What is the Mass-Radius Relation for a White Dwarf with 0.6 Solar Masses?

[Catria]

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.

h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ Jh\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s\ s

m_{e}\ =\ 9.10938188(72)\ \times\ 10^{-31}\ kg

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}

Mass of the white dwarf: \ 1.2 \times\ 10^{30}kg

Homework Equations

M = \frac{f}{R^{3}}

f = \frac{π}{3}\left(\frac{15C}{2πG} \right)^{3}

\frac{N}{V} = \frac{ρN_{0}}{2}, since Z/A = 1/2

P=Cρ^{\frac{5}{3}} = \left(\frac{N}{V}\right)^{\frac{5}{3}} \left(\frac{3h^{3}}{8π}\right)^{\frac{2}{3}}\frac{1}{5m}

The Attempt at a Solution

C = \frac{1.064\times10^{-67}}{5m}\left(\frac{N_{0}}{2}\right)^{\frac{5}{3}}

I took N_{0} = 6.02 x 10^{23} so C = 31.57

\frac{15C}{2πG}=\frac{473.53}{4.19\times 10^{-10}}=1.13\times10^{12}

Putting that into f, we get f = 1.511\times10^{36}

Now, 1.2\times10^{30} = \frac{1.511\times10^{36}}{R^{3}}

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 N_{0}.

 

[mfb]

Working with units would help to spot the error.
What is N₀? If it is the avogadro constant, why do you get it as factor between a density and another density? Which units do you use for ρ?

 

[Catria]

I hate the textbook and the class notes but, per class notes, I assume ρ is in g/cc, and N/V is in particles per cubic meter.

 

[mfb]

cc? cm³? Then you need an additional factor of 10^6 to convert between m³ and cm³.

 

[Catria]

But then I'd have to divide by 1,000 to go from grams to kilograms... so I'd need an additional factor of 1,000 rather than 10^6.

 

[mfb]

As you can see, it would be useful to work with units everywhere. It is easier to spot (or avoid) those prefactors if you know the units of your parameters.

 

[Catria]

And now it gives a more sensible size (somewhat larger than Earth in volume, yet on the same order of magnitude); I find that the Coulomb pressure wouldn't affect the radius very much.