Astrophysics Tutorial 5  Classification of Stars
Contents 
Stars at night appear to be white. This happens because the light intensity through our unaided eye is so low that the cells in out eyes that pick up colour are not activated. So we see in black and white. With good telescopes we can see the different colours of the stars. We can capture the colours using charged coupled devices or sensitive colour chemical film.
Stars glow in the same way as other glowing objects. If we turn the voltage up across a light bulb from zero volts up to its normal voltage, we see the filament glow a dull red, then to orange, to yellow to white. If we look at a spectrum as this happens, we see that:
there is a continuous range of colours,
but the relative intensity changes.
The light that we see is the resultant of that mixture of colours and other wavelengths. On this graph, the visible spectrum is to the left, between 300 and 600 nm. To the right are wavelengths of infrared radiation.
We do not see green stars because even if the peak wavelength were in the green region, 500 nm, there are also red and blue components as well. Therefore the star appears white, because red, green, and blue make white.
We look at the temperature of stars by looking at their colours. A lot of energy is given off as thermal radiation. Objects that are red hot have a temperature of about 1200 K. To understand how the colour of an object depends on its temperature, we need to understand the concept of a black body. A black body is a perfect absorber so that all radiation that falls on it is absorbed.
A black body is NOT the same as a black hole 
A perfect absorber is a perfect emitter. Therefore if we heat it up it will emit radiation including visible light. This is true (to a first approximation) for stars. Note the following for black bodies:
a hot object emits radiation across a wide range of wavelength;
there is a peak in intensity at a given wavelength;
the hotter the object the higher the peak;
the hotter the object the shorter the peak wavelength.
the area under the graph is the total energy radiated per unit time per unit surface area.
The peak wavelength is l_{max} which is the wavelength at which maximum energy is radiated. This is inversely proportional to the Kelvin temperature. It is called Wien's Displacement Law (as the peak is displaced towards shorter wavelengths). We write it as:
l_{max} T = constant = 0.00289 m K
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? 
Answer l_{max} = 0.00289 m K ÷ 2000 K l_{max} = 1.45 x 10^{6} m = 1450 nm This is in the infrared region. 
Betelgeuse appears to be red. If red light has a wavelength of about 600 nm, what would the surface temperature be? 
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 l_{max} in the green region will actually appear to be white.
The American astronomer Anne Cannon was the first to classify stars using the science of spectroscopy. Astronomers look for Balmer lines which arise from electron transitions in hydrogen atoms (See Module 1). As the electron drops from high levels to the second energy level (the one above the ground state), photons are emitted. Because they are emitted in random directions, against a complete spectrum of colours the emissions of photons would appear black. This is an absorption spectrum. Compare the emission spectrum:
with the absorption spectrum:
By studying the spectral absorption lines at wavelengths corresponding to the photons of the Balmer series, astronomers can get an idea of the temperature:
At low temperatures there are few violent collisions to excite electrons which remain in the ground state. Energy level changes are rare.
At high temperatures, there are many violent collisions between atoms. Electron transitions occur at higher levels so there are comparatively few Balmer transitions.
At intermediate levels many electrons are performing Balmer transitions, so there are strong absorption lines.
The Balmer lines are seen in the corona of the star. The Balmer lines are the transitions in between energy levels in the hydrogen atom that end at the energy level n = 2 (n = 1 is the ground state)
The temperatures can be related to the transitions. Each level has a definite energy level. Remember from Quantum Physics Tutorial 4 that:
So we can work out the photon wavelength for the transition n = 3 to n = 2.
Calculate the photon wavelength for the transition n = 3 (1.51 eV) to n = 2 (3.41 eV). What colour is this light? 
Now we have a wavelength, we can use the Wien equation to calculate the temperature:
l_{max} T = 0.00289 m K
Use your answer to Question 2 to calculate the temperature associated with this wavelength. 
The graph of intensity against temperature looks like this:
Notice that:
temperature decreases from left to right;
for a given intensity, two temperatures are possible.
To overcome this, the spectra of other elements are analysed. Peak intensities of different elements are found, and this can tie down the temperature. The idea is shown on the next graph:
Therefore in the Sun, the spectral lines would be seen for iron and calcium, indicating a surface temperature of about 6000 K. Very hot stars show spectral lines for light elements while cool stars will show up heavy elements, and spectra for molecules as well.
What would you not see when looking at the spectrum of the red giant Betelgeuse? What elements would you expect to see? 
The table shows the spectral classes for stars:
Spectral Class 
Surface temp (K) 
H Balmer Series 
Other elements 
O 
40 000 
weak 
ionised He 
B 
20 000 
medium 
He atoms 
A 
10 000 
strong 
weak ionised Ca 
F 
7500 
medium 
weak ionised Ca 
G 
5500 
weak 
medium ionised Ca 
K 
4500 
weaker 
strong ionised Ca 
M 
3000 
very weak 
strong TiO 
The graph shows part of the visible spectrum for the star Vega: (note that it is an absorption spectrum so the intensity dips to a minimum at the emitted wavelengths.)
The absorption lines are due to excited hydrogen atoms. The wavelength of each absorption is given in nm.
(a) Explain how Hydrogen atoms produce these absorption lines. (b) The diagram below shows the first six energy levels of a hydrogen atom. State which is the largest energy transition which produces an absorption line in the visible spectrum of Vega.
(c) State the value of the wavelength corresponding to this transition. (d) What is the name given to the series which gives rise to the visible region of the hydrogen spectrum? (e) For which spectral classes are these lines the dominant feature? (AQA Past Question)

The classifications of stars according to spectra are also subdivided further with numbers (e.g. A5) to allow the surface temperature to be determined within about 5 %.
