The surface temperature of the stars is determined using:

$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now,due to a change in the temperature of that body,it radiates maximum energy at wavelength $\frac{2\lambda}{3}$. At that new temperature,the emissive power is:

Which graph represents the $v_m - T$ relationship for a black body,where $v_m$ is the frequency corresponding to the maximum energy emission and $T$ is the absolute temperature?

$A$ black body has a maximum wavelength $\lambda_{m}$ at a temperature of $2000 \ K$. Its corresponding wavelength at a temperature of $3000 \ K$ will be:

The plots of intensity versus wavelength for three black bodies at temperatures $T_1$,$T_2$ and $T_3$ respectively are as shown. Their temperatures are such that

Two bodies $A$ and $B$ have emissivities $0.01$ and $0.81$ respectively. The outer surface areas of both bodies are the same. Both bodies emit total radiant power at the same rate. The wavelength $\lambda_B$ corresponding to the maximum spectral radiance of $B$ is $1.0 \mu m$. If the temperature of $A$ is $5802 \ K$,calculate the wavelength $\lambda_A$ in $\mu m$. (Note: The original question asked for $\lambda_B$ but provided $\lambda_B$ as $1.0 \mu m$ and asked for $\lambda_A$ based on the context of the solution provided).