JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    The potential difference applied to an X-ray tube is \[5\,KV\] and the current through it is \[3.2\,mA\]. Then number of electrons striking the target per second is:

    A) \[2\times {{10}^{16}}\]     

    B)                                    \[5\times {{10}^{6}}\]

    C) \[1\times {{10}^{17}}\]                 

    D) \[4\times {{10}^{15}}\]

    Correct Answer: A

    Solution :

    \[I=\frac{Q}{t}=\frac{ne}{t}\] \[or,\,\,n=\frac{It}{e}=\frac{3.2\times {{10}^{-3}}\times 1}{(1.6\times {{10}^{-19}})}=2\times {{10}^{-16}}\]


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