CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    An electron is accelerated through a potential difference of 45.5 volt. The velocity acquired by it is (in \[m{{s}^{-1}}\]):

    A)  \[{{10}^{6}}\]                                   

    B)  zero

    C)  \[4\times {{10}^{6}}\]                                  

    D)  \[4\times {{10}^{4}}\]

    Correct Answer: C

    Solution :

    When electron is accelerated through a potential difference of V volts, then Kinetic energy =eV \[i.e.\]  \[\frac{1}{2}m{{\upsilon }^{2}}=eV\] \[\Rightarrow \]               \[\upsilon =\sqrt{\left( \frac{2eV}{m} \right)}\] \[\therefore \]  \[\upsilon =\sqrt{\left( \frac{2\times 1.6\times {{10}^{-19}}\times 45.5}{9.1\times {{10}^{-31}}} \right)}\]                 \[=4\times {{10}^{6}}m/\sec \] 


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