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## Blackbody Radiation

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**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 black-body 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.**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):- • 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 • λmax = 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 – 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:**Inverse Square Law**• From Earth we can measure the intensity of the star:- 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 ×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