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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    A particle moving with kinetic energy E has de Broglie wavelength\[\lambda \]. If energy \[\Delta E\] is added to its energy, the wavelength become\[\lambda /2\]. Value of \[\Delta E\], is:   [JEE MAIN Held on 09-01-2020 Morning]

    A) 4E

    B) E

    C) 2E

    D) 3E

    Correct Answer: D

    Solution :

    [d] \[\lambda =\frac{h}{\sqrt{2mk}}\] \[\lambda '=\frac{\lambda }{2}\Rightarrow k'=4k\] \[\Rightarrow \Delta E=3E\]


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