Max Planck and Blackbody Radiation by xuyuzhu

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									                        Max Planck and Blackbody Radiation

Conventional physics would suggest that if energy was a continuous wave, then matter would absorb
and emit radiation continuously. But this does not happen. As steel is heated it glows red to white hot.
If continuous absorption occurred, then heated metal would pass through the entire visible light
spectrum (red, orange, yellow, green, blue, indigo, violet). However, some of the colors are not
emitted from a continuously heated metal. There was also the belief that a white hot metal might
become invisible as its energy reached the wavelength of ultraviolet light, but this does not happen.

A blackbody is a simplified, idealized scientist-created object that absorbs all EM radiation that falls
on it (no reflection or passing through).

However, in the real world, simplified models are not of much help. Many solids do emit light in the
same relative proportions as a black body, but not in the same amounts. Table 1 gives a summary of
visual temperature phenomena of solid bodies - for instance, a glowing piece of charcoal is a good
approximation to the blackbody.



Temperature
K         oC       Subjective color
750       480      faint red glow
850       580      dark red
1000      730      bright red, slightly orange
1200      930      bright orange
1400      1100     pale yellowish orange
1600      1300     yellowish white
> 1700   > 1400     white (yellowish if seen from a distance)




Table 1. The perceived color of heated solid bodies.

Due to the lack of continuous absorption and release of energy, Planck theorized that energy exists in
discrete, finite packets or bundles. He called these energy amounts “quanta” and he believed that
objects absorb and release energy in specific amounts. For example, our currency system is based
on the quantum of the cent (penny). Just as you cannot have 1.3 cents, energy only gets absorbed,
transferred, or emitted in 1, 2, 3, etc. quanta (you can’t have 1.3 quanta). Einstein built upon Planck’s
“quantum theory” with the photoelectric effect: light moves in physical packets called “photons” that
hold specific quanta energy values. Electrons could only be discharged from metals if, and only if, a
specific frequency of light was shone on the metal. For instance, violet light worked, while red light did
not no matter what the intensity and the duration of the red light was.


                    Electrons in Atoms: Introductory Comments

This unit is about where the electrons are located in an atom and how they behave.

For example, you will soon learn that the ground state electron configuration for sodium is 1s2 2s22p6
3s1. Once you learn the pattern, such configurations will come easy. However, the process by which
the above answer was discovered took decades and involved many, many people and lots of blind
alleys and wrong turns. The process will be simplified for you sot that you can better understand the
chemical behavior of substances without worrying about those wrong turns.

However, you will learn a little of the process. That is why units like this one typically start out with a
discussion of light and what a spectrum is. The reason for this is that the above configuration (and
many others) was discovered through intensive study of the spectra of the elements, principally
hydrogen's spectrum.

All these discoveries about how electrons in an atom behave must be able to account for the behavior
of the elements. Why are some elements very chemically reactive? Why are some not so reactive?
Why does one member of an otherwise similar-behaving group of elements engage in different
behavior from the others? Why, where certain chemicals react, is one substance produced, but not
another? Why cannot certain chemical compounds be made? The list goes on and on and on, as you
can well imagine.

I have no wise words that will help make this unit easy. It is not, because the ideas involved are very
complex. All I can say is, if you are stuck, do not give up. Take a break, eat some cake and go back
and try it again. You'll do just fine.
             Two Equations Governing Light's Behavior: Part One
                                                λν = c

There are two equations concerning light that are usually taught in high school. Typically, both are
taught without any derivation as to why they are the way they are. That is what I will do in the
following.


Equation Number One: λν = c

Brief historical note: I am not sure who wrote this equation (or its equivalent) first. The wave theory of
light has its origins in the late 1600's and was developed mathematically starting in the early 1800's. It
was James Clerk Maxwell, in the 1860's, who first predicted that light was an electromagnetic wave
and computed (rather than measured) its speed. By the way, the proof that light's speed was finite
was published in 1676 and the first reliable measurements of the speed of light, ones that were very
close to the modern value, took place in the late 1850's.

Each symbol in the equation is discussed below. Also, right before the example problems, there is a
mention of the two main types of problems teachers will ask using the equation. I encourage you to
take a close look at that section.


1) λ is the Greek letter lambda and it stands for the wavelength of light. Wavelength is defined as the
distance between two successive crests of a wave. When studying light, the most common units used
for wavelength are: meter, centimeter, nanometer, and Ångström. Even though the official unit used
by SI is the meter, you will see explanations and problems which use the other three. Less often, you
will see other units used; picometer is the most common one among the less-often used wavelength
units. Ångström is a non-SI unit commonly included in discussions of SI units because of its wide
usage.

