The
N a t u reo
f
L i 9 ht
99
Visible light
I~I
i
emission of a blackbody isJn~~iselyproportional
to its
According
the wavelength
maximum
temperature toinWien's
kelvins.law,
In 6thef
words, if theoftemperature
of the blackbody doubles, its wavelength of maximum emission is
halved, and vice versa. For example, Figure 5-10 shows blackbody
curves for temperatures of 3000 I~, 6000 K, and.12,QQO K. From
Wien's law, a blackbody with a temperature of 6000 K has a wavelength of maximum emission A
= (0.0029 K m)/(6000
K) =
4.8 X 10-7 m = 480 nm, in the visible,part of the electromagnetic
spectrum. At 12,000 K, or twice' the, temperature, the blackbody
has a wavelength of maximum emission halT as great, or Arnax =
240 nm; this is in the ultraviolet. At 3000 K, just half our original
temperature, the value of Arnax is twice the original value-960 nm,
which is an infrared wavelength. You can see that these wavelengths agree with the peaks of the curves in Figure 5-10.
•~t-J\M $",
Blackbody curve at 5800 K
Sun's intensity curve
o
500
1000
2000
3000
Wavelength (nrn) ~
fi?ure 5-11
The Sun as a Blackbody
This graph shows that the intensity of sunlight
over a wide range of wavelengths
(black curve) is a remarkably close match
to the intensity of radiation coming from a blackbody at a temperature
5800 K (red curve). The measurements
above the Earth's atmosphere
sunlight).
of
of the Sun's intensity were made
(which absorbs ce,rtain wavelengths
It's not surprising that the Sun's spectrum
of
peaks at visible
wavelengths; the human eye evolved to take advantage of the most plentiful
light available.
5-4 I Wit'n's law and tht' Stt'fan-Boltzmann
law art' useful
tools for analyzinq qlowinq objects like stars
The mathematical formula that describes the blackbody curves in
Figure 5-10 is a rather complicated one. But there are two simpler
formulas for blackbody radiation that prove to be very useful in
many branches of astronomy. They are used by astronomers who
investigate the stars as well as by those who study the planets
(which are dense, relatively cool objects that emit infrared radiation). One of these formulas relates the temperature of a blackbody
to its wavelength of maximum emission, and the other relates the
temperature to the amount of energy that the blackbody emits.
Figure 5-10 shows that the higher the temperature (T) of a blackbody, the shorter its wavelength of maximum emission (Arnaxl- In
1893 the German physicist Wilhelm Wien used ideas about both
heat and electromagnetism to make this relationship quantitative.
The formula that he derived, which today is called Wien's law, is
Wien's law for a blackbody
0.0029 K m
T
Arnax
= wavelength of maximum
emission of the object
(in meters)
T
=
temperature
of the object (in kelvins)
maximum emission in meters. If you want to convert the
c)\uD,Tt04Remember that
Wien's lawyou
involves
the wavelength
of
•
wavelength
to nanometers,
must multiply
the wavelength in meters by (109 nm)/(l m).
Wien's law is very useful for determining the surface temperatures of stars. It is not necessary to know how far away the star is,
how large it is, or how much energy it radiates into space. All we
need to know is the dominant wavelength of the star's electromagnetic radiation.
The other useful formula for the radiation from a blackbody involves the total amount of energy the blackbody radiates at all
wavelengths. (By contrast, the curves in Figure 5-10 show how
much energy a blackbody radiates at each individual wavelength.)
Energy is usually measured in joules (J), named after the
nineteenth-century English physicist James Joule. A joule is the
amount of energy contained in the motion of a 2-kilogram mass
moving at a speed of 1 meter per second. The joule is a convenient
unit of energy because it is closely related to the familiar watt (W):
1 watt is 1 joule per second, or 1 W = 1 ]Is = 1 J s-l. (The superscript -1 means you are dividing by that quantity.) For example, a
100-watt lightbulb uses energy at a rate of 100 joules per second, or
100 J/s. The energy content of food is also often measured in joules;
in most of the world, diet soft drinks are labeled as "low joule"
rather than "low calorie."
The amount of energy emitted by a blackbody depends both on
its temperature and on its surface area. This makes sense: A large
burning log radiates much more heat than a burning match, even
though the temperatures are the same. To consider the effects of
temperature alone, it is convenient to look at the amount of energy
emitted from each square meter of an object's surface in a second.
