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Question

# 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  $\left[\sqrt{10}=3.1622\right]$

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Solution

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

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