Manipal Medical Manipal Medical Solved Paper-2011

  • question_answer
    The time of revolution of an electron around a nucleus of charge\[Ze\]in nth Bohrs orbit is irectly proportional to

    A)  \[n\]

    B)  \[\frac{{{n}^{3}}}{{{Z}^{2}}}\]

    C)  \[\frac{{{n}^{2}}}{Z}\]

    D)  \[\frac{Z}{n}\]

    Correct Answer: B

    Solution :

     \[r\propto \frac{{{n}^{2}}}{Z}\]and \[v\propto \frac{X}{{{n}^{2}}}\] Time period of revolution of an electron around the nucleus of charge\[Ze\]is \[T=\frac{2\pi r}{v}=2\pi \frac{{{n}^{2}}}{Z}.\frac{n}{Z}\] \[\Rightarrow \] \[T\propto \frac{{{n}^{3}}}{{{Z}^{2}}}\]


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