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CGP EDU Academic Team
Published on: September 12, 2026
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 $5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{K}^{-4}$ , constant $\mathbf{b} = 2898 \mu \mathrm{mK})$ –
Text Solution
Verified by ExpertsThe correct answer is:
A
We know $\lambda_{\max}$
$\Rightarrow \mathbf{T} = \frac{b}{\frac{\lambda_{\max} \ 2898 \times 10^{-6^4}}{289.8 \times 10^{-9}}}$
According to Stefan’s Law
$E = \sigma T^{4} = (5.67 \times 10^{-8})(10^{4})^{4} = 5.67 \times 10^{8} \mathrm{W/m^{2}}$
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