Sun radiates thermal radiation with maximum intensity at the wavelength  $\lambda =0.5\text{\hspace{0.17em}μm}$ while its surface temperature is  $6000\text{\hspace{0.17em}K}$. If sun cools down to a temperature where it emits only 81% of its present power, the maximum intensity will then be emitted at wavelength $\lambda \text{'}$  (in micro metre )is equal to  $\left[\sqrt{10}=3.1622\right]$

1. When sun cools to the temperature T where it emits only 81% of its present radiating power, then
$\sigma T^4=\left(\frac{81}{100}\right)\sigma {\left(6000\right)}^4\left(\because E=\sigma T^4\right)$   $\sigma =5.67x {10}^{-8} unit$
$\Rightarrow T=\sqrt{0.9}\left(6000\right)$ 
$=0.95\times 6000=5700\text{\hspace{0.17em}K}$ 
2. By Wien’s displacement law,  ${\lambda }_{m}T=$constant
Hence if  ${\lambda }_m$ is the required wavelength  ${\lambda }_m\left(5700\right)=0.5\times 6000$
${\lambda }_m=\frac{6000}{5700}\times 0.5=\frac{10}{19}=0.53\text{\hspace{0.17em}μm}$ 
Therefore, the correct answer is 0.53