Unformatted text preview: SelfStudy 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 46 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, xrays, 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 lowercase 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 lowercase 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 multiterm 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 Xrays 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 metricprefix 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 nonSI 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 rewritten 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 blackbody 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 problemsolving 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 highenergy 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 bluegreen, one blueviolet, 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 bluegreen line consists of waves with a frequency of 615 THz. What is the
energy of these waves?
c. The blueviolet 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 Hatom 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 Hatom electron must be found on one of the energy levels of the staircase: it
cannot rest between steps.
The steps of the Hatom 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 Hatom 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 Hatom, such as from heat, energy waves, or high
voltage, the electron can be promoted up the staircase. If the Hatom electron is above the
bottom step, it is unstable and said to be in an “excited state.”
An Hatom 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 Hatom spectrum. To calculate
the energy of the lines in the spectrum, let’s add energy values to this Hatom model. The H Energy Levels
Using this equation: En = ― 21.8 x 10―19 J (per each atom)
n2
calculate and fillin each missing En value for the steps of the Hatom 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 Hatom 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 Hatom 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 Hatom 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 HAtom 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 HAtom Visible Spectrum
Color Wavelength (nm) Frequency (Hz) red 656 bluegreen 486 6.15 x 1014 blueviolet 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 Hatom 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 Hatom visible spectrum
a. using Planck’s equation.
b. Use the energylevel diagram for the Hatom to determine which transition
produces a photon with the energy of the violet line. 3. If sufficient energy is added to the Hatom 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 Hatom 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 Hatom 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. Blueviolet. See values in the table above.
d. For the blueviolet 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 highenergy ultraviolet lines of the Hatom 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 energylevel 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 energylevel 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 Hatom 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 energylevel 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 Hatom 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 Hatom electron in its ground state is in the 1s orbital. If sufficient energy is added
to an Hatom, 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
Hatom 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.
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