Star in a Box Exploring the lifecycle of stars. White slides are section headings, and are hidden from the presentation. Show or hide the slides in each.

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Star in a Box Exploring the lifecycle of stars

White slides are section headings, and are hidden from the presentation. Show or hide the slides in each section as appropriate to the level required. Rough guide to the levels: Beginner: KS3 Intermediate: KS4 (GCSE) GUIDE TO THIS PRESENTATION

INTRODUCTION Basics of what a star is and how we observe them. Level: Beginner +

Stars in the Night Sky

What is a Star?  A cloud of gas, mainly hydrogen and helium.  The core is so hot and dense that nuclear fusion can occur. −This is where the energy comes from that makes the stars shine.  The fusion converts light elements into heavier ones. −This is where all the atoms in your body have come from.

All the Stars in the Night Sky are Different  Luminosity: Tells us how bright the star is, i.e. How much energy is being produced in the core.  Colour: Tells us the surface temperature of the star. Rigel

Units of Luminosity WWe measure the luminosity of every day objects in Watts. –H–How bright is a light bulb? 10-20W BBy comparison, the Sun outputs: 380,000,000,000,000,000,000,000,000 Watts (380 million million million million Watts!) or 3.8 x Watts  This is the amount of energy it emits per second WWe measure the luminosity of other stars relative to the Sun.

Units of Temperature  Temperature is measured in Kelvin.  The Kelvin temperature scale is the same as the Celsius scale, but starts from -273 o. 0 K (or -273 o C) is known as “absolute zero” -273 o C-173 o C0 o C100 o C 0 K100 K273 K373 K 1000 o C 1273 K Kelvin = Celsius + 273

Measuring the Temperature  The colour of a star indicates its temperature.  Red stars are cold, and blue stars are hot.  The Sun is a yellow star, its temperature is 5800 K. Betelgeuse is a red supergiant. Its temperature is 3,000 K Rigel is a blue supergiant. Its temperature is 12,000K

Stefan-Boltzmann Law r

BLACK BODY RADIATION More detail about the colour and temperature of a star, using black body radiation. Level: Advanced +

Black Body radiation  A black body is… – A body that absorbs all wavelengths of EM radiation and can emit all wavelengths of EM radiation.  A star is a good approximation of a black body.  The intensity of each wavelength of radiation a star emits depends on its temperature.

Black Body Radiation

How hot is the Sun?  This is a graph of the Sun’s energy output – its ‘Blackbody Curve’  Its peak wavelength is around 0.5μm, −this is in the visible region of the Electromagnetic Spectrum Intensity

Wien’s Law  The peak intensity of the radiation is related to the surface temperature of the star. Temperature (K) = Wien’s constant (m.K) / peak wavelength (m) T = b max b = x m.K  Looking at the graph on the previous slide, can we determine how hot the surface of the Sun is?

Wien’s Law

HERTZSPRUNG-RUSSELL DIAGRAM An introduction to the H-R diagram, on which various stars will be plotted – try to get the students to suggest where they might appear before they are plotted. Level: Beginner +

The Hertzsprung Russell Diagram We can compare stars by showing a graph of their temperature and luminosity. Where do the stars in the night sky fit on this graph? Sirius Betelgeuse Rigel Aldebaran

Luminosity (relative to Sun) , Temperature (Kelvin) 25,00010,0007,0005,0003,000 We start by drawing the axes: Luminosity up the vertical axis (measured relative to the Sun) Temperature along the horizontal axis (measured in Kelvin) Where would you mark the Sun on the plot? It has Luminosity of 1 relative to itself Its temperature is 5800 K The stars Vega and Sirius are brighter than the Sun, and also hotter. Where would you put them? Some stars are much cooler and less luminous, such as the closest star to the Sun, Proxima Centauri. Where would you plot these? These stars are called red dwarfs. Sun Sirius Vega Proxima Centauri In fact, most stars can be found somewhere along a line in this graph. This is called the “Main Sequence”. Main Sequence

Luminosity (relative to Sun) , Temperature (Kelvin) 25,00010,0007,0005,0003,000 Sun Sirius Vega Proxima Centauri Main Sequence Rigel Betelgeuse Deneb Arcturus Aldebaran Sirius B But not all stars lie on the main sequence. Some, such as Arcturus and Aldebaran, are much brighter than the Sun, but cooler. Where would these lie on the diagram? These are red giant stars. The bright star Betelgeuse is even more luminous than Aldebaran, but has a cooler surface. This makes it a red supergiant. Even brighter than Betelgeuse are stars like Deneb and Rigel, which are much hotter. These are blue supergiants. Some of the hottest stars are actually much fainter than the Sun. Which corner would they be in? These are white dwarfs, such as Sirius B which orbits Sirius.

