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JCECE Medical JCECE Medical Solved Paper-2012

  • question_answer
    Consider a hydrogen like atom whose energy in  nth excited state is given by \[{{E}_{n}}=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}},\] when this excited atom makes a transition from excited state to ground state, most energetic photons have energy \[{{E}_{\max }}=52.224\,eV\]and least   energetic   photons   have   energy \[{{E}_{\min }}=1.224\,eV.\] The atomic number of atom is

    A)  2

    B)  4

    C)  5

    D)  None of the above

    Correct Answer: A

    Solution :

     Maximum energy is liberated for transition \[{{E}_{n}}\to 1\]and minimum energy for \[{{E}_{n}}\to {{E}_{n-1}}\] Hence,   \[\frac{{{E}_{1}}}{{{n}^{2}}}-{{E}_{1}}=52.224\,eV\] ?(i) and \[\frac{{{E}_{1}}}{{{n}^{2}}}-\frac{E}{{{(n-1)}^{2}}}=1.224\,eV\] ?(ii) Solving Eqs. (i) and (ii), we get \[{{E}_{1}}=-54.4\,eV\] and \[n=5\] But \[{{E}_{1}}=-\frac{13.6{{Z}^{2}}}{{{1}^{2}}}\] \[\therefore \] \[-54.4=\frac{13.6}{{{1}^{2}}}{{Z}^{2}}\] \[\Rightarrow \] \[Z=2\]


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