Sci Phy 119a Homework Solution Set # 4
1. 1. Saha Equation The problem asks you to calculate for three different elements the ratio of ions (atoms with an electron removed) to neutral atoms in stars of different temperatures. We need the Saha equation log Nr+1 Nr =- 5040 5 log T + - 5 log Pe - 0.48 T 2 (1)
where Pe is the electron pressure (100 g cm-1 ), T is the temperature of the star in Kelvin, and is the ionization potential (the energy required to liberate one electron from a neutral atom) in electron volts. The quantity Nr+1 is the number of ionized atoms and Nr is the number of neutral atoms, and the quantity Nr+t /N r is exactly the ratio we're asked to compute. Normally one is used to seeing the units of pressure as Force/Area, and it's not immediately apparent that the units of Pe are indeed force. Recall that force is given by F = ma. The unit of mass m is g and the units of acceleration a are cm s-2 , so the units of force are g cm s-2 . The units of Area are cm2 , so the units of Force/Area are (g cm s-2 )/cm2 = g cm-1 s-2 . The problem misstated the units of electron pressure, so kudos to you if you noticed this. Starting with helium which has an ionization potential of = 25 eV we calculate Nr+1 /Nr = NHe+ /NHe0 for a temperature of 30,000 K. Plugging in numbers to the right hand side of the Saha equation, we get
log
NHe+ NH e0
= -
5040 25 5 log(3 104 ) + - 5 log 100 - 0.48 3 104 2
= 4.51. Since 10log x = x, we have NHe+ = 104.51 = 3.3 104 NH e0 (2)
which means that for a star with a temperature of 30,000 K almost all of the helium is ionized. What about a star at 10,000 K? Putting in the numbers to the Saha equation we get
2
log
NHe+ NH e0
= -
5040 25 5 log(1 104 ) + - 5 log 100 - 0.48 1 104 2
= -5.08. NHe+ NH e0 = 10-5.08 = 8.3 10-6
which means that only a very small fraction of the helium in this star is ionized. We do the same to calculate this ratio for T= 5000 K and 3000 K. We follow the same procedure for the other two elements, hydrogen and calcium, using their respective ionization potentials of 13.6 eV and 6.1 eV for . We can then construct the following table: 3000 K NHe+ /NHe0 NH + /NH 0 NCa+ /NCa0 1.6 10-36 2.3 10-17 9.2 10-5 5000 K 3.7 10-19 1.1 10-7 4.2 10,000 K 8.3 10-6 4.6 2.8 104 30,000 K 3.3 104 2.7 106 4.9 107
Note that the element with the lowest ionization potential, Ca, has the highest fraction of ionization at a given temperature. It takes much less energy to free an electron from a calcium atom than it does from a helium atom so you might have expected this. 2. Spectral Line Temperature Dependence (a) There are two requirements that must be satisfied in order to see spectral lines from neutral hydrogen. First, there has to be enough neutral hydrogen to cause the lines. Second there must be high enough energy photons being generated to cause the transition of a bound electron to a state of higher energy. From the first condition you might expect that you would see stronger spectral lines of neutral hydrogen in a star at 5000 K. After all, according to the table there is quite a bit more H0 at this temperature than at 10, 000 K. However, the energy of the gas in a 5000 K star is E = kT = (1.38 10-16 erg s-1 )(5000K) = 6.90 10-13 erg. Here k is the Boltzmann constant. Convert ergs to eV using 1 eV = 1.60210-12 erg and we get that the gas energy in a 5000 K star is E = 0.43 eV. Compare this to the average energy in a 10, 000 K star, E = 0.86 eV. The latter energy is much closer to the transition energies for hydrogen quoted in the problem, so the 10, 000 K star has many more photons with enough energy to excite transitions compared to the 5000 K star. Because of this, the 10, 000 K star's spectral lines will be stronger than the 5000 K star's. You might wonder why we see any lines at all considering that neither of the two calculated gas energies are enough to excite a transition. From problem 1, we know that kT only has to be 10% or so of the required energy before you start to see spectral lines. In
3 practice, it is the photons generated in the gas that cause the transitions of electrons. If you remember back to the blackbody curve you'll know that photons have a large range of energies. Even if the average photon energy is only 0.86 eV there will still be a large number of photons with the required energy. However, at higher temperatures there will be proportionally more photons with enough energy to cause transitions and the spectral lines will be stronger. (b) Now we need condition one for seeing spectral lines. The spectral lines of hydrogen are caused by the bound electron absorbing a photon, so an ionized hydrogen atom will not be able to absorb anything and will hence show no spectral lines. From the table in problem 1, the hydrogen in a 30, 000 K star is almost completely ionized. Thus a very hot star will not show any spectral lines due to neutral hydrogen. (c) As in part (a), compute kT for a cool star. The 3000 K star we used in problem 1 will suffice. In this case, the gas energy is k T = 0.26 eV. This is 100 times less than the energy required to excite transition a between the first and second energy levels of a helium atom (21 eV). One will never see this spectral line in cool stars. 3. White Dwarf Density, or, Does One Teaspoon's Worth of Sirius B Really Weigh More Than My 1987 Volvo 240 DL?
