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J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    Solar radiation emitted by sun correspond to that emitted by black body at a temperature of 6000 K. Maximum intensity is emitted at wavelength of 4800\[\overset{0}{\mathop{A}}\,\] If the sun was to cool down from 6000 K to 3000 K, then the peak intensity of emitted radiation would occur at a wavelength :

    A)  4300\[\overset{0}{\mathop{A}}\,\]                                       

    B)  9600\[\overset{0}{\mathop{A}}\,\]

    C)  2400\[\overset{0}{\mathop{A}}\,\]                                       

    D)  19200\[\overset{0}{\mathop{A}}\,\]

    Correct Answer: B

    Solution :

                     We ins displacement law states that there is an inverse relationship between the wavelength\[({{\lambda }_{m}})\]of the peak of the emission of a black body and its temperature (T). \[\therefore \]  \[{{\lambda }_{m}}=\frac{b}{T}\]                                       (b = constant) \[\therefore \]  \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\Rightarrow \]               \[{{\lambda }_{2}}=\frac{{{\lambda }_{1}}{{T}_{1}}}{{{T}_{2}}}\] Given, \[{{\lambda }_{1}}=4800\text{ }{\AA},\text{ }{{T}_{1}}=6000\text{ }K,\] \[{{T}_{2}}=3000\text{ }K.\] \[\therefore \]  \[{{\lambda }_{2}}=\frac{4800\times 6000}{3000}=9600{\AA}\]


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