Keep in mind these definitions:

one centimeter equals 10¯2 meter
one nanometer equals 10¯9 meter
one Ångström equals 10¯8 centimeter

The symbol for the Ångström is Å.
Most certainly, you will need to move easily from one unit to the other. For example, notice that 1 Å =
10¯10 meter. This means that 10 Å = 1 nm. So, if you are given an Ångström value for wavelength
and a nanometer value is required, divide the Ångström value by 10.


2) ν is the Greek letter nu. It is NOT the letter v, it is the Greek letter nu!!! It stands for the frequency
of the light wave. Frequency is defined as the number of wave cycles passing a fixed reference point
in one second. When studying light, the unit for frequency is called the Hertz (its symbol is Hz). One
Hertz is when one complete cycle passes the fixed point, so a million Hz is when the millionth cycle
passes the fixed point.

There is an important point to make about the unit on Hz. It is NOT commonly written as cycles per
second (or cycles/sec), but only as sec¯1 (more correctly, it should be written as s¯1; you need to
know both ways). The "cycles" part is deleted, although you may see an occasional problem which
uses it.

A brief mention of cycle: imagine a wave, frozen in time and space, where a wave crest is exactly
lined up with our fixed reference point. Now, allow the wave to move until the following crest is exactly
lined up with the reference point, then freeze the wave in place. This is one cycle of the wave and if
all that took place in one second, then the frequency of the wave is 1 Hz.

In any event, the only scientifically useful part of the unit is the denominator and so "per second"
(remember, usually as s¯1) is what is used. The numerator "cycles" is not needed and so its presence
is simply understood and, if writing a fraction is necessary, a one would be used, as in 1/sec.

3) c is the symbol for the speed of light, the speed at which all electromagnetic radiation moves when
in a perfect vacuum. (Light travels slower when passing through objects such as water, but it never
travels faster than when in a perfect vacuum.)

Both ways shown below are used to write the value. You need to be aware of both:

3.00 x 108 m/s
3.00 x 1010 cm/s

The actual value is just slightly less, but the above values are the one generally used in introductory
classes. (sometimes you'll see 2.9979 rather than 3.00.) Be careful about using the exponent and unit
combination. Meters are longer than centimeters, so there are less of them used above.
Since there are two variables (λ and ν), we can have two types of calculations: (a) given wavelength,
calculate frequency and (b) given frequency, calculate wavelength. By the way, these are the two
types of problems teachers generally ask on the test.

Note that we can easily rearrange the main equation to fit these two types of problems:

(a) given the wavelength, calculate the frequency; use this equation: ν = c / λ

(b) given the frequency, calculate the wavelength; use this equation: λ = c / ν




Example problem #1: What is the frequency of red light having a wavelength of 7000 Å?

The solution below depends on converting Å into cm. This means you must remember that the
conversion is 1 Å = 10¯8 cm. The solution:

(7000 x 10¯8 cm) (x) = 3.00 x 1010 cm/sec

Notice how I did not bother to convert 7000 x 10¯8 into scientific notation. If I had done so, the value
would have been 7.000 x 10¯5.

Note also that I effectively consider 7000 to be 4 significant figures. I choose to do this because I
know wavelength measurements are very accurate and that 6, 7, or even 8 sig figs are possible. At
an introductory level, you will not know this, so that is why I am telling you here. Also, the value for
the speed of light is known to nine significant figures, as in 299,792,458 m s¯ 1. However, 3.00 is good
enough for introductory work.

The answer is 4.29 x 1014 s¯1

Example problem #2: What is the frequency of violet light having a wavelength of 4000 Å?

The solution below depends on converting Å into cm. This means you must remember that the
conversion is 1 Å = 10¯8 cm. The solution:

(4000 x 10¯8 cm) (x) = 3.00 x 1010 cm/sec

Notice how I did not bother to convert 4000 x 10¯8 into scientific notation. If I had done so, the value
would have been 4.000 x 10¯5. Note also that I effectively consider 4000 to be 4 significant figures.
The correct answer is 7.50 x 1014 s¯1

Be aware that the range of 4000 to 7000 Å is taken to be the range of visible light. Notice how the
frequencies stay within more-or-less the middle area of 1014, ranging from 4.29 to 7.50, but always
being 1014. If you are faced with this calculation and you know the wavelength is a visible one (say
555 nm), then you know the exponent on the frequency MUST be 10 14. If it isn't, then YOU (not the
teacher) have made a mistake.

Example problem #3: What is the frequency of EMR having a wavelength of 555 nm? (EMR is an
abbreviation for electromagnetic radiation.)

