Wavelength ratio at peak radiation emission for two spherical stars

According to the Stefan-Boltzmann law, the total power $P$ emitted by a spherical blackbody of radius $R$ and surface temperature $T$ is given by: $P = 4\pi R^2 \sigma T^4$

where $\sigma$ is Stefan's constant.

For two spherical stars $A$ and $B$, the ratio of their emitted power is: $\frac{P_A}{P_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4$

From the given data:

• $\frac{R_A}{R_B} = 400$
• $\frac{P_A}{P_B} = 10^4$

Substituting these values into the power ratio formula: $10^4 = (400)^2 \left(\frac{T_A}{T_B}\right)^4$

$10^4 = 16 \times 10^4 \left(\frac{T_A}{T_B}\right)^4$

Dividing both sides by $10^4$: $1 = 16 \left(\frac{T_A}{T_B}\right)^4$

$\left(\frac{T_A}{T_B}\right)^4 = \frac{1}{16}$

Taking the fourth root on both sides gives: $\frac{T_A}{T_B} = \frac{1}{2}$

According to Wien's displacement law, the peak wavelength $\lambda$ is inversely proportional to the absolute temperature $T$: $\lambda \propto \frac{1}{T}$

Therefore, the ratio of the peak wavelengths $\frac{\lambda_A}{\lambda_B}$ is: $\frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} = \frac{1}{T_A / T_B} = \frac{1}{1/2} = 2$