RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    The atomic mass of ionic lithium atoms are 0.06 amu. If they are motion with energy 400 eV in magnetic field of 0.08 T, then radius of circular path will be

    A)  8.83 cm         

    B)  9.23 cm

    C)  10.5 cm         

    D)  11.25 cm

    Correct Answer: A

    Solution :

     \[r=\frac{mv}{Bq}=\sqrt{\frac{2mE}{Bq}}\] \[=\sqrt{\frac{2\times 6.06\times 1.66\times {{10}^{-27}}\times 400\times 1.6\times {{10}^{-19}}}{1.6\times {{10}^{-19}}\times 0.08}}\] \[=8.83\,cm\]


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