Luminosity (relative to Sun) , Temperature (Kelvin) 25,00010,0007,0005,0003,000 Main Sequence Giants Supergiants White Dwarfs Almost all stars we see are in one of these groups, but they change groups during their lives. As stars evolve they change in luminosity and temperature. This makes them change position on the Hertzprung-Russell diagram. Sun Sirius Vega Proxima Centauri Betelgeuse Arcturus Rigel Deneb Sirius B

Luminosity (relative to Sun) , Temperature (Kelvin) 25,00010,0007,0005,0003,000 Sun The Sun has been on the Main Sequence for billions of years, and will remain there for billions more. But eventually it will swell into a giant star, becoming more luminous but cooler.

Luminosity (relative to Sun) , Temperature (Kelvin) 25,00010,0007,0005,0003,000 Sun At this point it is a red giant star. It will get then hotter and slightly brighter.

Luminosity (relative to Sun) , Temperature (Kelvin) 25,00010,0007,0005,0003,000 Sun Finally nuclear fusion in the core will cease. The Sun will become a white dwarf, far less luminous, but with a hotter surface temperature.

STAR IN A BOX At this point, run star in a box to explore the Hertzsprung-Russell diagram for different mass stars. Level: Beginner +

NUCLEAR FUSION The processes taking place in the centre of a star. Level: Intermediate +

Nuclear Fusion The luminosity of a star is powered by nuclear fusion taking place in the centre of the star converting hydrogen into helium. – The temperature and density must be high enough to allow nuclear fusion to occur. – Stars are primarily composed of hydrogen, with small amounts of helium.

The proton-proton chain At temperatures above 4 million Kelvin hydrogen nuclei fuse into helium

Stable Stars  While the star is on the Main Sequence, it is in a stable state.  The inward force of gravity trying to collapse it, and the radiation pressure outwards from Hydrogen fusion are balanced.

Running out of Hydrogen  As the hydrogen runs out, the energy released from fusion decreases which reduces the outward force.  The forces are now unbalanced, and the larger force of gravity causes the star core to collapse.  If the star is massive enough, the core temperature increases until helium fusion starts.

Helium Fusion When the temperature is greater than 100 million Kelvin, 3 Helium nuclei can be fused together to produce 1 carbon nucleus.

Running out of Helium  The carbon nucleus can then fuse with another helium nucleus to make oxygen.  Eventually the helium supply will run out and the star will collapse.  If the star has enough mass, the temperature in the collapsed core will increase enough to allow carbon- carbon fusion. Fusion

Heavier Elements  This cycle continues, fusing heavier elements each time the core collapses. e.g. neon, magnesium, silicon and iron  The more massive a star is, the heavier the elements it can create in its core.  Iron is the heaviest element that can be created through nuclear fusion without adding extra energy.

EFFICIENCY OF FUSION Level: Advanced

Fusion Reactions

Mass  Mass of a Hydrogen nucleus (H): u  Mass of a positron (e + ): u  Mass of a helium nucleus (He): u  Mass of Carbon nucleus (C) : 12u How much mass is lost in each reaction?

Mass lost in a Hydrogen fusion reaction

 From last slide, Mass lost = Δm = 4.43 x kg ΔE = Δm c 2  Energy lost = ΔE = 4.43 x kg x (3 x 10 8 ms -1 ) 2 = 3.99 x J = 24.9 MeV Energy lost in a Hydrogen fusion reaction

Mass lost in a Helium fusion reaction

Energy lost in a Helium fusion reaction  From last slide, Mass lost = Δm = 7.50 x kg ΔE = Δm c 2  Energy lost = ΔE = 7.50 x kg x (3 x 10 8 ms -1 ) 2 = 6.75 x J = 4.22 MeV