- #1
Catria
- 152
- 4
Homework Statement
Compute the numerical constant C for an electron gas (take Z = 6 and A = 12) and determine the radius of a white dwarf whose mass is 0.6 solar masses.
[itex]h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ Jh\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s\ s[/itex]
[itex]m_{e}\ =\ 9.10938188(72)\ \times\ 10^{-31}\ kg[/itex]
[itex]G\ =\ 6.673(10)\ \times\ 10^{-11}G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}\ m^{3} kg^{-1} s^{-2}[/itex]
Mass of the white dwarf: [itex]\ 1.2 \times\ 10^{30}kg[/itex]
Homework Equations
M = [itex]\frac{f}{R^{3}}[/itex]
f = [itex]\frac{π}{3}[/itex][itex]\left(\frac{15C}{2πG} \right)^{3}[/itex]
[itex]\frac{N}{V}[/itex] = [itex]\frac{ρN_{0}}{2}[/itex], since Z/A = 1/2
[itex]P=Cρ^{\frac{5}{3}}[/itex] = [itex]\left(\frac{N}{V}\right)^{\frac{5}{3}}[/itex] [itex]\left(\frac{3h^{3}}{8π}\right)^{\frac{2}{3}}[/itex][itex]\frac{1}{5m}[/itex]
The Attempt at a Solution
C = [itex]\frac{1.064\times10^{-67}}{5m}[/itex][itex]\left(\frac{N_{0}}{2}\right)^{\frac{5}{3}}[/itex]
I took [itex]N_{0}[/itex] = 6.02 x [itex]10^{23}[/itex] so C = 31.57
[itex]\frac{15C}{2πG}=\frac{473.53}{4.19\times 10^{-10}}=1.13\times10^{12}[/itex]
Putting that into f, we get f = [itex]1.511\times10^{36}[/itex]
Now, [itex]1.2\times10^{30}[/itex] = [itex]\frac{1.511\times10^{36}}{R^{3}}[/itex]
and finally R = 107.98m, which doesn't make any sense to me. The only place where I think I might have it wrong is the value of [itex]N_{0}[/itex].