If the ionization energy of a hydrogen atom is ― 21.8 x 10―19 J, calculate the energy of level n = 3 ? ANSWERS 1. a. E5 ― E2 = ― 0.872 x 10―19 J ― (― 5.45 x 10―19 J) = + 4.58 x 10―19 J b. This problem involves E (from part a) and λ . The equation that relates those variables is E= h ∙c λ DATA: h = 6.63 x 10―34 J ∙ s c = 3.00 x 108 m ∙ s―1 (list the two constants, convert DATA to their units) E = ? J = 3.03 x 10―19 J λ =? m then convert to nm WANTED ( c uses meters ) nano means x 10―9 Page 399 Module 19 — Light and Spectra SOLVE: λ (in m) = h ∙ c = ( 6.63 x 10―34 J ∙ s ) ( 3.00 x 108 m ∙ s―1 ) = 4.34 x 10―7 m E 4.58 x 10―19 J = 434 x 10―9 m = 434 nm c. Blue-violet. See values in the table above. d. For the blue-violet wave, E = 4.58 x 10―19 J , and the electron is falling from step 5 2. 2. a. Only λ is known in the table; E is WANTED. The form of Planck’s equation that relates λ and E is E= h ∙c λ DATA: c = 3.00 x 108 m ∙ s―1 h = 6.63 x 10―34 J ∙ s (list constants first, use their units to solve) E = ? J = WANTED ( h uses joules ) λ = ? m = 410 nm = 410 x 10―9 m (convert to the meters used in c ) E = h ∙ c = ( 6.63 x 10―34 J ∙ s ) ( 3.00 x 108 m ∙ s―1 ) = 4.85 x 10―19 J λ 410 x 10―9 m b. Based on the pattern in the spectrum data table, the violet line should represent the 6 2 transition. Check it: E6 ― E2 = ― 0.606 x 10―19 J ― (― 5.45 x 10―19 J) = + 4.84 x 10―19 J Allowing for rounding, this agrees with the 2a answer. 3. a. Enough energy must be added to promote the electron from n = 1 to n = ∞, E∞ ― E1 = 0 J ― (― 21.8 x 10―19 J) = + 21.8 x 10―19 J b. From n = ∞ down to n = 1, the energy difference is the same: E∞ ― E1 = + 21.8 x 10―19 J The energy put in to pull away (ionize) the electron, starting from n = 1, must equal the total energy the electron releases when it returns to n = 1. c. + 21.8 x 10―19 J is a larger energy than those for the visible waves listed in the table. d. Part (d) involves E (from part c) and λ . The equation that relates those variables is E= h∙c λ DATA: h = 6.63 x 10―34 J ∙ s c = 3.00 x 108 m ∙ s―1 (list the two constants, convert DATA to those units) E in J = 21.8 x 10―19 J λ in m = ? then convert to nm nano means x 10―9 (for units consistent with c, solve in m first ) SOLVE: λ (in m) = h ∙ c = ( 6.63 x 10―34 J ∙ s ) ( 3.00 x 108 m ∙ s―1 ) = 0.912 x 10―7 m E 21.8 x 10―19 J = 91.2 x 10―9 m = 91.2 nm Page 400 Module 19 — Light and Spectra e. No. This wave has a shorter wavelength (and a higher frequency and energy) than the eye can see. 91.2 nm is in the ultraviolet (UV) region of the spectrum. The high-energy ultraviolet lines of the H-atom spectrum can be seen by special films. Prolonged exposure to UV radiation is dangerous to the eyes and skin. The sun produces high amounts of UV radiation, but most is absorbed by the ozone layer in the earth’s upper atmosphere before it can reach the earth’s surface. If the ozone layer decays, the incidence of skin cancer on earth would likely increase, among other harmful effects. 4. To find the transition using the energy-level diagram, the energy of the line is needed. The equation that relates E and υ is h = 6.63 x 10―34 J ∙ s DATA: E=hυ (list constants first, use their units for WANTED and DATA) E=?J = ? υ in s―1 = 2.46 x 1015 Hz s―1 ( h uses seconds ) E (in J) = h υ = ( 6.63 x 10―34 J ∙ s ) (2.46 x 1015 s―1 ) = 1.63 x 10―18 J SOLVE: Which transition has this energy? ***** Convert this E to 16.3 x 10―19 J for easier comparison to the numbers in the energy-level diagram. ***** E2 ― E1 = ―5.45 x 10―19 J ― (― 21.8 x 10―19 J) = + 16.4 x 10―19 J That matches the energy value calculated from the frequency above, allowing for rounding. For all of the transitions where the electron falls to n = 1, the H-atom will emit UV radiation. 5. WANTED: DATA: ? kJ mol 13.6 eV = 1 atom 1 eV = 1.60 x 10―19 J 6.02 x 1023 atoms = 1 mole atoms (mixes atoms with moles of atoms) A ratio is WANTED. The data is two ratios/equalities/conversions. Try conversions (Lesson 11B). ***** ? kJ = 1.60 x 10―19 J • 1 kJ • mol 1 eV 103 J 13.6 eV • 6.02 x 1023 atoms = 1.31 x 103 kJ 1 atom 1 mole mol 6. The ionization energy is the energy need to promote the electron from n = 1 to n = ∞. That energy is also the value in the energy-level equation En = ― 21.8 x 10―19 J n2 For energy level n = 3, E3 = ― 21.8 x 10―19 J = ― 2.42 x 10―19 J (per atom) 32 ***** Page 401 Module 19 — Light and Spectra Lesson 19E: The Wave Equation Model Schrődinger’s Wave Equation Though the Bohr model explained the spectrum of hydrogen, other aspects of the model were less successful at explaining the behavior of the hydrogen electron. In 1926, the German physicist Erwin Schrődinger developed equations which described the electron as if it were a wave, similar to the “standing waves” created by stringed instruments. Solutions to these equations are termed the quantum mechanical (wave) model for the atom. This model remains today our best explanation for the behavior of the hydrogen electron, and it is the basis for predicting the behavior of electrons in all other atoms as well. Solutions to the wave equation generally involve complex mathematics, and one question may have multiple solutions. In many respects, however, the wave equation produces a model for the hydrogen atom based on mathematical patterns that are elegant in their simplicity. The following points are part of the description of the hydrogen atom based on quantum mechanics. Predictions of the Wave Equation For Hydrogen 1. Inside the hydrogen atom are the levels described by Bohr: n = 1, 2, 3, …., ∞. When used with the wave equation, these numbers are called principal quantum numbers. Higher quantum numbers represent higher energy levels. These energy levels have the En values calculated by Bohr’s En equation. 