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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are  charge and mass of an electron respectively, then the value of\[h/\lambda \](where \[\lambda \]is wavelength associated with electron wave) is given by : [JEE Main Solved Paper-2016 ]

    A) \[\sqrt{2meV}\]                              

    B)  2meV

    C) 2meV                   

    D) \[\sqrt{meV}\]

    Correct Answer: A

    Solution :

                    As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron becomes = 'eV' As wavelength of \[{{e}^{-}}\] wave \[(\lambda )=\frac{h}{\sqrt{2mK.E.}}\] \[\lambda =\frac{h}{\sqrt{2meV}}\]                       \[\therefore \]\[\frac{h}{\lambda }=\sqrt{2meV}\]


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