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JEE Main & Advanced Sample Paper JEE Main Sample Paper-3

  • question_answer
    A hydrogen-like atom has one electron revolving around a stationary nucleus. The energy required to excite the electron from the second orbit to the third orbit is 47.2 eV. The atomic number of the atom is

    A)  3                            

    B)  4         

    C)  5                                            

    D)  6

    Correct Answer: C

    Solution :

     \[E=\frac{13.6{{Z}^{2}}}{{{n}^{2}}};\]According to question, \[\frac{13.6{{Z}^{2}}}{4}-\frac{13.6{{Z}^{2}}}{9}=47.2\Rightarrow 13.6{{Z}^{2}}\left( \frac{5}{36} \right)=47.2\]\[\Rightarrow \]\[{{Z}^{2}}=25\Rightarrow Z=5\]


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