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JCECE Engineering JCECE Engineering 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 these

    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        \[\frac{{{E}_{1}}}{{{n}^{2}}}=\frac{{{E}_{1}}}{{{(n-1)}^{2}}}=1.224\,\,eV\]              ... (ii) Solving Eqs. (i) and (ii), we get                 \[{{E}_{1}}=-54.4\,\,eV\] and        \[n=5\] But         \[{{E}_{1}}=\frac{13.6{{Z}^{2}}}{{{1}^{2}}}\] \[\Rightarrow \]               \[Z=2\]


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