(a) To calculate the radii of the stars, we'll use the relationship between radius, luminosity, and temperature Ra = Rb La Lb
1/2
Tb 2 . Ta
(3)
We've already found that the luminosities of the two stars, L1 = 22.7L L2 = 2.27 10-3 L . Remember that the symbol means the sun, so 3.7M means 3.7 solar masses, 2.9L means 2.9 times the solar luminosity, "why won't my landlord turn off the heat it's as hot as the friggin in here" means...you get the picture. Since both stars have the same color they both have the same temperature, T = 104 K. So we can compute
1/2 2 2
R1 R R2 R
=
L1 L
T T1
= (22.7)1/2 5600 K 104 K
2
5600 K 104 K
= 1.5
= (2.27 10-3 )1/2
= 0.015.
So we have R1 = 1.5 R and R2 = 0.015 R .
4 (b) Using Rearth /R = 109.0, we convert R1 R R Rearth R1 = 1.5 109 = 163 Rearth R2 = 0.015 109 = 1.63. Rearth (c) We know the radii of the stars. To find their respective densities, we need to know their masses. From Homework #3, problem 3g, we know that the total mass of the binary system is M = M1 + M2 = 3.7M . Using the relation between the masses in a binary system and their orbital radii r2 M1 = M2 r1 and the fact given in the problem that r2 /r1 = 2, we find that M1 =2 M2 M1 = 2M2 . (6) (5) (4)
=
M1 + M2 = 3.7M = 2M2 + M2 = 3M2
M2 = 1.23M and M1 = 2.47M . (7)
Mean density is mass/volume, = M/(4/3)R3 , so M1 2.47M (2.47)(2 1033 g) = 1.0 g cm-3 = = 3 3 (4/3)(1.5 6.96 1010 cm)3 (4/3)R1 (4/3)(1.5R ) 2 = 1.23M = 5.2 105 g cm-3 . (4/3)(0.015R )3
1 =
(8)
(9)
If you convert 2 to lbs., this density works out to 1,140 lb/cm3 . A totally sweet 1987 Volvo 240DL weighs 2,840 lbs. So it would take a little over two sugar cubes' worth of matter of a white dwarf to match this. That's still pretty impressive. (d) The mean density of earth is earth = 5.97 1027 g Mearth = 5.5 g cm-3 . = 3 (4/3)(6.38 108 cm)3 (4/3)Rearth (10)
Star 1 is five times less dense than the earth, and star 2 is 100,000 more dense.
5 4. Fusion as an Energy Source
(a) When four hydrogen atoms fuse to form a helium atom, the mass of the final product (helium) is less than the mass of the constituents (4 hydrogen atoms) The difference in mass between 4 hydrogen atoms and 1 helium atom is m = 4mH - mHe = 4(1.673 10-24 g) - (6.645 10-24 g) = 4.700 10-26 g (b) (c) Where does this missing mass go? It becomes energy. Using Einstein's famous result from special relativity, the mass-energy equivalence relation, we can calculate the amount of energy into which the missing mass is converted.
E = mc2 = (4.700 10-26 g)(3.00 1010 cm) = 4.23 10-5 g cm2 s-2 = 4.23 10-5 erg. It is oftentimes more convenient to express the energy in electron volts, 1 eV = 1.602 10-12 erg. So 1 eV 1.602 10-12 erg
E =
4.23 10-5 erg
= 2.64 107 eV = 26.4 106 eV = 26.4 MeV. This is an enormous amount o...