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JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Cathode Rays and Positive Rays

  • question_answer
    An a particle is accelerated through a p.d of \[{{10}^{6}}\] volt then K.E. of particle will be                                            [Pb. PET 2003]

    A)            8 MeV                                     

    B)            4 MeV

    C)            2 MeV                                     

    D)            1 MeV

    Correct Answer: C

    Solution :

               \[K=Q.\Delta V=(2e)\times {{10}^{6}}V=2\times {{10}^{6}}eV=2M\,eV\]


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