At a temperature of $2000 \; K$,the maximum energy of a black body is emitted at a wavelength of $14 \; \mu m$. When its temperature is reduced to $1000 \; K$,the wavelength corresponding to the maximum energy emitted is ....... $\mu m$.

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 energy distribution $E$ with the wavelength $\lambda$ for black body radiation at temperature $T \ K$ is shown in the figure. As the temperature is increased,the maxima will:

The wavelength of the radiation emitted by a black body is $6 \ mm$ and Wien's constant is $3 \times 10^{-3} \ mK$. Then the temperature of the black body is (in $K$)

The radiation energy density per unit wavelength at a temperature $T$ has a maximum at a wavelength $\lambda_0$. At temperature $2T$,it will have a maximum at a wavelength:

$A$ black body is at a temperature of $2880\;K$. The energy of radiation emitted by this object with wavelength between $499\;nm$ and $500\;nm$ is ${U_1}$,between $999\;nm$ and $1000\;nm$ is ${U_2}$ and between $1499\;nm$ and $1500\;nm$ is ${U_3}$. The Wien's constant $b = 2.88 \times {10^6}\;nm\,K$. Then