The area under the graph above is related to the rate at which a black body radiates energy. 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. Formally this is stated as:
The total energy per unit time radiated by a black body is proportional to the fourth power of its absolute temperature.
In other words double the temperature and the power goes up sixteen times. In physics code we write:
[P Power (W); s  Stefan's constant; A  area (m^{2}); T  temperature (K)]
The strange looking symbol s is "sigma", a Greek letter lower case 's'. It Stefan's Constant.
s = 5.67 x 10^{8} W m^{2} K^{4}
We can treat a star as a perfect sphere (A = 4pr^{2}) and a perfect black body. So for any star, radius r, we can write:
P = 4pr^{2}sT^{4}
(Note: in some text books the power may be represented as luminosity with the physics code L)
Stars with the same absolute magnitude have the same power output. We can justify this statement by considering stars P and Q:
Power of P = P_{P} = A_{P}sT_{P}^{4} where A_{P} is the area of P and T_{P} is the surface temperature of P.
Power of Q = P_{Q} = A_{Q}sT_{Q}^{4} where A_{Q} is the area of Q and T_{Q} is the surface temperature of Q.
We can equate the two expressions to give:
A_{P}sT_{P}^{4} = A_{Q}sT_{Q}^{4}
So we can write:
So if the temperatures are the same, the areas will be the same. Therefore the radii will be the same.
If the Sun has a radius of 6.96 x 10^{8} 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? 
The intensity of the Sun's radiation decreases by an inverse square law. Therefore Saturn, about 10 times further from the Sun (i.e. 10 AU) receives only 1 % of the intensity of the Sun's radiation as Earth does.
Inverse Square Law and Luminosity
As the energy leaves a star, it becomes more spread out. The intensity of the radiation is the power per unit area, and is sometimes called the flux, given the physics code F. The luminosity of the star is the power of the star, given the physics code L.
We are a distance d from a star of luminosity L. The radiation propagates radially as shown:
The total area of the sphere is given by:
A = 4pd^{ 2}
The radiation flux is the radiation energy per unit area, so the flux is given by the equation:
Of course we can use the code r for the distance.
On the equator, the average intensity of the Sun's rays is about 1400 W m^{2}. In practice, some is absorbed by the atmosphere, and some is reflected as heat, but we will use this in a calculation to work out the power given out by the Sun.
We can work out the power of the Sun by working out the total area of a sphere that has the radius of the Earth's orbit.
A = 4pr^{2} = 4 × p × (1.50 × 10^{11})^{2} = 2.83 × 10^{23} m^{2}
Since each square metre receives 1400 W, the total power of the Sun is:
2.83 × 10^{23} m^{2} × 1400 W m^{2} = 3.96 × 10^{26} W = 4.0 × 10^{26} W (to 2 s.f.)
You may have noticed that this figure is slightly lower than the answer worked out in question 6, but some energy is absorbed and reflected, so 1400 W m^{2} is slightly too low.
The Sun has an absolute magnitude of +4.8.
It is not possible to measure the diameter of a star directly, but we can use the absolute magnitude of the Sun (+4.8) and the other star to get an idea of the distance. Consider Star PLC A2.47 which has an absolute magnitude of 0.2.
Show that PLC A2.47 has a power that is 100 times greater than the Sun. 
The power of PLC A2.47 is therefore 4.0 × 10^{28} W.
We can use Stefan's Law to work out the diameter of our star and compare it to the Sun. We use
P = AsT^{4}
Rearrange the equation for s:
We can equate the expressions for the Sun and PLC A2.47:
The Sun has a diameter of 1.4 × 10^{9} m and a surface temperature of 5800 K (a) Calculate the area of the Sun. (b) Star PLC A2.47 is a G class star with a surface temperature of 5500 K. Calculate the diameter of PLC A2.47. (c) Compare the size of PLC A2.47 with the Sun 
A dwarf star is one that is much smaller than the Sun;
A giant star is one that is much bigger than the Sun. PLC A2.47 is a giant star.
Stefan's Law applies across the whole spectrum while magnitudes relate to the visible spectrum only. Some stars release their radiation in wavelengths that are NOT in the visible region. The magnitudes of such stars have to be modified to take this into account, which is not on the syllabus.
The study of binary stars can tell us much about why the main sequence stars (see Astrophysics 6) have the range of masses and luminosities that they have. Using Kepler III, the orbital periods of binary stars can be worked out, hence the masses. The luminosity can be measured as we have seen above. If we plot the luminosity against the mass, we see that the two are related as below:
We can turn this into an equation:
Where:
L  luminosity of the star;
L_{sun}  luminosity of the Sun;
M  mass of the star;
M_{sun}  mass of the Sun.
If we plot the log of the luminosity (compared to that of the Sun) against the mass (compared to that of the Sun), we get a graph like this:
This is true for the main sequence stars.
A star has a mass of 3 times that of the Sun. What will its luminosity be compared with the Sun? 

Star PLC A2.47 has a power (luminosity) 100 times greater than the Sun. What is its mass compared with the Sun? 
The upper limit on the mass of a star is about 80 solar masses.
The Eddington limit (Arthur Stanley Eddington (1882  1944)) of a star is determined by the balance of the outward (radiation) pressure and the inward force of gravity. The photons that come from the star have relativistic momentum which can cause atoms to be removed from the photosphere of the star. Above 80 solar masses, addition of mass causes an increase in luminosity, so any extra material is simply blown away.
The lowest mass possible mass for a star is about 0.08 solar masses. Below this mass gravitational pressure is too low for nuclear fusion to occur. The body will heat up as it accretes, but the temperature is too low to initiate fusion reactions. The body radiates energy, but there is no selfsustaining fusion reaction. These are called brown dwarfs.