$A$ 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:

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 $b = 2898 \, \mu m K$).

In a certain planetary system,it is observed that one of the celestial bodies having a surface temperature of $200 \; K$,emits radiation of maximum intensity near the wavelength $12 \; \mu m$. The surface temperature (in $K$) of a nearby star which emits light of maximum intensity at a wavelength $\lambda = 4800 \; \mathring A$ is

The wavelength of maximum energy released during an atomic explosion was $2.93 \times 10^{-10} \ m$. Given that Wien's constant is $b = 2.93 \times 10^{-3} \ m \cdot K$,the maximum temperature attained must be of the order of:

Three black discs $x, y, z$ have radii $1 \ m, 2 \ m$ and $3 \ m$ respectively. The wavelengths corresponding to maximum intensity are $200 \ nm, 300 \ nm$ and $400 \ nm$ respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is:

$A$ black body is at a temperature of $5780 \text{ K}$. The energy of radiation emitted by the body at wavelength $300 \text{ nm}$ is $U_1$, at wavelength $500 \text{ nm}$ is $U_2$, and at wavelength $900 \text{ nm}$ is $U_3$. Wien's constant $b = 2.89 \times 10^6 \text{ nm K}$. This shows that: