BHU PMT BHU PMT Solved Paper-2002

  • question_answer
    The maximum wavelength of radiation emitted at 900 K is \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\, \right)\times \left( \overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{b}}\, \right)\]. The maximum wavelength of Radiation emitted at 1200 K, will be:                                                                                                    [BHU PMT-2002]

    A)                  \[2\left( \overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{a}}\, \right)\]                                       

    B)                  \[-2\left( \overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{a}}\, \right)\]     

    C)                  \[\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{a}}\,\]                                   

    D)                  \[\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,\]

    Correct Answer: D

    Solution :

                     From Wien?s displacement law                 \[2\,\,Ml\]                 Where \[Ml\] is the maximum wavelength emitted and \[x\] is absolute temperature.                                 \[y=7\,\sin \left[ 7\,\,\pi t-0.4\,\pi x+\frac{\pi }{3} \right]\]                 \[\frac{2\pi }{49}m/s\]   \[\frac{49}{2\pi }m/s\]                 \[49\,\pi \,m/s\]               \[17.5\,m/s\]


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