CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • question_answer
    The specific charge of a proton is 9.6 x 107 C kg-1. The specific charge of an alpha panicle will be:

    A)  \[9.6\times {{10}^{7}}C\,\,k{{g}^{-1}}~~\]

    B)  \[19.2\times {{10}^{7}}C\text{ }k{{g}^{-1}}\]

    C)  \[4.8\times {{10}^{7}}C\text{ }k{{g}^{-1}}\]

    D)  \[2.4\times {{10}^{7}}C\text{ }k{{g}^{-1}}\]

    Correct Answer: C

    Solution :

    Specific charge on proton \[={{\left( \frac{e}{m} \right)}_{p}}\]\[=9.6\times {{10}^{7}}C/kg\] specific charge on \[\text{ }\!\!\alpha\!\!\text{ -}\] particle, \[{{\left( \frac{q}{m} \right)}_{\alpha }}=\frac{2e}{4m}=\frac{1}{2}{{\left( \frac{e}{m} \right)}_{p}}=\frac{1}{2}\times 9.6\times {{10}^{7}}\] \[=4.8\times {{10}^{7}}C/kg\]


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