For this question we will try to apply the ideal gas law to the center of the su

Question

For this question we will try to apply the ideal gas law to the center of the sun.

a) The core of the sun is 36% hydrogen and 64% helium by mass and has a density of 158 g/cm3. At such high temperatures there is 100% ionization, thus we can assume no bound electrons. How many Hydrogen nuclei, Helium nuclei, and free electrons?

b) Assuming that the particles are perfect spheres, what percentage of the volume is from that of the actual particles? Use the approximation that the radius of a nuclear particle can be estimated from r = (1.4 x 10-14 cm)(atomic number)1/3 and assume that the radius of an electron is 1.4 x 10-15 cm.

Explanation / Answer

a. radius of core of sun, R = 0.25*695700,000 m = 173,935,000 m

now, volume, V = 4*pi*R^3/3

given, 36% is hydrogen, 64% is helium by mass

density of core , rho = 158,000 kg/m^3

hence

mass of core m = rho*V

so mass of hydrogen, Mh = 0.36m

numbner of atoms of hydrogen Nh = Mh/1.6*10^-27 = 0.36*rho*4*pi*R^3/3*1.6*10^-27

Nh = 0.36*rho*4*pi*R^3/3*1.6*10^-27 = 0.36*158000*4*pi*(173,935,000)^3/3*1.6*10^-27

Nh = 7.8359*10^56 atoms

similiarly

mass of helium = Mhe = 0.64m

hence

Nhe = Mhe/4*1.6*10^-27 = 3.4826*10^56 atoms

so number of hydrogen nuclei, Nh =7.8359*10^56

Number of helium nuclei = Nhe = 3.4826*10^56

number of free electrons = Nh + 2Nhe = 14.8011*10^56

b. total volume V = 4*pi*R^3/3 = 2.20419*10^25 m^3

volume of hydrogen nucleus, Vh = 4*pi*(1.4*10^-16)^3/3 = 1.1494*10^-47 m^3

volume of helium nucleus, Vhe = 4*pi*(1.4*10^-16)^3*2/3 = 2.298*10^-47 m^3

hence percentage of volume occupied by ions = (Nh*Vh + Nhe*Vhe)*100/V = 7.7182*10^-14 %

now volume of electron Ve = 4*pi*(1.4*10^-17)^3/3 m^3

hence percentage volume occupied by electrons = (Nh + 2Nhe)*Ve*100/V = 7.71822*10^-17 %

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