91Ó°ÊÓ

The formula used is $R=\sqrt{\left(\frac{L}{T^4}\right)}$to calculate the desired result.

$\text{The surface temperature of Sirius}A\text{is given by}$

$\left(\frac{2.4\mathrm{eV}}{1.41\mathrm{eV}}\right)=1.702$

role="math" localid="1647766605943" $\text{So the radius of Sirius A shouldis}$

role="math" localid="1647766683798" $=\sqrt{\left[\frac{24}{(1.70{)}^4}\right]}$

$=1.69\text{(In units of suns radius)}$

The radius of Sirius A is$1.69$.

$\text{The surface temperature of SiriusBis given by}$

$\left(\frac{7\mathrm{eV}}{1.41\mathrm{eV}}\right)=4.96$

$\text{Whichis nearly five times the suns temperature, so the radius of Sirius B should be}$

$R=\sqrt{\left(\frac{L}{T^4}\right)}$

$=\sqrt{\left(\frac{0.03}{(4.96{)}^4}\right)}$

$=0.007$

As it is less than $1\%$of suns radius and just slightly smaller than the earth's radius. This result is in rough agreement with that of$7.23\left(d\right)$ where we calculated that, one solar - mass white decay should have a radius first slightly larger than earth's.

$\text{The surface temperature ofBetalgeuseis given by}$

$\frac{(0.8\mathrm{eV})}{(1.41\mathrm{eV})}=0.57$

$\text{the radius of Betelgeuse will be}$

$R=\sqrt{\left(\frac{L}{T^4}\right)}$

$=\sqrt{\left[\frac{10,000}{(0.57{)}^4}\right]}$

$=310$

As the radius is larger than the radius of earths orbit, and nearly as large as the orbit of mass. Super giant is certainly an approximate term. As for "red" the spectrum of Betelgeuse is certainly redder than the sun, due to its lower temperature of $3300\mathrm{K}$ which makes its spectrum peak well into the infrared and fall of considerably at the blue end of the visible range. But this temperature is still slightly hotter than the filament of an incandescent bulb, so the color of Betelgeuse shouldn't be any redder than that of incandescent light, "yellow-orange' would be a more accurate description.