Two spherical stars $A$ and $B$ emit blackbody radiation. The radius of $A$ is $400$ times that of $B$ and $A$ emits $10^4$ times the power emitted from $B$. The ratio $(\lambda_A / \lambda_B)$ of their wavelengths $\lambda_A$ and $\lambda_B$ at which the peaks occur in their respective radiation curves 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?

The surface temperature of the sun, which has maximum energy emission at $500 \, nm$, is $6000 \, K$. The temperature of a star, which has maximum energy emission at $400 \, nm$, will be ........ $K$.

$A$ 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

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 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?