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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    A source \[{{S}_{1}}\] is producing, \[{{10}^{15}}\]photons per second of avelength\[5000\overset{o}{\mathop{A}}\,\]. Another source \[{{S}_{2}}\] is producing \[1.02\times {{10}^{15}}\] photons per second of wavelength \[5100\overset{o}{\mathop{A}}\,\] Then, (power of \[{{S}_{2}}\]) to the (power of \[{{S}_{1}}\]) is equal to :

    A) \[1.00\]

    B)                  \[1.02\]

    C) \[1.04\]                   

    D)   \[0.98\]

    Correct Answer: A

    Solution :

    Energy emitted/sec by \[{{S}_{1}},{{P}_{1}}={{n}_{1}}\frac{hc}{{{\lambda }_{1}}}\] Energy emitted/sec by \[{{S}_{2}},{{P}_{2}}={{n}_{2}}\frac{hc}{{{\lambda }_{2}}}\] \[\therefore \,\,\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{{{n}_{2}}}{{{n}_{1}}}.\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{1.02\times {{10}^{15}}}{{{10}^{15}}}.\frac{5000}{5100}=1.0\]


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