ASTR 1040 Spring, 2005
1. Let's consider what might happen when hydrogen in the Sun's core is all used up via the proton-proton chain, and the Sun is left with a pure helium core. In the following, assume that the luminosity of the Sun remains at a constant level of L = 3.83 X 1026 J/sec (W) and that nuclear reactions take place only in the innermost 10 percent of the Sun's mass. The form of helium that remains after the p-p reaction is helium 4, aka 4He, which has two protons and two neutrons in its nucleus. The mass of this nucleus is 4.002603 u, where the u is the universal mass unit (formerly known as the atomic mass unit, or amu) and has the value 1 u = 1.67 X 10-27 kg. Here is the brief table of isotopic masses that are relevant to this problem:
1H (hydrogen: mass = 1.007825 u)
2H (deuterium: mass = 2.014102 u)
3H (tritium: mass = 3.016049 u)
3He (helium 3: mass = 3.016029 u)
4He (helium 4: mass = 4.002603 u)
6Li (lithium 6: mass = 6.015121 u)
7Li (lithium 7: mass = 7.016003 u)
7Be (beryllium 7: mass = 7.016928 u)
8Be (beryllium 8: mass = 8.005305 u)
10B (boron 10: mass = 10.012938 u)
12C (carbon12: mass = 12.000000 u)
13C (carbon 13: mass = 13.003355 u)
14C (carbon 14: mass = 14.003241 u)
14N (nitrogen 14: mass = 14.003074 u)
16O (oxygen 16: mass = 15.994915 u)
(a) What element is likely to be formed when two 4He nuclei merge? Specify the particular isotope (combination of mass number and atomic number) that you choose.
The most likely reaction would be the straightforward combination of the two helium nuclei to form beryllium (8Be), which has the same number of nucleons. No other potential products appear on the list given.
(b) Calculate the mass difference between the ingredients (the two helium nuclei) and the product (the isotope you chose in part a). What fraction of the ingredient mass is this?
The sum of the masses of the two helium nuclei is 2 X 4.002603 u = 8.005206 u. The mass of the 8Be nucleus is given as 8.005305, which is slightly greater than the sum of masses of the two 4He particles. The mass difference is -0.000099 u = -9.9 X 10^-5 u. Thus this reaction is endothermic (i.e., requires more energy than it creates).
The fractional mass change (in this case an addition of mass, representing energy that has to be added) is -0.000099/8.0005206 = -1.24 X 10-5.
(c) Calculate the energy released in a single reaction.
It is actually energy lost, which you can calculate as follows:
E = mc^2 = (-9.9 X 10^-5 u)(1.67 X 10^-27 kg/u)(3 X 10^8 m/sec)2 = -1.48 X 10^-14 J
(d) Do you think that the reaction you chose in part a is viable? In other words, is this reaction likely to be the next source of energy for a star after its hydrogen has been converted to helium? Explain.
This reaction requires energy rather than producing it, making it technically an endothermic reaction. So it will not be an important source of energy.
(a) Calculate the relative brightness for this star pair: Star A has m = 4.83; star B has m = 1.97;
Use the magnitude equation m_1 - m_2 = 2.5log(I_2/I_1), or I_2/I_1 = 10(m_1 - m_2)/2.5:
I_B/I_A = 10(4.83-1.97)/2.5 = 102.86 = 13.93
(b) Find the visual absolute magnitudes M_v for the following: A star with apparent visual magnitude m_v = +16.84 and distance d = 518 pc
Use the expression m_v - M_v = 5log(d/10); i.e. M_v = m_v - 5log(d/10): M_v= 16.84 - 5log(51.8) = 16.84 - 5(1.71) = 8.27
(c) Find the distances to the following: A star with m_v = +18.42 and M_v = +4.85
Solve for d: d/10 = 10(m_v - M_v)/5:
d/10 = 10(18.41 - 4.85)/5 = 102.71 = 518 pc; d = 5180 pc
3. What is the luminosity (in solar units) of a star whose bolometric absolute magnitude is M_bol = -6.83? (The Sun's bolometric absolute magnitude is Mbol)sun) = +4.75). If lambda(max) = 1000Å, what is the diameter of the star? What is its angular diameter, if its parallax is π = 0.00048"? Is this angular diameter detectable with the Hubble Space Telescope when observing at visible wavelengths? (For this last part you can use your value for Hubble's visible-wavelength resolving power from question 1.d., above.)
Use the magnitude equation to compare bolometric absolute magnitudes: L*/L(sun) = 10^0.4(M(sun) - M*) = 10^0.4[4.75 - (-6.83)] = 104.63 = 4.29 X 10^4. The temperature, from Wien's law, is T = 0.0029/1000(10^-10) = 29,000 K = 5.02 T(sun). As usual, it is easiest to work with ratios to find the radius: R*/R(sun) = sqrt[(L*/L(sun)) (T(sun)/T*)^2] = sqrt[4.29 X 10^4 (1/5.02)^2] = 8.22 R(sun) = 5.72 X ^9 m. The distance to the star is d = 1/π = 1/0.00048 = 2083 pc = 6.45 X 10^19 m. The angular diameter is (theta) = 2R/d = 2(5.72 X 10^9/6.45 X 10^19 = 1.77 X 10^-10 rad = 3.65 X 10^-5 arcseconds. This is not detectable by current technology.
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