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NEET Physics Simple Harmonic Motion NEET PYQ-Simple Harmonic Motion

  • question_answer
    A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is \[{{U}_{1}},\] at wavelength 500 nm is \[{{U}_{2}}\] and that at 1000 nm is \[{{U}_{3}}\]. Wien's constant, \[b=2.88\times {{10}^{6}}\text{ }nmK\]. Which of the following is correct?                                                           [NEET - 2016]

    A)  \[{{U}_{1}}=0\]         

    B)       \[{{U}_{3}}=0\]

    C)  \[{{U}_{1}}>\text{ }{{U}_{2}}\]        

    D)       \[{{U}_{2}}>\text{ }{{U}_{1}}\]

    Correct Answer: D

    Solution :

    [d] Maximum amount of emitted radiation corresponding to \[{{\lambda }_{m}}=\frac{b}{T}\]
    \[{{\lambda }_{m}}=\frac{2.88\times {{10}^{6}}\,nmK}{5760K}=500\,nm\]
    From the graph \[{{U}_{1}}<\text{ }{{U}_{2}}>\text{ }{{U}_{3}}\]


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