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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    A proton and a deuteron are both accelerated through the same potential difference and enter in a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius R, assuming the neutron and proton masses are nearly equal, the radius of the proton's path will be     JEE Main  Online Paper (Held On 19  May  2012)

    A) \[\sqrt{2}R\]                                     

    B)                        \[\frac{R}{\sqrt{2}}\]

    C)                        \[\frac{R}{2}\]  

    D)                         R

    Correct Answer: B

    Solution :

                    As charge on both proton and deuteron is same i.e. 'e' Energy acquired by both, E = eV For Deuteron. Kinetic energy, \[=\frac{1}{2}m{{V}^{2}}=eV\] [V is the potential difference]\[v=\sqrt{\frac{2eV}{{{m}_{d}}}}\] But\[{{m}_{d}}=2m\] Therefore, \[v=\sqrt{\frac{2eV}{2m}}=\sqrt{\frac{eV}{m}}\] Radius of path, \[R=\frac{mv}{eB}\] Substituting value of V we get\[R=\frac{2m\sqrt{\frac{ev}{m}}}{eB}\] \[\frac{R}{2}=\frac{m\sqrt{\frac{ev}{m}}}{eB}\]                                                ?(i) For proton : \[\frac{1}{2}m{{V}^{2}}=eV\]     \[V=\sqrt{\frac{2eV}{m}}\] Radius of path, \[R'=\frac{mV}{eB}=\frac{m\sqrt{\frac{2eV}{m}}}{eB}\] \[R'=\sqrt{2}\times \frac{R}{2}\][From eq. (i)]\[R'=\frac{R}{2}\]


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