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BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    \[{{\text{H}}^{\text{+}}}\]\[\text{H}{{\text{e}}^{2+}}\] and \[{{\text{O}}^{\text{2-}}}\] all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of \[{{\text{H}}^{\text{+}}}\text{,}\] \[\text{H}{{\text{e}}^{\text{2+}}}\] and \[{{\text{O}}^{\text{2-}}}\] are 1 amu, 4 amu and 16 amu, respectively. Then:

    A) \[{{\text{H}}^{\text{+}}}\] will be deflected most

    B)        \[{{\text{O}}^{\text{2}}}\] will be deflected most

    C)        \[H{{e}^{\text{2+}}}\]and \[{{\text{O}}^{\text{2-}}}\] will be deflected most

    D)         all will be deflected most

    Correct Answer: A

    Solution :

    When a charged particle enters magnetic field perpendicularly, then it moves on circular path under magnetic force providing centripetal force. Magnetic force = centripetal force i.e.          \[Bqv=\frac{m{{v}^{2}}}{r}\]                 or            \[r=\frac{mv}{Bq}\]                 But         \[E=\frac{1}{2}m{{v}^{2}}={{p}^{2}}/2m\]                 \[\therefore \]  \[r=\frac{\sqrt{2ME}}{Bq}\]                 or            \[r\propto \frac{\sqrt{m}}{q}\]                 \[\therefore \]  \[{{r}_{{{H}^{+}}}}:{{r}_{H{{e}^{2+}}}}:{{r}_{{{O}^{2-}}}}=\frac{\sqrt{m}}{e}:\frac{\sqrt{4m}}{e}:\frac{\sqrt{16m}}{2e}\]                                 \[=1:2:2\] Thus, \[H{{e}^{2+}}\]and\[{{O}^{2-}}\]are deflected equally while \[{{H}^{+}}\] is deflected most.


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