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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Critical Thinking

  • question_answer
    Consider a hydrogen like atom  whose energy  in nth exicited 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 Emax = 52.224 eV and least energetic photons have energy Emin = 1.224 eV. The atomic number of atom is

    A)            2    

    B)            5

    C)            4    

    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 equations (i) and (ii) we get                    \[{{E}_{1}}=-54.4\,eV\] and \[n=5\]                    Now \[{{E}_{1}}=-\frac{13.6\,{{Z}^{2}}}{{{1}^{2}}}=-54.4\,eV.\]Hence \[Z=2\]


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