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

The following graph represents the radiant power versus wavelength of a black body. The area under the curve represents:

$A$ 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:

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

$A$ particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^4 \ K$. The wavelength in nanometers at which its radiation becomes maximum is $(b = 0.0029 \ mK)$.

The wavelength of maximum emitted energy $(\lambda_m)$ of a body at $700 \ K$ is $4.08 \ \mu m$. If the temperature of the body is raised to $1400 \ K$,then the value of $\lambda_m$ will be (in $\mu m$)