First, let us convert nm into meters. Since one meter contains 109 nm, we have the following
conversion:

555 nm x (1 m / 109 nm)

555 x 10¯9 m = 5.55 x 10¯7 m

Inserting into λν = c, gives:

(5.55 x 10¯7 m) (x) = 3.00 x 108 m s¯1

x = 5.40 x 1014 s¯1

Example problem #4: What is the wavelength (in nm) of EMR with a frequency of 4.95 x 10 14 s¯1?

Substitute into λν = c, as follows:

(x) (4.95 x 1014 s¯1) = 3.00 x 108 m s¯1

x = 6.06 x 10¯7 m

Now, we convert meters to nanometers:

6.06 x 10¯7 m x (109 nm / 1 m) = 606 nm

Example problem #5: What is the wavelength (in both cm and Å) of light with a frequency of 6.75 x
1014 Hz?
The fact that cm is asked for in the problem allows us to use the cm/s value for the speed of light:

(x) (6.75 x 1014 s¯1) = 3.00 x 1010 cm s¯1

x = 4.44 x 10¯5 cm

Next, we convert to Å:

(4.44 x 10¯5 cm) x (1 Å / 10¯8 cm) = 4440 Å



                         Calculate the frequency when given the wavelength.

1) Calculate the frequency of EMR that has a wavelength of 1.315 micrometers.

2) Calculate the frequency of radiation with a wavelength of 442 nm.

3) Calculate the frequency of radiation with a wavelength of 4.92 cm.

4) Calculate the frequency of radiation with a wavelength of 8973 Å.

5) Calculate the frequency of radiation with a wavelength of 4.55 x 10¯9 cm.




1) Light with a frequency of 7.26 x 1014 Hz lies in the violet region of the visible spectrum. What is the
wavelength of this frequency of light? Answer in units of nm.

2) When an electron beam strikes a block of copper, x-rays of frequency 1.07 x 1019 Hz are emitted.
What is the wavelength of these x-rays? Answer in units of pm.

3) Calculate the wavelengths (in meters) of radiation a frequency of 5.00 x 1015 s¯1.

4) Calculate the wavelengths (in meters) of radiation a frequency of 2.11 x 10 14 s¯1.

5) Calculate the wavelengths (in meters) of radiation a frequency of 5.44 x 10 12 s¯1.

6) What is the wavelength of sound waves having a frequency of 256.0 sec¯1 at 20 °C? Speed of
sound = 340.0 m/sec. (The problem is about sound, but this does not change the basic idea of the
equation "wavelength times frequency = speed."
7) The radio station KUSC (in Southern California) broadcasts at 91.5 MHz. Calculate its wavelength
in meters.

An interesting little light trivia: light travels about one foot every nanosecond. You might try and work
out the proper calculation
             Two Equations Governing Light's Behavior: Part Two
                                                E = hν

There are two equations concerning light that are usually taught in high school. Typically, both are
taught without any derivation as to why they are the way they are. That is what I will do in the
following.


Equation Number Two: E = hν

Brief historical note: It is well-known who first wrote this equation and when it happened. Max Planck
is credited with the discovery of the "quantum," the discovery of which took place in December 1900.
It was he who first wrote the equation above in his announcement of the discovery of the quantum.

1) E is the energy of the particular quantum of energy under study. When discussing electromagnetic
quanta (of which light is only one example, x-rays and radio waves being two other examples), the
word photon is used. A photon (the word is due to Albert Einstein) is a quantum of electromagnetic
energy. The word quantum (quanta is the plural) is usually used in a more general sense, to describe
various ideas of quantum theory or even, as I just did, to describe the entire theory itself.

2) h stands for a fundamental constant of nature now known as Planck's Constant. By the way, the
discovery of the quantum had, and continues to have, many profound effects. Enough so that all of
science (especially physics) before 1900 is referred to as "classical" and the science since 1900 is
called "modern."

The value for Planck's Constant is 6.6260755 x 10¯34 Joule second. Please note that the unit is Joule
MULTIPLIED BY second. It is not a division, both Joule and second are in the numerator.

3) ν is the frequency of the particular photon being studied. The discussion about frequency above
applies here.

Before going on, I want to discuss one little issue (heh, heh, heh). Frequency is a wave-based idea.
What is it doing in a particle-based idea like the quantum? Good question. So much so that the term
"wave packet" is often used in discussing these ideas. Indeed, modern science now speaks of "wave-
particle duality" rather than "Light is a wave" or "Light is a particle." This whole area is profound and
can lead to years of probing discussion. Albert Einstein and Niels Bohr (who were great friends)
discussed these issues (and more) often over a period of many years, especially in the late 1920's
and early 1930's. Their discussions are still important enough to merit historical study today. Every
year, several books are published which delve into one or more of the implications of "wave-particle
duality."


