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NEET Sample Paper NEET Sample Test Paper-66

  • question_answer
    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\[5.67\times {{10}^{-8}}\,W{{m}^{-2}}\,{{K}^{-\,4}}\], constant\[\operatorname{b} =2898 \mu km\].

    A) \[5.67\times {{10}^{8}}\,W{{m}^{-2}}\]        

    B) \[5.67\times {{10}^{12}}\,W{{m}^{-2}}\]

    C) \[5.67\times {{10}^{7}}\,W{{m}^{-2}}\]        

    D) \[5.67\times {{10}^{14}}\,W{{m}^{-2}}\]

    Correct Answer: A

    Solution :

    We know \[{{\lambda }_{\max }}T=b\] \[\Rightarrow \,\,T=\frac{b}{{{\lambda }_{\max }}}\,\,=\,\,\frac{2898\times {{10}^{-6}}}{289.8\times {{10}^{-9}}}\,\,\,=\,\,{{10}^{4}}\,K\] According to Stefan?s law \[E=\sigma {{T}^{4}}=\,\,(5.67\times {{10}^{-8}})\,{{({{10}^{4}})}^{4}}\] \[=\,\,\,5.67\times {{10}^{\text{8}}}\,\,W/{{m}^{2}}\]


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