The wavelength corresponding to the maximum energy emitted by a body at a certain temperature is ${\lambda _0}$. If the temperature of the body is increased such that the new maximum wavelength becomes $\frac{3{\lambda _0}}{4}$, then by what factor does the emissive power increase?

In the figure,the distribution of energy density of the radiation emitted by a black body at a given temperature is shown. The possible temperature of the black body is ....... $K$.

The distribution of relative intensity $I(\lambda)$ of blackbody radiation from a solid object versus the wavelength $\lambda$ is shown in the figure. If the Wien displacement law constant is $2.9 \times 10^{-3} \ mK$,what is the approximate temperature of the object in $K$?

If $\lambda_{m}$ denotes the wavelength at which the radioactive emission from a black body at a temperature $T \; K$ is maximum,then

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 emission from a body at a temperature of $200 \, K$ is $14 \, \mu m$. If the temperature of the body is increased to $1000 \, K$,find the new wavelength of maximum emission.