Answer
$$\frac{d_{480}}{d_{660}}=1.413 \approx 1.4 $$
Work Step by Step
The temperature of each star can be found from Wien’s law, Eq. $27–2.$
The peak wavelength is used as a subscript to designate each star’s properties.
$$\lambda_{\mathrm{P}} T=2.90 \times 10^{-3} \mathrm{m} \cdot \mathrm{K} \rightarrow$$
$$T_{660}=\frac{2.90 \times 10^{-3} \mathrm{m} \cdot \mathrm{K}}{660 \times 10^{-9} \mathrm{m}}=4394 \mathrm{K} \quad $$ $$T_{480}=\frac{2.90 \times 10^{-3} \mathrm{m} \cdot \mathrm{K}}{480 \times 10^{-9} \mathrm{m}}=6042 \mathrm{K}$$
The luminosity of each star can be found from the H-R diagram.
$$
L_{660} \approx 7 \times 10^{25} \mathrm{W} \quad $$$$ L_{480} \approx 5 \times 10^{26} \mathrm{W}
$$
we know that $\quad P=\beta A T^{4} \quad$from the Stefan-Boltzmann equation , where
$\beta$ is a constant and
$A$ is the radiating area.
The $P$ in the Stefan-Boltzmann equation is the same as the luminosity $L$ given in Eq. $33-1 .$ Form the ratio of the two luminosities.
$$\frac{L_{480}}{L_{660}}=\frac{\beta A_{480} T_{480}^{4}}{\beta A_{660} T_{660}^{4}}=\frac{4 \pi r_{480}^{2} T_{480}^{4}}{4 \pi r_{660}^{2} T_{660}^{4}} \rightarrow $$
$$\frac{r_{480}}{r_{660}} \times \frac{T_{660}^{2}}{T_{480}^{2}}=\sqrt{\frac{5 \times 10^{26} \mathrm{W}}{7 \times 10^{25} \mathrm{W}}} \times \frac{(4394 \mathrm{K})^{2}}{(6042 \mathrm{K})^{2}}=1.413$$
$\text{The diameters are in the same ratio as the radii.}$
$$\frac{d_{480}}{d_{660}}=1.413 \approx \left[ 1.4 \right]$$
