Wien's displacement law expresses the relationship between:

The wavelength of maximum intensity of radiation emitted by a star is $289.8 \, nm$. The radiation intensity for the star is: (Stefan's constant $\sigma = 5.67 \times 10^{-8} \, W m^{-2} K^{-4}$,Wien's constant $b = 2898 \, \mu m \cdot K$)

Solar radiation emitted by the Sun resembles that of a black body at a temperature of $6000 \, K$. The maximum intensity of radiation is emitted at a wavelength of $4800 \, \mathring{A}$. If the temperature of the Sun decreases from $6000 \, K$ to $3000 \, K$,then at what wavelength (in $\mathring{A}$) will the maximum intensity of radiation be emitted?

$Assertion :$ For higher temperature, the peak emission wavelength of a blackbody shifts to lower wavelengths.
$Reason :$ Peak emission wavelength of a blackbody is proportional to the fourth power of temperature.

If the wavelengths of maximum intensity of radiations emitted by the sun and the moon are $0.5 \times 10^{-6} \ m$ and $10^{-4} \ m$ respectively,the ratio of their temperatures is:

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