JEE Main & Advanced Physics Electrostatics & Capacitance NEET PYQ - Electrostatics and Capacitance

  • question_answer
    The mean free path of electrons in a metal is \[4\times {{10}^{-8}}\,m\]. The electric field which can give on an average 2 eV energy to an electron in the metal will be in unit of \[V{{m}^{-1}}\]               [AIPMT (S) 2009]

    A) \[8\times {{10}^{7}}\]

    B) \[5\times {{10}^{-11}}\]

    C) \[8\times {{10}^{-11}}\]

    D) \[5\times {{10}^{7}}\]

    Correct Answer: D

    Solution :

    [d] Energy \[=2\,eV\]
    \[e{{V}_{0}}=2\,eV\]
    \[\Rightarrow \]   \[{{V}_{0}}=2\]
    Now, electric field \[E=\frac{2}{4\times {{10}^{-8}}}\]
    \[=0.5\times {{10}^{8}}\]
    \[=5\times {{10}^{7}}\,V{{m}^{-1}}\]


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