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JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    The energy of a photon is equal to the Kinetic energy of a photon. The energy of photon is E. Let \[{{\lambda }_{1}}\] be the de-Broglie wavelength of the proton and \[{{\lambda }_{2}}\] be the wavelength of photon. The ratio \[{{\lambda }_{1}}/{{\lambda }_{2}}\] is proportional to:

    A) \[{{E}^{o}}\]         

    B)                                    \[{{E}^{1/2}}\]

    C) \[{{E}^{-1}}\]                                     

    D) \[{{E}^{-2}}\]

    Correct Answer: B

    Solution :

    \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=[h/{{(2mE)}^{1/2}}]/[hc/E]\]                           Thus, \[{{\lambda }_{1}}/{{\lambda }_{2}}\propto {{E}^{1/2}}\]


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