Transcript Today`s Objectives - RanelaghALevelPhysics

Blackbody Radiation
Astrophysics Lesson 9
Homework
 None today, module exams?
Learning Objectives
 Define the term blackbody.
 Sketch and describe how shape of black body
curves change as temperature is increased.
 Use Wien’s displacement law to estimate blackbody temperature of sources.
 Use Stefan’s law to estimate area needed for
sources to have same power output as the sun.
 Recap inverse square law, state assumptions in
its application.
Definition of a Blackbody
• A body that absorbs all wavelengths of
electromagnetic radiation and can emit all
wavelengths of electromagnetic radiation.
Pure black surfaces emit radiation strongly and in
a well-defined way – this is called blackbody
radiation.
It is a reasonable approximation to assume that
stars behave as black bodies.
Black Body Radiation
• The blackbody radiation of stars produces a
continuous spectrum.
• A graph of intensity against wavelength for
black body radiation is known as a black body
curve.
• The blackbody curve is dependent on the
temperature.
Black Body Curve Shape
Note the following for black bodies:
a hot object emits radiation across a wide range of
wavelength;
As the temperature of the object increases:peak of the graph moves towards the shorter
wavelengths.
the peak is higher
the area under the graph is the total energy radiated
per unit time per unit surface area.
Wien’s Displacement Law
• The peak wavelength, λmax, is the wavelength at
which maximum energy is radiated.
• This is inversely proportional to the temperature, T,
in Kelvin.
• This is called Wien's Displacement Law (as the peak is
displaced towards shorter wavelengths):•
max T  0.0029 mK
• Note the units of mK means a metrekelvin.
Worked Example
• What is the peak wavelength of a black body
emitting radiation at 2000 K? In what part of
the electromagnetic spectrum does this lie?
• λmax = 0.0029 mK ÷ 2000 K
• λ max = 1.45 x 10-6 m = 1450 nm
• This is in the infra-red region.
Question
• Betelgeuse appears to be red. If red light has a
wavelength of about 600 nm, what would the
surface temperature be?
• Why no green stars?
Answer
• Betelgeuse appears to be red. If red light has a
wavelength of about 600 nm, what would the
surface temperature be?
• T = 0.0029 mK ÷ 600 x 10-9 m
• T = 4800 K
• Why no green stars?
Why no green stars?
• You don't get green stars because the light from
stars is emitted at a range of wavelengths, so
there is mixing of colours. So those stars with a
λmax in the green region will actually appear to be
white.
Luminosity of Stars
• The luminosity of a star is the total energy
given out per second, so it's the power.
• From the graph the luminosity increases rapidly
with temperature, which gives rise to Stefan's
Law.
• The total energy per unit time radiated by a
black body is proportional to the fourth
power of its absolute temperature.
•
Stefan’s Law
• In other words double the temperature and the power goes up
sixteen times. In symbols:
L  AT
•
•
•
•
4
L – Luminosity of the star (W)
σ – Stefan's constant = 5.67 x 10-8 W m-2 K-4
A – surface area (m2)
T – surface temperature (K)
Applied to Stars
• We can treat a star as a perfect sphere (A = 4πr2) and a
perfect black body. So for any star, radius r, we can
write:
•
2
4
•
L  4r T
Inverse Square Law
• From Earth we can measure the intensity of the
star:-
L
I
2
4d
Where L is the luminosity of the star
d is the distance from the star
Question
• If the Sun has a radius of 6.96 x 108 m and a
surface temperature of about 6000 K, what is its
total power output?
• What is the power per unit area?
• What is the peak wavelength?
Answer
L  4r T
2
4
• L = 4 × π × (6.96 x 108 m)2 × 5.67 × 10-8 W m-2 K-4 x (6000
K)4
• L = 4.47 × 1026 W
•
• The power per unit area = 4.47 × 1026 W ÷ 6.09 × 1018 m2 =
7.34 × 107 W/m2
• Peak wavelength λmax = 0.0029 mK ÷ 6000 K = 4.82 × 10-7 m
= 482 nm