The sun is the only star whose size we can easily measure

Step 1 of 3

(a)  The luminosity is given by:

\(L=A e \sigma T^{4}=4 \pi R^{2} \sigma T^{4}\)

where \(R\) is the radius and \(T\) is the temperature. We need to write this relation in terms of the sun units, the luminosity of the sun is:

\(L_{s}=4 \pi \sigma R_{s}^{2} T_{s}^{4}\)

to convert this to sun's unit, we need to let \(L_{s}=1, T_{s}=1 \text { and } R_{s}=1\) this means the luminosity of the sun is 1 times the unit of the luminosity of the sun and thus for the radius and the temperature, substituting with these values implies that:

\(4 \pi \sigma=1\)

so if we substitute with this in the first expression, the expression will measure the luminosity in terms of the sun's units, so:

\(L=R^{2} T^{4}\)

solve for \(R\) to get:

\(R=\frac{\sqrt{L}}{T^{2}}\)

now consider the spectrum of Sirius \(A\), the peak photon energy \(2.4 \mathrm{eV}\), given that the peak energy of the photon for the sun to be \(1.41 \mathrm{eV}\), this means the temperature of the Sirius \(A\) in units of the sun is \(2.4 \mathrm{eV} / 1.41 \mathrm{eV}=1.70\), substitute to get:

\(\begin{aligned} R & =\frac{\sqrt{24}}{(1.70)^{2}} \\ & =1.695 \\ R & =1.695 \end{aligned}\)

This means the radius of Sirius \(A\) is 1.695 times the radius of the sun (note that we used the fact that Sirius \(A\) has a luminosity of 24 times the luminosity of the sun).