Inverse Square Law, Blackbody Radiation The Inverse Square Law for Radiation The amount of energy emitted in one second by a source of light is called...

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The Inverse Square Law for Radiation The amount of energy emitted in one second by a source of light is called its luminosity and is measured in watts. A source of light with a luminosity of 1 watt emits one joule of energy per second. The luminosity of the Sun is 3.86×1026 watts. As light travels away from a spherical source, its energy spreads out over the surfaces of spheres of increasing radius as shown in the figure. How bright the light appears to a detector (such as the eye or a photometer) depends on how fast the radiant energy enters the detector through the aperture (the pupil in the case of an eye). This depends on the flux (energy crossing a unit of area in a unit of time). time)

Flux =

Energy Luminosity = Area × Time Area

F1 ⎛ r2 ⎞ =⎜ ⎟ F2 ⎝ r1 ⎠

2

F=

Area of a sphere = 4πr2

L A

The flux due to a spherical source is inversely proportional to the square of the distance from the source.

r1 r2

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Blackbody Radiation Blackbody = something that absorbs all electromagnetic radiation incident on it. A blackbody does not necessarily look black. Its color depends p on its temperature. p The Sun and other stars behave approximately like blackbodies. The amount of electromagnetic radiation, with a given wavelength, emitted by a square meter of a blackbody’s surface in one second is given by the Planck blackbody radiation law. It is displayed as a graph on the following slide. The temperature of an object is a measure of its random kinetic energy.

The temperature scale used by physicists and astronomers is the absolute Kelvin scale. The temperature at which a body can emit no energy is absolute zero. One Kelvin = 1 Celsius degree = 1.8 Fahrenheit degree. The freezing point of water is about 273 K = 32oF.

flux in watts per square m meter

Comparison of the Solar (T = 5800 K) Continuous Spectrum with that of a 3500 K Star

wavelength in nanometers

The 3500 K star is brightest at a wavelength around 828 nm. This is infrared radiation, so we don’t see it. Since the visible part of this star’s spectrum is strongest at red and orange wavelengths, it has a faint red - orange color.

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flux in watts per square meter

Planck’s Law for Blackbody Radiation For a star with T = 7500 K

For a star with T = 5800 K The visible spectrum (about 400 to 700 nm) is superimposed on the graph). wavelength in nanometers

The peak of the graph for the star with a surface temperature of 7500 K is at a wavelength of 386 nm (violet). In the visible light from this star, there would be a bit more blue and violet than the other colors. The star would have a faint blue color. Our Sun has a surface temperature of 5800 K. The wavelength at which it is brightest is 500 nm (green). It has a faint yellow color.

Wien’s Law and the Stefan-Boltzman Law 7000 K

Notice that (a) the higher the temperature, the shorter the wavelength at which the star is brightest and (b) the higher the temperature the greater the area under the graph.

5800 K

F

3500 K

(a) is Wien’s law. Stated more precisely, it

says that the wavelength λmax at which the blackbody is brightest is inversely proportional to its absolute temperature T. If T is in Kelvins and λ is in nm, then λ=

2.898 × 106 T

(b) is the Stefan-Boltzman Law: The flux from the blackbody’s surface (F) is directly proportional to the fourth power of its temperature. If F is in watts per square meter and T is in Kelvins, F = σT 4

W ⎞ ⎛ −8 ⎜ σ = 5.67040 × 10 ⎟ m2K 4 ⎠ ⎝

4

FA ⎛ TA ⎞ =⎜ ⎟ . FB ⎝ TB ⎠

3

Example 2: The surface temperature of the Sun is about 5800 K, and the surface temperature of Sirius is about 11,000 K. Compare the surface flux (the electromagnetic energy emitted by 1 m2 of surface in 1 second) of the star Sirius with the surface flux of the Sun.

A = Sirius and B = Sun FA ⎛ 11000 ⎞ =⎜ ⎟ FB ⎝ 5800 ⎠

4

TA = 11,000 K

FA 4 = (1.897 ) FB

TB = 5800 K

FA = 12.9 FB

Sirius is about 13 times brighter than the Sun.

Example 3: At what wavelength does Sirius emit the most electromagnetic energy? λ=

2.898 × 106 . T

T = 11,000 K

λ=

2.898 × 106 . 11000

λ = 263 nm

This is a bit shorter than violet, so Sirius is brightest at ultraviolet wavelengths. The visible wavelengths at which it is brightest are violet and blue, so Sirius has a pale blue color.

Estimating the Temperature of a Planet rate of absorption of solar energy rate of emission of blackbody radiation = by the planet by the planet

L: = Solar luminosity Sunlight

a = albedo = fraction of incident light that is reflected.

1 − a = fraction that is absorbed

πR 2 = cross section of the planet illuminated by the Sun. L: = flux of solar energy at distance d from the Sun. 4πd 2

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πR 2

L: = solar energy hitting the planet's surface in a unit of time 4πd 2

(1 − a ) πR 2

L: = solar energy absorbed by the planet in a unit of time 4πd 2

σT 4 = energy emitted by a unit of the planet's surface in a unit of time. 4πR 2 = surface area of the planet.

4πR 2σT 4 = energy emitted by the planet's in a unit of time. 4πR 2 σT 4 = (1 − a ) πR 2

L: ⎤ ⎡ T = ⎢(1 − a ) 16 σπd 2 ⎥⎦ ⎣

L: 4πd 2

1/ 4

4σT 4 = (1 − a )

1/ 4

L: 4πd 2

T 4 = (1 − a )

L: 16σπd 2

⎛ L ⎞ ⎛1− a ⎞ =⎜ : ⎟ ⎜ 2 ⎟ ⎝ 16σπ ⎠ ⎝ d ⎠

1/ 4

1/ 2

⎛ 1− a ⎞ T = 279K ⎜ ⎟ ⎝ d AU ⎠

d AU = distance from the Sun in AU.

Example 4: Earth’s average albedo is approximately 0.3. Calculate its average temperature in (a) Kelvin, (b) Celsius, and (c) Fahrenheit. 1/ 2

⎛ 1 − 0.3 ⎞ T = 279K ⎜ ⎟ 1 ⎝ ⎠

= 233K

⎛9 ⎞ = ( 233 − 273) C = −40°C = ⎜ ( −40 ) + 32 ⎟ = −40°F 5 ⎝ ⎠

Example 5: Repeat Example 4 for Venus (a = 0.76, dAU =0.723). Example 6: Repeat Example 4 for Mercury (a = 0.1, dAU =0.387).

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