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JEE Main & Advanced Sample Paper JEE Main Sample Paper-1

  • question_answer
    A\[\alpha -\] particle passes through a potential difference of \[2\times {{10}^{6}}V\] and then it becomes incident on a silver foil. The charge number of silver is 47. The energy of incident particles will be: (in joule)

    A)  \[5\times {{10}^{-12}}\]                              

    B)  \[6.4\times {{10}^{-13}}\]

    C)  \[5.8\times {{10}^{-14}}\]                           

    D)  \[9.1\times {{10}^{-15}}\]

    Correct Answer: B

    Solution :

    Energy of incident particles ie, alpha particles will be \[U=qV=2e\times V\] \[=2\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{6}}\] \[=6.4\times {{10}^{-13}}\text{J}\]


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