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:

The maximum energy in the thermal radiation from a hot source occurs at a wavelength of $11 \times 10^{-5} \ cm$. According to Wien's law, the temperature of the source (on Kelvin scale) will be $n$ times the temperature of another source (on Kelvin scale) for which the wavelength at maximum energy is $5.5 \times 10^{-5} \ cm$. The value of $n$ is

The plots of intensity $(I)$ versus wavelength $(\lambda)$ for three black bodies at temperatures $T_1, T_2$ and $T_3$ respectively are as shown. Their temperatures are such that:

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 $\mu m$)

$A$ 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:

Two stars $A$ and $B$ radiate maximum energy at wavelengths $3.9 \times 10^{-7} \text{ m}$ and $5.2 \times 10^{-7} \text{ m}$ respectively. The ratio of the temperature of star $A$ to that of star $B$ will be