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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    The relationship between the energy\[{{E}_{1}}\]of the radiation with a wavelength \[8000\overset{\text{o}}{\mathop{\text{A}}}\,\] and the energy\[{{E}_{2}}\]of the radiation with a wavelength \[16000\overset{\text{o}}{\mathop{\text{A}}}\,\] is:

    A) \[{{E}_{1}}=6{{E}_{2}}\]                

    B) \[{{E}_{1}}=2{{E}_{2}}\]

    C) \[{{E}_{1}}=4{{E}_{2}}\]       

    D) \[{{E}_{1}}=1/2{{E}_{2}}\]

    E) \[{{E}_{1}}={{E}_{2}}\]

    Correct Answer: B

    Solution :

    \[E=\frac{hc}{\lambda },\] h and c for both cases are same so, \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{16000}{8000}\] \[{{E}_{1}}=2{{E}_{2}}\]


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