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Manipal Medical Manipal Medical Solved Paper-2015

  • question_answer
    The first orbital of H is represented by \[\psi =\frac{1}{\sqrt{\pi }}{{\left( \frac{1}{{{a}_{0}}} \right)}^{3/2}}{{e}^{-\pi /{{a}_{0}}}}\], where \[{{a}_{0}}\] is Bohr's radius. The probability of finding the electron at a distance r from the nucleus in the region \[dV\]is

    A) \[{{\psi }^{2}}dr\]

    B) \[\int {{\psi }^{2}}4\pi {{r}^{2}}dV\]

    C) \[{{\psi }^{2}}4\pi {{r}^{2}}dr\]

    D) \[\int \psi dV\]

    Correct Answer: C

    Solution :

    This is obtained by the solution of Schrodinger wave equation. Probability \[={{\psi }^{2}}dV\]                 Its orbital is spherically symmetrical. Thus,                 Volume,                       \[V=\frac{4}{3}\pi {{r}^{3}}\]                 \[\therefore \]                             \[\frac{dV}{dr}=4\pi {{r}^{2}}\]                 \[\because \]                Probability \[={{\psi }^{2}}4\pi {{r}^{2}}dr\]


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