2. Around the H-atom nucleus are orbitals that describe the space where an electron is likely to be found. The orbital has an energy and a shape, but the location of an electron in an orbital is described in terms of probability. The shape of the orbital describes where the electron will be 90% of the time. 3. At each principal quantum number n, there are n2 total orbitals, and n different types of orbitals. a. At level n = 1, there is one orbital, the 1s orbital. An s orbital has a spherical symmetry around the nucleus: in an s orbital, at a given distance from the nucleus in all directions, there is an equal chance of finding an electron. b. At level n = 2, there are two types of orbitals and four total orbitals: one 2s orbital with spherical symmetry, and three 2p orbitals. The three p orbitals are perpendicular to each other: they can be described as falling on x, y, and z axes around the nucleus. c. At level n = 3, there are three types of orbitals and nine total orbitals: one spherical 3s orbital, three perpendicular 3p orbitals, and five 3d orbitals. Most (but not all) of the d orbitals are diagonal to the p orbitals. Page 402 Module 19 — Light and Spectra d. At level n = 4, there are four types of orbitals and 16 total orbitals: one 4s, three 4p , five 4d, and seven 4f orbitals. (It may help to remember the spdf order of the orbitals as “stupid pirates die fighting.”) 4. The H-atom electron in its ground state is in the 1s orbital. If sufficient energy is added to an H-atom, its electron can be promoted into one of the higher energy orbitals. The above points can be summarized by a diagram. Below is the model for the H-atom predicted by the wave equation for the first four principal quantum numbers. Each line ( ___ ) represents an orbital. The Hydrogen Atom: Orbitals For the First Four Energy Levels 4s __ 4p __ __ __ 4d __ __ __ __ __ 4f __ __ __ __ __ __ __ ( = 16 orbitals) 3s __ 3p __ __ __ 3d __ __ __ __ __ ( = 9 orbitals) 2s __ 2p __ __ __ ( = 4 orbitals ) 1s __ ( = 1 orbital ) Practice Commit the diagram above to memory, then do the problems below. 1. At level n = 4 of the hydrogen atom are a. how many types of orbitals? b. How many total orbitals? c. Write the number and letter used to identify the types of orbitals, and list the number of orbitals there will be of each type, at n = 4. 2. At level n = 5, the H atom has orbitals designated 5s, 5p, 5d, 5f, and 5g. How many orbitals are there of each type? ANSWERS 1. a. 4 types of orbitals b. 16 total orbitals c. 1 4s, 3 4p, 5 4d, and 7 4f orbitals 2. At n = 4, there are n2 = 16 total orbitals: 1 s, 3 p, 5 d, and 7 f . At level n = 5, there must be n2 = 25 total orbitals. 25 – 16 = 9 additional orbitals at n = 5. At level n = 5, there must be one 5s, three 5p, five 5d, seven 5f, and nine 5g orbitals. ***** Page 403 Module 19 — Light and Spectra Summary: Light and Spectra 1. Wavelength: the distance between the crests of a wave. The symbol is λ (lambda). The units are distance units: either the base unit meters, or nanometers, etc. 2. Frequency is a number of events per unit of time. The units are 1/time. For waves, frequency is the number of wave crests that pass a point per unit of time. In wave equations, the symbol for frequency is υ (nu). SI frequency units: In equations, use 1/seconds = s―1 = hertz (Hz). In conversions, use wave cycles or wave lengths/second In equation calculations, write hertz as s―1 so that units will cancel properly. 3. The speed of a wave is equal to its frequency times its wavelength. Wave speed = λ υ = (lambda)(nu). 4. Electromagnetic waves travel at the speed of light (symbol c). For all electromagnetic waves, c = λ υ = 3.00 x 108 m ∙ s―1 in vacuum or air. 5. To simplify solving wave calculations using equations, a. In the DATA table, list the constants first. Convert the DATA to consistent units: those of the constant of the equation if there is one, or those used in the WANTED unit, or an SI unit, in that order. b. SOLVE for the consistent unit, then convert to a different WANTED unit if needed. 6. Planck’s equation that relates frequency and electromagnetic energy is E=hυ or E= h∙ c λ where h = Planck’s constant = 6.63 x 10―34 J ∙ s Higher frequency waves have higher energy and lower wavelength. 7. The energy levels inside a hydrogen atom can be calculated by En = ― 21.8 x 10―19 J (per each atom) n2 The spectrum of the hydrogen atom can be explained by assuming the hydrogen atom electron moves between these energy levels. ##### Page 404 ...
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