Example problem #1: (a) Identify λν = c as either a direct or inverse mathematical relationship. (b) Do
the same for E = hν. (c) Write a mathematical equation for the relationship between energy and
wavelength. (d) Identify the equation for (c) as either direct or inverse.

The answers:


(a) λν = c is an inverse relationship. As one value (say the wavelength) goes up, the other value (the
frequency) must go down. Why? Because the product of the two must always equal the same value,
c, which is a constant.


(b) E = hν is a direct relationship. As the frequency increases, so does E. Why? Because h remains
constant. Notice that in this equation the two quantities which can change (ν and E) are on
DIFFERENT sides of the equation whereas in the inverse relationship in (a), both λ and ν are on the
same side, with the constant on the opposite side.

(c) Rearrange the light equation thusly:


ν=c/λ

Substitute for ν in E = hν:


E = hc / λ

Rearrange to group the variables and the constants:


Eλ = hc

(d) The mathematical relationship is an inverse one. Note the equation's similarity to λν = c, with two
values that can vary on the left side and a constant (h times c) on the right.


Example problem #2: How many Joules of energy are contained in a photon with λ = 550 nm? How
many kJ of energy is this?
I will first do the problem step by step, then in a more combined way.


Use ν = c / λ to get the frequency:

x = (3.00 x 108 m s¯1) / (550 x 10¯9 m)

This equals 5.4508 x 1014 s¯1. I left a couple extra digits in the answer and notice that the wavelength
is not in scientific notation. Why? Think about how I got from nm to m for the value used. (I also used
299,792,458 in the calculation, not 3.00 x 108. Sorry!! Using 3.00 x 108 gives 5.454 x 1014 s¯1.)


Now use E = hν to get the energy:

x = (6.6260755 x 10¯34 J s) (5.4508 x 1014 s¯1)

This equals 3.612 x 10¯19 J. Most emphatically, this is not the final answer. This is the energy for one
photon!!

Dividing the answer by 1000 to make the change to kilojoules, we get 217.5. Yes, this is my final
answer!

Example problem #3: How many kJ (of photons) of energy is contained in light with a wavelength of
496.36 nm?

The answer is 241.00 kJ.




Three more problems:

1) What is the energy of a photon of green light with a frequency of 5.76 x 10 14 s¯1.

2) A particular x-ray has a wavelength of 1.2 Å. Calculate the energy of 1000 photons with this
wavelength.

3) When excited, some atoms produce an emssion with a frequency of 7.25 x 10 12 Hz.
(a) calculate the energy, in Joules, for one photon with this frequency.
(b) calculate the energy, in kJ.
(c) Is this light visible? Why or why not?



Example problem #4: When it decays by beta decay, cobalt-60 also produces two gamma rays with
energies of 1.17 MeV and 1.33 MeV. For the 1.17 MeV photon, determine (a) the energy in Joules of
one photon.

Solution:

To solve this problem, you must know the relationship between electron-volts (eV) and Joules. I will
use 1 eV = 1.6022 x 10¯19 J for this problem.

(a) convert MeV to Joules:

1.17 MeV = 1.17 x 106 eV

(1.17 x 106 eV) x (1.6022 x 10¯19 J/eV) = 1.87 x 10¯13 J (to three sig. figs.)

(b) convert J of one photon to kJ/mol of photons:

(1.874574 x 10¯13 J) x (6.022 x 1023 mol¯1) = 1.129 x 1011 J/mol = 1.129 x 108 kJ/mol

Note the use of the unrounded off value for the energy of one photon.




"Eine neue wissenschaftliche Wahrheit pflegt sich nicht in der Weise durchzusetzen, dass ihre
Gegner überzeugt werden und sich als belehrt erklären, sondern vielmehr dadurch, dass die Gegner
allmählich aussterben und dass die heranwachsende Generation von vornherein mit der Wahrheit
vertraut gemacht ist." --- Max Planck


[A new scientific truth does not, generally speaking, succeed because the opponents are convinced
or declare themselves educated, however because they die and the new generations from the
beginning learn about it as the truth.]
Visible Colors in the hydrogen spectrum. Can you calculate the frequencies and energies equivalent
to the wavelengths?
                               Wavelength (nm)           Color
                                     656.2                red
                                     486.1                blue
                                     434.0             blue-violet
                                     410.1               violet

								
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