Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius\[r\]. The coulomb force F between the two is

    A) \[k\frac{{{e}^{2}}}{{{r}^{3}}}r\]                 

    B) \[-k\frac{{{e}^{2}}}{{{r}^{3}}}r\]

    C) \[k\frac{{{e}^{2}}}{{{r}^{2}}}r\] 

    D)        \[-k\frac{{{e}^{2}}}{{{r}^{2}}}r\]

    Correct Answer: A

    Solution :

    \[F=-\frac{1}{4\pi \,\,{{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=r\] \[F=-k\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=r\] \[F=-k\frac{e.e}{{{r}^{2}}}\widehat{\mathbf{r}}=-k\frac{{{e}^{2}}}{{{r}^{2}}}\widehat{\mathbf{r}}\] \[\widehat{\mathbf{r}}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{r}{r}\] \[F=-k\frac{{{e}^{2}}}{{{r}^{2}}}\cdot \frac{\mathbf{r}}{r}=-k\frac{{{e}^{2}}}{{{r}^{3}}}\mathbf{r}\]


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