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JEE Main & Advanced Sample Paper JEE Main Sample Paper-42

  • question_answer
    The ratio of kinetic energy and potential energy of an electron in Bohr orbit of a hydrogen like species is

    A)  1/2                             

    B)  -1/2

    C)  1                                

    D)  - 1

    Correct Answer: B

    Solution :

    \[KE=\frac{m{{v}^{2}}}{2},PE=\frac{-Z{{e}^{2}}}{r}\] But electrostatic force of attraction \[\left( \frac{Z{{e}^{2}}}{{{r}^{2}}} \right)=\]centripetal force\[\left( \frac{m{{v}^{2}}}{r} \right)\]or\[\frac{Z{{e}^{2}}}{r}=m{{v}^{2}}\] Therefore,\[PE=-m{{v}^{2}}\]\[\frac{KE}{PE}=\frac{-1}{2}\]


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