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JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    Orbits of a particle moving in a circle are such that the perimeter of orbit equals an integer number of de - Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to:     JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[{{n}^{2}}\]                                         

    B)  \[n\]

    C)  \[{{n}^{1/2}}\]                                     

    D)  \[{{n}^{1/4}}\]

    Correct Answer: C

    Solution :

     According to the question, \[2\pi r=n\lambda =\frac{nh}{p}=\frac{nh}{mv}\] or \[mvr=\frac{nh}{2\pi }\]or \[mv=\frac{nh}{2\pi r}\] \[F=q{{v}_{B}}=\frac{m{{v}^{2}}}{r}\] or,          \[{{q}_{B}}=\frac{mv}{r}=\frac{nh}{2\pi r.r}\] or,          \[{{r}^{2}}=\frac{nh}{2\pi qB}\] or,          \[r=\sqrt{\frac{nh}{2\pi qB}}\] i.e. \[r\propto {{n}^{1/2}}\]


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