This quantity is called the energy flux (F). Flux means "rate of
flow," and thus F is a measure of how rapidly energy is flowing out
of the object. It is measured in joules per square meter per second,
usually written as ]lm2/s or J m-2 S-l. Alternatively, because 1 watt
equals 1 joule per second, we can express flux in watts per square
meter (W/m2, or W m-2).
In 1879 the Austrian physicist Josef Stefan reported experimental evidence that the flux from a blackbody is proportional to the
fourth power of the object's temperature (measured in kelvins). Five
years after Stefan announced his law, another Austrian physicist,
11
00
I
CHAPTER
5
_TOOLsoii~H~(isTR
0 N 0 M[R' S
Using the Laws of Blackbody Radiation
Wien's law and the Stefan-Boltzmann law can therefore be
The
Sun
nearly perfect
used
to and
relatestars
the behave
surface like
temperature
of theblackbodies.
Sun or a
distant star to the energy flux and wavelength of maximum
emission of its radiation. The following examples show how
to do this.
~
'.
At a dIstance of 1 AU
j
from the Sun, this square meter
of area receives 1370 watts of
I
,
light power from the Sun.
EXAMPLE: The maximum intensity of sunlight is at a
wavelength of roughly 500 nm = 5.0 x 10-7 m. Use this
information to determine the surface temperature of the Sun.
Situation: We are given the Sun's wavelength of maximum
emission Amax' and our goal is to find the Sun's surface
temperature, denoted by T('). (The symbol (') is the standard
astronomical symbol for the Sun.)
Tools: We use Wien's law to relate the values of
Amax
and T(').
As written, Wien's law tells how to find A max if we
know the surface temperature. To find the surface
temperature from \nax' we first rearrange the formula, then
substitute the value of Amax:
AnswC?r:
T(') = 0.0029 K m
Amax
0.0029 K m = 5800 K
5.0 x 10-7 m
This is a very high temperature by Earth standards,
about the same as an iron welding arc.
RC?viC?w:
EXAMPLE: Using detectors above the Earth's atmosphere,
astronomers have measured the average flux of solar energy
arriving at Earth. This value, called the solar constant, is
equal to 1370 W m-2. Use this information to calculate the
Sun's surface temperature. (This provides a check on our
result from the preceding example.)
Situation: The solar constant is the flux of sunlight as
measured at the Earth. We want to use the value of the solar
constant to calculate T(').
Tools: It may seem that all we need is the Stefan-Boltzmann
law, which relates flux to surface temperature. However, the
quantity F in this law refers to the flux measured at the Sun's
surface, not at the Earth. Hence we will first need to calculate
F from the given information.
To determine the value of F, we first imagine a
huge sphere of radius 1 AU with the Sun at its center, as
shown in the figure. Each square meter of that sphere receives
1370 watts of power from the Sun, so the total energy
radiated by the Sun per second is equal to the solar constant
multiplied by the sphere's surface area. The result, called the
luminosity of the Sun and denoted by the symbol L('), is
AnswC?r:
L(') = 3.90 X 1026 W That is, in 1 second the Sun radiates
3.90 x 1026 joules of energy into space. Because we know the
size of the Sun, we can compute the energy flux (energy
emitted per square meter per second) at its surface. The
radius of the Sun is R(') = 6.96 X 108 m, and the Sun's surface
area is 4nR(') 2. Therefore, its energy flux F(') is the Sun's
luminosity (total energy emitted by the Sun per second)
divided by the Sun's surface area (the number of square
meters of surface):
L(')
F (') -
-2
4nR(')
3.90
X
1026W
6.41
X
107 W m-2
4n(6.96 x 108 m)2
Once we have the Sun's energy flux F('), we can use the
Stefan-Boltzmann law to find the Sun's surface temperature
T('):
Taking the fourth root (the square root of the square root) of
this value, we find the surface temperature of the Sun to be
T(') = 5800 K.
Our result for T(') agrees with the value we computed
in the previous example using Wien's law. Notice that the
solar constant of 1370 W m-2 is very much less than F('), the
flux at the Sun's surface. By the time the Sun's radiation
reaches Earth, it is spread over a greatly increased area.
