$A$ black disc has a radius $R$ and the wavelength corresponding to the maximum intensity is $\lambda$. The emissive power $(E)$ for the different radii and maximum wavelengths is directly proportional to:

Two bodies $A$ and $B$ of equal surface area have thermal emissivities of $0.01$ and $0.81$ respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies $A$ and $B$ at wavelengths $\lambda_A$ and $\lambda_B$ respectively. The difference in these two wavelengths is $1 \mu m$. If the temperature of body $A$ is $5802 \ K$, then the value of $\lambda_B$ is:

$A$ black body at a temperature of $3000 \ K$ cools down. The wavelength corresponding to maximum energy density shifts by $\Delta \lambda = 9 \ \mu m$. What is the new temperature of the black body (in $K$)? (Given $b = 3 \times 10^{-3} \ m \cdot K$)

If a graph is plotted by taking spectral emissive power along $y$-axis and wavelength along $x$-axis,then the area under the graph above the wavelength axis is ...........

Two bodies $A$ and $B$ radiate maximum energy with a wavelength difference of $4 \mu m$. The absolute temperature of body $A$ is $3$ times that of body $B$. The wavelength at which body $B$ radiates maximum energy is: (in $\mu m$)

The wavelength of maximum intensity of radiation emitted by a star is $289.8 \, nm$. The radiation intensity of the star is (Stefan's constant $\sigma = 5.67 \times 10^{-8} \, W m^{-2} K^{-4}$, Wien's constant $b = 2898 \, \mu m K$).