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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    There are \[{{n}_{1}}\] photons of frequency \[{{v}_{1}}\] in a beam of light. In an equally energetic beam there are \[{{n}_{2}}\] photons of frequency \[{{v}_{2}}\]. Then the correct relation:

    A)  \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{v}_{1}}}{{{v}_{2}}}\]                             

    B)  \[\frac{{{n}_{1}}}{{{n}_{2}}}=1\]

    C)  \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{v}_{2}}}{{{v}_{1}}}\]                             

    D)  \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{v}_{2}}^{2}}{{{v}_{1}}^{2}}\]

    Correct Answer: C

    Solution :

    Here : \[{{E}_{1}}={{E}_{2}}\]                 \[{{n}_{1}}h{{v}_{1}}={{n}_{2}}h{{v}_{2}}\] So,          \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{v}_{2}}}{{{v}_{1}}}\]


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