RC?viC?w:
The
Nature
of
Light
1101
I
,··c
TOOLS OfTH[
ASTRONOM[R'S
TRAD[ (ontinul'd
EXAMPLE: Sirius, the brightest star in the night sky, has a
surface temperature of about 10,000 K. Find the wavelength at
which Sirius emits most intensely.
Our goal is to calculate the wavelength of maximum
emission of Sirius (A,maxl from its surface temperature T.
Situation:
Tools: We use Wien's law to relate the values of
A,max
and T.
Answer: Using Wien's law,
A,
max
= 0.0029 K m
T
0.0029 K m = 2.9 x 10-7 m = 290 nm
10000
, K
Situation: To compare the energy fluxes from the two stars, we
want to find the ratio of the flux from Sirius to the flux from
the Sun.
Tools: We use the Stefan-Boltzmann
law to'End the flux from
Sirius and from the Sun, which from the preceding examples
have surface temperatures 10,000 K and 5800 K, respectively.
Answer: For the Sun, the Stefan-Boltzmann law is Fa = aT a 4,
and for Sirius we can likewise write F" = aT,:4, where the
subscripts a and ':' refer to the Sun and Sirius, respectively. If
we divide one equation by the other to find the ratio of fluxes,
the Stefan-Boltzmann constants cancel out and we get
Our result shows that Sirius emits light most intensely
in the ultraviolet. In the visible part of the spectrum, it emits
more blue light than red light (like the curve for 12,000 K in
Figure 5-10), so Sirius has a distinct blue color.
Review:
EXAMPLE: How does the energy flux from Sirius compare to
the Sun's energy flux?
Ludwig Boltzmann, showed how it could be derived mathematically
from basic assumptions about atoms and molecules. For this reason,
Stefan's law is commonly known as the Stefan-Boltzmann law. Written as an equation, the Stefan-Boltzmann law is
Stefan-Boltzmann
law for a blackbody
F = energy flux, in joules per square meter of surface
per second
a
= a constant = 5.67 x 10-8 W m-2 K-4
T = object's temperature,
in kelvins
The value of the constant a (the Greek letter sigma) is known
from laboratory experiments.
The Stefan-Boltzmann law says that if you double the temperature of an object (for example, from 300 K to 600 K), then the
energy emitted from the object's surface each second increases by
a factor of 24 = 16. If you increase the temperature by a factor of
10 (for example, from 300 to 3000 K), the rate of energy emission
increases by a factor of 104 = 10,000. Thus, a chunk of iron at
room temperature (around 300 K) emits very little electromagnetic
radiation (and essentially no visible light), but an iron bar heated
to 3000 K glows quite intensely.
Box 5-2 gives several examples of applying Wien's law and the
Stefan-Boltzmann law to typical astronomical problems.
(5800 K)4
5800
Fa = Ta4
T",4 = (10,000
K)4 = (10,000)4=
Po"
8.8
Review: Because Sirius has such a high surface temperature, each
square meter of its surface emits 8.8 times more energy per
second than a square meter of the Sun's relatively cool surface.
I 5-51
Light has properties
of both waves
and particles
At the end of the nineteenth century, physicists mounted a valiant
effort to explain all the characteristics of blackbody radiation. To
this end they constructed theories based on Maxwell's description
of light as electromagnetic waves. But all such theories failed to
explain the characteristic shapes of blackbody curves shown in
Figure 5-10.
In 1900, however, the German physicist Max Planck discovered
that he could derive a formula that correctly described blackbody
curves if he made two radical assumptions-that
electromagnetic
energy is emitted in discrete, particlelike packets, and that the
energy of each such packet-today
called a photon-is
related to
the wavelength of light: The greater the wavelength, the lower the
energy of a photon associated with that wavelength. Thus, a photon
of red light (wavelength A, = 700 nm) has less energy than a photon
of violet light (A, = 400 nm). In this picture, light has a dual personality; it behaves as a stream of particlelike photons, but each photon
has wavelike properties. In this sense, the best answer to the question "Is light a wave or a stream of particles?" is "Yes!"
It was soon realized that Planck's photon hypothesis explains
more than just the detailed shape of blackbody curves. For
example, it explains why only ultraviolet light causes suntans
and sunburns. The reason is that tanning or burning involves a
chemical reaction in the skin. High-energy, short-wavelength
ultraviolet photons can trigger these reactions, but the lowerenergy, longer-wavelength photons of visible light cannot. Similarly,